build

en/docs/chapter_appendix/contribution.md

@@ -16,7 +16,7 @@ The GitHub IDs of all [contributors](https://github.com/krahets/hello-algo/graph
 
 ### 1.   Content fine-tuning
 
-As shown in the Figure 16-3 , there is an "edit icon" in the upper right corner of each page. You can follow these steps to modify text or code.
+As shown in Figure 16-3, there is an "edit icon" in the upper right corner of each page. You can follow these steps to modify text or code.
 
 1. Click the "edit icon". If prompted to "fork this repository", please agree to do so.
 2. Modify the Markdown source file content, check the accuracy of the content, and try to keep the formatting consistent.
@@ -26,7 +26,7 @@ As shown in the Figure 16-3 , there is an "edit icon" in the upper right corner
 
 <p align="center"> Figure 16-3 &nbsp; Edit page button </p>
 
-Images cannot be directly modified and require the creation of a new [Issue](https://github.com/krahets/hello-algo/issues) or a comment to describe the problem. We will redraw and replace the images as soon as possible.
+Figures cannot be directly modified and require the creation of a new [Issue](https://github.com/krahets/hello-algo/issues) or a comment to describe the problem. We will redraw and replace the figures as soon as possible.
 
 ### 2. &nbsp; Content creation
 

en/docs/chapter_appendix/installation.md

@@ -12,7 +12,7 @@ We recommend using the open-source, lightweight VS Code as your local Integrated
 
 <p align="center"> Figure 16-1 &nbsp; Download VS Code from the official website </p>
 
-VS Code has a powerful extension ecosystem, supporting the execution and debugging of most programming languages. For example, after installing the "Python Extension Pack," you can debug Python code. The installation steps are shown in the following figure.
+VS Code has a powerful extension ecosystem, supporting the execution and debugging of most programming languages. For example, after installing the "Python Extension Pack," you can debug Python code. The installation steps are shown in Figure 16-2.
 
 ![Install VS Code Extension Pack](installation.assets/vscode_extension_installation.png){ class="animation-figure" }
 

en/docs/chapter_appendix/terminology.md

@@ -4,7 +4,7 @@ comments: true
 
 # 16.3   Glossary
 
-The Table 16-1  lists the important terms that appear in the book, and it is worth noting the following points.
+Table 16-1 lists the important terms that appear in the book, and it is worth noting the following points.
 
 - It is recommended to remember the English names of the terms to facilitate reading English literature.
 - Some terms have different names in Simplified and Traditional Chinese.

en/docs/chapter_array_and_linkedlist/array.md

@@ -4,7 +4,7 @@ comments: true
 
 # 4.1   Array
 
-An "array" is a linear data structure that operates as a lineup of similar items, stored together in a computer's memory in contiguous spaces. It's like a sequence that maintains organized storage. Each item in this lineup has its unique 'spot' known as an "index". Please refer to the Figure 4-1  to observe how arrays work and grasp these key terms.
+An "array" is a linear data structure that operates as a lineup of similar items, stored together in a computer's memory in contiguous spaces. It's like a sequence that maintains organized storage. Each item in this lineup has its unique 'spot' known as an "index". Please refer to Figure 4-1 to observe how arrays work and grasp these key terms.
 
 ![Array definition and storage method](array.assets/array_definition.png){ class="animation-figure" }
 
@@ -537,7 +537,7 @@ It's important to note that due to the fixed length of an array, inserting an el
 
 ### 4.   Deleting elements
 
-Similarly, as depicted in the Figure 4-4 , to delete an element at index $i$, all elements following index $i$ must be moved forward by one position.
+Similarly, as depicted in Figure 4-4, to delete an element at index $i$, all elements following index $i$ must be moved forward by one position.
 
 ![Array element deletion example](array.assets/array_remove_element.png){ class="animation-figure" }
 

en/docs/chapter_array_and_linkedlist/linked_list.md

@@ -1249,7 +1249,7 @@ Traverse the linked list to locate a node whose value matches `target`, and then
 
 ## 4.2.2   Arrays vs. linked lists
 
-The Table 4-1  summarizes the characteristics of arrays and linked lists, and it also compares their efficiencies in various operations. Because they utilize opposing storage strategies, their respective properties and operational efficiencies exhibit distinct contrasts.
+Table 4-1 summarizes the characteristics of arrays and linked lists, and it also compares their efficiencies in various operations. Because they utilize opposing storage strategies, their respective properties and operational efficiencies exhibit distinct contrasts.
 
 <p align="center"> Table 4-1 &nbsp; Efficiency comparison of arrays and linked lists </p>
 

en/docs/chapter_array_and_linkedlist/ram_and_cache.md

@@ -26,7 +26,7 @@ There are three types of storage devices in computers: "hard disk," "random-acce
 
 </div>
 
-We can imagine the computer storage system as a pyramid structure shown in the Figure 4-9 . The storage devices closer to the top of the pyramid are faster, have smaller capacity, and are more costly. This multi-level design is not accidental, but the result of careful consideration by computer scientists and engineers.
+We can imagine the computer storage system as a pyramid structure shown in Figure 4-9. The storage devices closer to the top of the pyramid are faster, have smaller capacity, and are more costly. This multi-level design is not accidental, but the result of careful consideration by computer scientists and engineers.
 
 - **Hard disks are difficult to replace with memory**. Firstly, data in memory is lost after power off, making it unsuitable for long-term data storage; secondly, the cost of memory is dozens of times that of hard disks, making it difficult to popularize in the consumer market.
 - **It is difficult for caches to have both large capacity and high speed**. As the capacity of L1, L2, L3 caches gradually increases, their physical size becomes larger, increasing the physical distance from the CPU core, leading to increased data transfer time and higher element access latency. Under current technology, a multi-level cache structure is the best balance between capacity, speed, and cost.
@@ -41,7 +41,7 @@ We can imagine the computer storage system as a pyramid structure shown in the F
 
 Overall, **hard disks are used for long-term storage of large amounts of data, memory is used for temporary storage of data being processed during program execution, and cache is used to store frequently accessed data and instructions** to improve program execution efficiency. Together, they ensure the efficient operation of computer systems.
 
-As shown in the Figure 4-10 , during program execution, data is read from the hard disk into memory for CPU computation. The cache can be considered a part of the CPU, **smartly loading data from memory** to provide fast data access to the CPU, significantly enhancing program execution efficiency and reducing reliance on slower memory.
+As shown in Figure 4-10, during program execution, data is read from the hard disk into memory for CPU computation. The cache can be considered a part of the CPU, **smartly loading data from memory** to provide fast data access to the CPU, significantly enhancing program execution efficiency and reducing reliance on slower memory.
 
 ![Data flow between hard disk, memory, and cache](ram_and_cache.assets/computer_storage_devices.png){ class="animation-figure" }
 

en/docs/chapter_array_and_linkedlist/summary.md

@@ -46,7 +46,7 @@ From a garbage collection perspective, for languages with automatic garbage coll
 
 If an element is searched first and then deleted, the time complexity is indeed `O(n)`. However, the `O(1)` advantage of linked lists in insertion and deletion can be realized in other applications. For example, in the implementation of double-ended queues using linked lists, we maintain pointers always pointing to the head and tail nodes, making each insertion and deletion operation `O(1)`.
 
-**Q**: In the image "Linked List Definition and Storage Method", do the light blue storage nodes occupy a single memory address, or do they share half with the node value?
+**Q**: In the figure "Linked List Definition and Storage Method", do the light blue storage nodes occupy a single memory address, or do they share half with the node value?
 
 The diagram is just a qualitative representation; quantitative analysis depends on specific situations.
 

en/docs/chapter_backtracking/backtracking_algorithm.md

@@ -538,7 +538,7 @@ Based on the code from Example One, we need to use a list `path` to record the v
 
 In each "try", we record the path by adding the current node to `path`; before "retreating", we need to pop the node from `path` **to restore the state before this attempt**.
 
-Observe the process shown below, **we can understand trying and retreating as "advancing" and "undoing"**, two operations that are reverse to each other.
+Observe the process shown in Figure 13-2, **we can understand trying and retreating as "advancing" and "undoing"**, two operations that are reverse to each other.
 
 === "<1>"
     ![Trying and retreating](backtracking_algorithm.assets/preorder_find_paths_step1.png){ class="animation-figure" }
@@ -1887,7 +1887,7 @@ Compared to the implementation based on preorder traversal, the code implementat
 
 ## 13.1.4 &nbsp; Common terminology
 
-To analyze algorithmic problems more clearly, we summarize the meanings of commonly used terminology in backtracking algorithms and provide corresponding examples from Example Three as shown in the Table 13-1 .
+To analyze algorithmic problems more clearly, we summarize the meanings of commonly used terminology in backtracking algorithms and provide corresponding examples from Example Three as shown in Table 13-1.
 
 <p align="center"> Table 13-1 &nbsp; Common backtracking algorithm terminology </p>
 

en/docs/chapter_backtracking/n_queens_problem.md

@@ -8,13 +8,13 @@ comments: true
 
     According to the rules of chess, a queen can attack pieces in the same row, column, or on a diagonal line. Given $n$ queens and an $n \times n$ chessboard, find arrangements where no two queens can attack each other.
 
-As shown in the Figure 13-15 , when $n = 4$, there are two solutions. From the perspective of the backtracking algorithm, an $n \times n$ chessboard has $n^2$ squares, presenting all possible choices `choices`. The state of the chessboard `state` changes continuously as each queen is placed.
+As shown in Figure 13-15, when $n = 4$, there are two solutions. From the perspective of the backtracking algorithm, an $n \times n$ chessboard has $n^2$ squares, presenting all possible choices `choices`. The state of the chessboard `state` changes continuously as each queen is placed.
 
 ![Solution to the 4 queens problem](n_queens_problem.assets/solution_4_queens.png){ class="animation-figure" }
 
 <p align="center"> Figure 13-15 &nbsp; Solution to the 4 queens problem </p>
 
-The following image shows the three constraints of this problem: **multiple queens cannot be on the same row, column, or diagonal**. It is important to note that diagonals are divided into the main diagonal `\` and the secondary diagonal `/`.
+Figure 13-16 shows the three constraints of this problem: **multiple queens cannot be on the same row, column, or diagonal**. It is important to note that diagonals are divided into the main diagonal `\` and the secondary diagonal `/`.
 
 ![Constraints of the n queens problem](n_queens_problem.assets/n_queens_constraints.png){ class="animation-figure" }
 
@@ -26,7 +26,7 @@ As the number of queens equals the number of rows on the chessboard, both being
 
 This means that we can adopt a row-by-row placing strategy: starting from the first row, place one queen per row until the last row is reached.
 
-The image below shows the row-by-row placing process for the 4 queens problem. Due to space limitations, the image only expands one search branch of the first row, and prunes any placements that do not meet the column and diagonal constraints.
+Figure 13-17 shows the row-by-row placing process for the 4 queens problem. Due to space limitations, the figure only expands one search branch of the first row, and prunes any placements that do not meet the column and diagonal constraints.
 
 ![Row-by-row placing strategy](n_queens_problem.assets/n_queens_placing.png){ class="animation-figure" }
 
@@ -40,7 +40,7 @@ To satisfy column constraints, we can use a boolean array `cols` of length $n$ t
 
 How about the diagonal constraints? Let the row and column indices of a cell on the chessboard be $(row, col)$. By selecting a specific main diagonal, we notice that the difference $row - col$ is the same for all cells on that diagonal, **meaning that $row - col$ is a constant value on that diagonal**.
 
-Thus, if two cells satisfy $row_1 - col_1 = row_2 - col_2$, they are definitely on the same main diagonal. Using this pattern, we can utilize the array `diags1` shown below to track whether a queen is on any main diagonal.
+Thus, if two cells satisfy $row_1 - col_1 = row_2 - col_2$, they are definitely on the same main diagonal. Using this pattern, we can utilize the array `diags1` shown in Figure 13-18 to track whether a queen is on any main diagonal.
 
 Similarly, **the sum $row + col$ is a constant value for all cells on a secondary diagonal**. We can also use the array `diags2` to handle secondary diagonal constraints.
 

en/docs/chapter_backtracking/permutations_problem.md

@@ -6,13 +6,13 @@ comments: true
 
 The permutation problem is a typical application of the backtracking algorithm. It is defined as finding all possible arrangements of elements from a given set (such as an array or string).
 
-The Table 13-2  lists several example data, including the input arrays and their corresponding permutations.
+Table 13-2 lists several example data, including the input arrays and their corresponding permutations.
 
 <p align="center"> Table 13-2 &nbsp; Permutation examples </p>
 
 <div class="center-table" markdown>
 
-| Input array | Permutations                                                   |
+| Input array | Permutations                                                       |
 | :---------- | :----------------------------------------------------------------- |
 | $[1]$       | $[1]$                                                              |
 | $[1, 2]$    | $[1, 2], [2, 1]$                                                   |
@@ -30,7 +30,7 @@ From the perspective of the backtracking algorithm, **we can imagine the process
 
 From the code perspective, the candidate set `choices` contains all elements of the input array, and the state `state` contains elements that have been selected so far. Please note that each element can only be chosen once, **thus all elements in `state` must be unique**.
 
-As shown in the following figure, we can unfold the search process into a recursive tree, where each node represents the current state `state`. Starting from the root node, after three rounds of choices, we reach the leaf nodes, each corresponding to a permutation.
+As shown in Figure 13-5, we can unfold the search process into a recursive tree, where each node represents the current state `state`. Starting from the root node, after three rounds of choices, we reach the leaf nodes, each corresponding to a permutation.
 
 ![Permutation recursive tree](permutations_problem.assets/permutations_i.png){ class="animation-figure" }
 
@@ -43,13 +43,13 @@ To ensure that each element is selected only once, we consider introducing a boo
 - After making the choice `choice[i]`, we set `selected[i]` to $\text{True}$, indicating it has been chosen.
 - When iterating through the choice list `choices`, skip all nodes that have already been selected, i.e., prune.
 
-As shown in the following figure, suppose we choose 1 in the first round, 3 in the second round, and 2 in the third round, we need to prune the branch of element 1 in the second round and elements 1 and 3 in the third round.
+As shown in Figure 13-6, suppose we choose 1 in the first round, 3 in the second round, and 2 in the third round, we need to prune the branch of element 1 in the second round and elements 1 and 3 in the third round.
 
 ![Permutation pruning example](permutations_problem.assets/permutations_i_pruning.png){ class="animation-figure" }
 
 <p align="center"> Figure 13-6 &nbsp; Permutation pruning example </p>
 
-Observing the above figure, this pruning operation reduces the search space size from $O(n^n)$ to $O(n!)$.
+Observing Figure 13-6, this pruning operation reduces the search space size from $O(n^n)$ to $O(n!)$.
 
 ### 2. &nbsp; Code implementation
 
@@ -532,7 +532,7 @@ After understanding the above information, we can "fill in the blanks" in the fr
 
 Suppose the input array is $[1, 1, 2]$. To differentiate the two duplicate elements $1$, we mark the second $1$ as $\hat{1}$.
 
-As shown in the following figure, half of the permutations generated by the above method are duplicates.
+As shown in Figure 13-7, half of the permutations generated by the above method are duplicates.
 
 ![Duplicate permutations](permutations_problem.assets/permutations_ii.png){ class="animation-figure" }
 
@@ -542,7 +542,7 @@ So, how do we eliminate duplicate permutations? Most directly, consider using a
 
 ### 1. &nbsp; Pruning of equal elements
 
-Observing the following figure, in the first round, choosing $1$ or $\hat{1}$ results in identical permutations under both choices, thus we should prune $\hat{1}$.
+Observing Figure 13-8, in the first round, choosing $1$ or $\hat{1}$ results in identical permutations under both choices, thus we should prune $\hat{1}$.
 
 Similarly, after choosing $2$ in the first round, choosing $1$ and $\hat{1}$ in the second round also produces duplicate branches, so we should also prune $\hat{1}$ in the second round.
 
@@ -1061,7 +1061,7 @@ Please note, although both `selected` and `duplicated` are used for pruning, the
 - **Repeated choice pruning**: There is only one `selected` throughout the search process. It records which elements are currently in the state, aiming to prevent an element from appearing repeatedly in `state`.
 - **Equal element pruning**: Each round of choices (each call to the `backtrack` function) contains a `duplicated`. It records which elements have been chosen in the current traversal (`for` loop), aiming to ensure equal elements are selected only once.
 
-The following figure shows the scope of the two pruning conditions. Note, each node in the tree represents a choice, and the nodes from the root to the leaf form a permutation.
+Figure 13-9 shows the scope of the two pruning conditions. Note, each node in the tree represents a choice, and the nodes from the root to the leaf form a permutation.
 
 ![Scope of the two pruning conditions](permutations_problem.assets/permutations_ii_pruning_summary.png){ class="animation-figure" }
 

en/docs/chapter_backtracking/subset_sum_problem.md

@@ -490,7 +490,7 @@ Unlike the permutation problem, **elements in this problem can be chosen an unli
 
 Inputting the array $[3, 4, 5]$ and target element $9$ into the above code yields the results $[3, 3, 3], [4, 5], [5, 4]$. **Although it successfully finds all subsets with a sum of $9$, it includes the duplicate subset $[4, 5]$ and $[5, 4]$**.
 
-This is because the search process distinguishes the order of choices, however, subsets do not distinguish the choice order. As shown in the following figure, choosing $4$ before $5$ and choosing $5$ before $4$ are different branches, but correspond to the same subset.
+This is because the search process distinguishes the order of choices, however, subsets do not distinguish the choice order. As shown in Figure 13-10, choosing $4$ before $5$ and choosing $5$ before $4$ are different branches, but correspond to the same subset.
 
 ![Subset search and pruning out of bounds](subset_sum_problem.assets/subset_sum_i_naive.png){ class="animation-figure" }
 
@@ -503,7 +503,7 @@ To eliminate duplicate subsets, **a straightforward idea is to deduplicate the r
 
 ### 2.   Duplicate subset pruning
 
-**We consider deduplication during the search process through pruning**. Observing the following figure, duplicate subsets are generated when choosing array elements in different orders, for example in the following situations.
+**We consider deduplication during the search process through pruning**. Observing Figure 13-11, duplicate subsets are generated when choosing array elements in different orders, for example in the following situations.
 
 1. When choosing $3$ in the first round and $4$ in the second round, all subsets containing these two elements are generated, denoted as $[3, 4, \dots]$.
 2. Later, when $4$ is chosen in the first round, **the second round should skip $3$** because the subset $[4, 3, \dots]$ generated by this choice completely duplicates the subset from step `1.`.
@@ -1029,7 +1029,7 @@ Besides, we have made the following two optimizations to the code.
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-The following figure shows the overall backtracking process after inputting the array $[3, 4, 5]$ and target element $9$ into the above code.
+Figure 13-12 shows the overall backtracking process after inputting the array $[3, 4, 5]$ and target element $9$ into the above code.
 
 ![Subset sum I backtracking process](subset_sum_problem.assets/subset_sum_i.png){ class="animation-figure" }
 
@@ -1043,7 +1043,7 @@ The following figure shows the overall backtracking process after inputting the
 
 Compared to the previous question, **this question's input array may contain duplicate elements**, introducing new problems. For example, given the array $[4, \hat{4}, 5]$ and target element $9$, the existing code's output results in $[4, 5], [\hat{4}, 5]$, resulting in duplicate subsets.
 
-**The reason for this duplication is that equal elements are chosen multiple times in a certain round**. In the following figure, the first round has three choices, two of which are $4$, generating two duplicate search branches, thus outputting duplicate subsets; similarly, the two $4$s in the second round also produce duplicate subsets.
+**The reason for this duplication is that equal elements are chosen multiple times in a certain round**. In Figure 13-13, the first round has three choices, two of which are $4$, generating two duplicate search branches, thus outputting duplicate subsets; similarly, the two $4$s in the second round also produce duplicate subsets.
 
 ![Duplicate subsets caused by equal elements](subset_sum_problem.assets/subset_sum_ii_repeat.png){ class="animation-figure" }
 
@@ -1616,7 +1616,7 @@ At the same time, **this question stipulates that each array element can only be
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     <div style="margin-top: 5px;"><a href="https://pythontutor.com/iframe-embed.html#code=def%20backtrack%28%0A%20%20%20%20state%3A%20list%5Bint%5D,%20target%3A%20int,%20choices%3A%20list%5Bint%5D,%20start%3A%20int,%20res%3A%20list%5Blist%5Bint%5D%5D%0A%29%3A%0A%20%20%20%20%22%22%22%E5%9B%9E%E6%BA%AF%E7%AE%97%E6%B3%95%EF%BC%9A%E5%AD%90%E9%9B%86%E5%92%8C%20II%22%22%22%0A%20%20%20%20%23%20%E5%AD%90%E9%9B%86%E5%92%8C%E7%AD%89%E4%BA%8E%20target%20%E6%97%B6%EF%BC%8C%E8%AE%B0%E5%BD%95%E8%A7%A3%0A%20%20%20%20if%20target%20%3D%3D%200%3A%0A%20%20%20%20%20%20%20%20res.append%28list%28state%29%29%0A%20%20%20%20%20%20%20%20return%0A%20%20%20%20%23%20%E9%81%8D%E5%8E%86%E6%89%80%E6%9C%89%E9%80%89%E6%8B%A9%0A%20%20%20%20%23%20%E5%89%AA%E6%9E%9D%E4%BA%8C%EF%BC%9A%E4%BB%8E%20start%20%E5%BC%80%E5%A7%8B%E9%81%8D%E5%8E%86%EF%BC%8C%E9%81%BF%E5%85%8D%E7%94%9F%E6%88%90%E9%87%8D%E5%A4%8D%E5%AD%90%E9%9B%86%0A%20%20%20%20%23%20%E5%89%AA%E6%9E%9D%E4%B8%89%EF%BC%9A%E4%BB%8E%20start%20%E5%BC%80%E5%A7%8B%E9%81%8D%E5%8E%86%EF%BC%8C%E9%81%BF%E5%85%8D%E9%87%8D%E5%A4%8D%E9%80%89%E6%8B%A9%E5%90%8C%E4%B8%80%E5%85%83%E7%B4%A0%0A%20%20%20%20for%20i%20in%20range%28start,%20len%28choices%29%29%3A%0A%20%20%20%20%20%20%20%20%23%20%E5%89%AA%E6%9E%9D%E4%B8%80%EF%BC%9A%E8%8B%A5%E5%AD%90%E9%9B%86%E5%92%8C%E8%B6%85%E8%BF%87%20target%20%EF%BC%8C%E5%88%99%E7%9B%B4%E6%8E%A5%E7%BB%93%E6%9D%9F%E5%BE%AA%E7%8E%AF%0A%20%20%20%20%20%20%20%20%23%20%E8%BF%99%E6%98%AF%E5%9B%A0%E4%B8%BA%E6%95%B0%E7%BB%84%E5%B7%B2%E6%8E%92%E5%BA%8F%EF%BC%8C%E5%90%8E%E8%BE%B9%E5%85%83%E7%B4%A0%E6%9B%B4%E5%A4%A7%EF%BC%8C%E5%AD%90%E9%9B%86%E5%92%8C%E4%B8%80%E5%AE%9A%E8%B6%85%E8%BF%87%20target%0A%20%20%20%20%20%20%20%20if%20target%20-%20choices%5Bi%5D%20%3C%200%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20break%0A%20%20%20%20%20%20%20%20%23%20%E5%89%AA%E6%9E%9D%E5%9B%9B%EF%BC%9A%E5%A6%82%E6%9E%9C%E8%AF%A5%E5%85%83%E7%B4%A0%E4%B8%8E%E5%B7%A6%E8%BE%B9%E5%85%83%E7%B4%A0%E7%9B%B8%E7%AD%89%EF%BC%8C%E8%AF%B4%E6%98%8E%E8%AF%A5%E6%90%9C%E7%B4%A2%E5%88%86%E6%94%AF%E9%87%8D%E5%A4%8D%EF%BC%8C%E7%9B%B4%E6%8E%A5%E8%B7%B3%E8%BF%87%0A%20%20%20%20%20%20%20%20if%20i%20%3E%20start%20and%20choices%5Bi%5D%20%3D%3D%20choices%5Bi%20-%201%5D%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20continue%0A%20%20%20%20%20%20%20%20%23%20%E5%B0%9D%E8%AF%95%EF%BC%9A%E5%81%9A%E5%87%BA%E9%80%89%E6%8B%A9%EF%BC%8C%E6%9B%B4%E6%96%B0%20target,%20start%0A%20%20%20%20%20%20%20%20state.append%28choices%5Bi%5D%29%0A%20%20%20%20%20%20%20%20%23%20%E8%BF%9B%E8%A1%8C%E4%B8%8B%E4%B8%80%E8%BD%AE%E9%80%89%E6%8B%A9%0A%20%20%20%20%20%20%20%20backtrack%28state,%20target%20-%20choices%5Bi%5D,%20choices,%20i%20%2B%201,%20res%29%0A%20%20%20%20%20%20%20%20%23%20%E5%9B%9E%E9%80%80%EF%BC%9A%E6%92%A4%E9%94%80%E9%80%89%E6%8B%A9%EF%BC%8C%E6%81%A2%E5%A4%8D%E5%88%B0%E4%B9%8B%E5%89%8D%E7%9A%84%E7%8A%B6%E6%80%81%0A%20%20%20%20%20%20%20%20state.pop%28%29%0A%0A%0Adef%20subset_sum_ii%28nums%3A%20list%5Bint%5D,%20target%3A%20int%29%20-%3E%20list%5Blist%5Bint%5D%5D%3A%0A%20%20%20%20%22%22%22%E6%B1%82%E8%A7%A3%E5%AD%90%E9%9B%86%E5%92%8C%20II%22%22%22%0A%20%20%20%20state%20%3D%20%5B%5D%20%20%23%20%E7%8A%B6%E6%80%81%EF%BC%88%E5%AD%90%E9%9B%86%EF%BC%89%0A%20%20%20%20nums.sort%28%29%20%20%23%20%E5%AF%B9%20nums%20%E8%BF%9B%E8%A1%8C%E6%8E%92%E5%BA%8F%0A%20%20%20%20start%20%3D%200%20%20%23%20%E9%81%8D%E5%8E%86%E8%B5%B7%E5%A7%8B%E7%82%B9%0A%20%20%20%20res%20%3D%20%5B%5D%20%20%23%20%E7%BB%93%E6%9E%9C%E5%88%97%E8%A1%A8%EF%BC%88%E5%AD%90%E9%9B%86%E5%88%97%E8%A1%A8%EF%BC%89%0A%20%20%20%20backtrack%28state,%20target,%20nums,%20start,%20res%29%0A%20%20%20%20return%20res%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20nums%20%3D%20%5B4,%204,%205%5D%0A%20%20%20%20target%20%3D%209%0A%20%20%20%20res%20%3D%20subset_sum_ii%28nums,%20target%29%0A%0A%20%20%20%20print%28f%22%E8%BE%93%E5%85%A5%E6%95%B0%E7%BB%84%20nums%20%3D%20%7Bnums%7D,%20target%20%3D%20%7Btarget%7D%22%29%0A%20%20%20%20print%28f%22%E6%89%80%E6%9C%89%E5%92%8C%E7%AD%89%E4%BA%8E%20%7Btarget%7D%20%E7%9A%84%E5%AD%90%E9%9B%86%20res%20%3D%20%7Bres%7D%22%29&codeDivHeight=800&codeDivWidth=600&cumulative=false&curInstr=16&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">Full Screen ></a></div>
 
-The following figure shows the backtracking process for the array $[4, 4, 5]$ and target element $9$, including four types of pruning operations. Please combine the illustration with the code comments to understand the entire search process and how each type of pruning operation works.
+Figure 13-14 shows the backtracking process for the array $[4, 4, 5]$ and target element $9$, including four types of pruning operations. Please combine the illustration with the code comments to understand the entire search process and how each type of pruning operation works.
 
 ![Subset sum II backtracking process](subset_sum_problem.assets/subset_sum_ii.png){ class="animation-figure" }
 

en/docs/chapter_computational_complexity/iteration_and_recursion.md

@@ -1220,7 +1220,7 @@ Observe the following code, where simply calling the function `recur(n)` can com
     <div style="height: 459px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20recur%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E9%80%92%E5%BD%92%22%22%22%0A%20%20%20%20%23%20%E7%BB%88%E6%AD%A2%E6%9D%A1%E4%BB%B6%0A%20%20%20%20if%20n%20%3D%3D%201%3A%0A%20%20%20%20%20%20%20%20return%201%0A%20%20%20%20%23%20%E9%80%92%EF%BC%9A%E9%80%92%E5%BD%92%E8%B0%83%E7%94%A8%0A%20%20%20%20res%20%3D%20recur%28n%20-%201%29%0A%20%20%20%20%23%20%E5%BD%92%EF%BC%9A%E8%BF%94%E5%9B%9E%E7%BB%93%E6%9E%9C%0A%20%20%20%20return%20n%20%2B%20res%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%205%0A%20%20%20%20res%20%3D%20recur%28n%29%0A%20%20%20%20print%28f%22%5Cn%E9%80%92%E5%BD%92%E5%87%BD%E6%95%B0%E7%9A%84%E6%B1%82%E5%92%8C%E7%BB%93%E6%9E%9C%20res%20%3D%20%7Bres%7D%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
     <div style="margin-top: 5px;"><a href="https://pythontutor.com/iframe-embed.html#code=def%20recur%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E9%80%92%E5%BD%92%22%22%22%0A%20%20%20%20%23%20%E7%BB%88%E6%AD%A2%E6%9D%A1%E4%BB%B6%0A%20%20%20%20if%20n%20%3D%3D%201%3A%0A%20%20%20%20%20%20%20%20return%201%0A%20%20%20%20%23%20%E9%80%92%EF%BC%9A%E9%80%92%E5%BD%92%E8%B0%83%E7%94%A8%0A%20%20%20%20res%20%3D%20recur%28n%20-%201%29%0A%20%20%20%20%23%20%E5%BD%92%EF%BC%9A%E8%BF%94%E5%9B%9E%E7%BB%93%E6%9E%9C%0A%20%20%20%20return%20n%20%2B%20res%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%205%0A%20%20%20%20res%20%3D%20recur%28n%29%0A%20%20%20%20print%28f%22%5Cn%E9%80%92%E5%BD%92%E5%87%BD%E6%95%B0%E7%9A%84%E6%B1%82%E5%92%8C%E7%BB%93%E6%9E%9C%20res%20%3D%20%7Bres%7D%22%29&codeDivHeight=800&codeDivWidth=600&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">Full Screen ></a></div>
 
-The Figure 2-3  shows the recursive process of this function.
+Figure 2-3 shows the recursive process of this function.
 
 ![Recursive process of the sum function](iteration_and_recursion.assets/recursion_sum.png){ class="animation-figure" }
 
@@ -1243,7 +1243,7 @@ Every time a recursive function calls itself, the system allocates memory for th
 - The function's context data is stored in a memory area called "stack frame space" and is only released after the function returns. Therefore, **recursion generally consumes more memory space than iteration**.
 - Recursive calls introduce additional overhead. **Hence, recursion is usually less time-efficient than loops.**
 
-As shown in the Figure 2-4 , there are $n$ unreturned recursive functions before triggering the termination condition, indicating a **recursion depth of $n$**.
+As shown in Figure 2-4, there are $n$ unreturned recursive functions before triggering the termination condition, indicating a **recursion depth of $n$**.
 
 ![Recursion call depth](iteration_and_recursion.assets/recursion_sum_depth.png){ class="animation-figure" }
 
@@ -1447,7 +1447,7 @@ For example, in calculating $1 + 2 + \dots + n$, we can make the result variable
     <div style="height: 423px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20tail_recur%28n,%20res%29%3A%0A%20%20%20%20%22%22%22%E5%B0%BE%E9%80%92%E5%BD%92%22%22%22%0A%20%20%20%20%23%20%E7%BB%88%E6%AD%A2%E6%9D%A1%E4%BB%B6%0A%20%20%20%20if%20n%20%3D%3D%200%3A%0A%20%20%20%20%20%20%20%20return%20res%0A%20%20%20%20%23%20%E5%B0%BE%E9%80%92%E5%BD%92%E8%B0%83%E7%94%A8%0A%20%20%20%20return%20tail_recur%28n%20-%201,%20res%20%2B%20n%29%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%205%0A%20%20%20%20res%20%3D%20tail_recur%28n,%200%29%0A%20%20%20%20print%28f%22%5Cn%E5%B0%BE%E9%80%92%E5%BD%92%E5%87%BD%E6%95%B0%E7%9A%84%E6%B1%82%E5%92%8C%E7%BB%93%E6%9E%9C%20res%20%3D%20%7Bres%7D%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
     <div style="margin-top: 5px;"><a href="https://pythontutor.com/iframe-embed.html#code=def%20tail_recur%28n,%20res%29%3A%0A%20%20%20%20%22%22%22%E5%B0%BE%E9%80%92%E5%BD%92%22%22%22%0A%20%20%20%20%23%20%E7%BB%88%E6%AD%A2%E6%9D%A1%E4%BB%B6%0A%20%20%20%20if%20n%20%3D%3D%200%3A%0A%20%20%20%20%20%20%20%20return%20res%0A%20%20%20%20%23%20%E5%B0%BE%E9%80%92%E5%BD%92%E8%B0%83%E7%94%A8%0A%20%20%20%20return%20tail_recur%28n%20-%201,%20res%20%2B%20n%29%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%205%0A%20%20%20%20res%20%3D%20tail_recur%28n,%200%29%0A%20%20%20%20print%28f%22%5Cn%E5%B0%BE%E9%80%92%E5%BD%92%E5%87%BD%E6%95%B0%E7%9A%84%E6%B1%82%E5%92%8C%E7%BB%93%E6%9E%9C%20res%20%3D%20%7Bres%7D%22%29&codeDivHeight=800&codeDivWidth=600&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">Full Screen ></a></div>
 
-The execution process of tail recursion is shown in the following figure. Comparing regular recursion and tail recursion, the point of the summation operation is different.
+The execution process of tail recursion is shown in Figure 2-5. Comparing regular recursion and tail recursion, the point of the summation operation is different.
 
 - **Regular recursion**: The summation operation occurs during the "returning" phase, requiring another summation after each layer returns.
 - **Tail recursion**: The summation operation occurs during the "calling" phase, and the "returning" phase only involves returning through each layer.

en/docs/chapter_computational_complexity/space_complexity.md

@@ -22,7 +22,7 @@ Temporary space can be further divided into three parts.
 - **Stack frame space**: Used to save the context data of the called function. The system creates a stack frame at the top of the stack each time a function is called, and the stack frame space is released after the function returns.
 - **Instruction space**: Used to store compiled program instructions, which are usually negligible in actual statistics.
 
-When analyzing the space complexity of a program, **we typically count the Temporary Data, Stack Frame Space, and Output Data**, as shown in the Figure 2-15 .
+When analyzing the space complexity of a program, **we typically count the Temporary Data, Stack Frame Space, and Output Data**, as shown in Figure 2-15.
 
 ![Space types used in algorithms](space_complexity.assets/space_types.png){ class="animation-figure" }
 
@@ -1444,7 +1444,7 @@ Linear order is common in arrays, linked lists, stacks, queues, etc., where the
     <div style="height: 477px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20linear%28n%3A%20int%29%3A%0A%20%20%20%20%22%22%22%E7%BA%BF%E6%80%A7%E9%98%B6%22%22%22%0A%20%20%20%20%23%20%E9%95%BF%E5%BA%A6%E4%B8%BA%20n%20%E7%9A%84%E5%88%97%E8%A1%A8%E5%8D%A0%E7%94%A8%20O%28n%29%20%E7%A9%BA%E9%97%B4%0A%20%20%20%20nums%20%3D%20%5B0%5D%20*%20n%0A%20%20%20%20%23%20%E9%95%BF%E5%BA%A6%E4%B8%BA%20n%20%E7%9A%84%E5%93%88%E5%B8%8C%E8%A1%A8%E5%8D%A0%E7%94%A8%20O%28n%29%20%E7%A9%BA%E9%97%B4%0A%20%20%20%20hmap%20%3D%20dict%5Bint,%20str%5D%28%29%0A%20%20%20%20for%20i%20in%20range%28n%29%3A%0A%20%20%20%20%20%20%20%20hmap%5Bi%5D%20%3D%20str%28i%29%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%205%0A%20%20%20%20print%28%22%E8%BE%93%E5%85%A5%E6%95%B0%E6%8D%AE%E5%A4%A7%E5%B0%8F%20n%20%3D%22,%20n%29%0A%0A%20%20%20%20%23%20%E7%BA%BF%E6%80%A7%E9%98%B6%0A%20%20%20%20linear%28n%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=20&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
     <div style="margin-top: 5px;"><a href="https://pythontutor.com/iframe-embed.html#code=def%20linear%28n%3A%20int%29%3A%0A%20%20%20%20%22%22%22%E7%BA%BF%E6%80%A7%E9%98%B6%22%22%22%0A%20%20%20%20%23%20%E9%95%BF%E5%BA%A6%E4%B8%BA%20n%20%E7%9A%84%E5%88%97%E8%A1%A8%E5%8D%A0%E7%94%A8%20O%28n%29%20%E7%A9%BA%E9%97%B4%0A%20%20%20%20nums%20%3D%20%5B0%5D%20*%20n%0A%20%20%20%20%23%20%E9%95%BF%E5%BA%A6%E4%B8%BA%20n%20%E7%9A%84%E5%93%88%E5%B8%8C%E8%A1%A8%E5%8D%A0%E7%94%A8%20O%28n%29%20%E7%A9%BA%E9%97%B4%0A%20%20%20%20hmap%20%3D%20dict%5Bint,%20str%5D%28%29%0A%20%20%20%20for%20i%20in%20range%28n%29%3A%0A%20%20%20%20%20%20%20%20hmap%5Bi%5D%20%3D%20str%28i%29%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%205%0A%20%20%20%20print%28%22%E8%BE%93%E5%85%A5%E6%95%B0%E6%8D%AE%E5%A4%A7%E5%B0%8F%20n%20%3D%22,%20n%29%0A%0A%20%20%20%20%23%20%E7%BA%BF%E6%80%A7%E9%98%B6%0A%20%20%20%20linear%28n%29&codeDivHeight=800&codeDivWidth=600&cumulative=false&curInstr=20&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">Full Screen ></a></div>
 
-As shown below, this function's recursive depth is $n$, meaning there are $n$ instances of unreturned `linear_recur()` function, using $O(n)$ size of stack frame space:
+As shown in Figure 2-17, this function's recursive depth is $n$, meaning there are $n$ instances of unreturned `linear_recur()` function, using $O(n)$ size of stack frame space:
 
 === "Python"
 
@@ -1869,7 +1869,7 @@ Quadratic order is common in matrices and graphs, where the number of elements i
     <div style="height: 405px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20quadratic%28n%3A%20int%29%3A%0A%20%20%20%20%22%22%22%E5%B9%B3%E6%96%B9%E9%98%B6%22%22%22%0A%20%20%20%20%23%20%E4%BA%8C%E7%BB%B4%E5%88%97%E8%A1%A8%E5%8D%A0%E7%94%A8%20O%28n%5E2%29%20%E7%A9%BA%E9%97%B4%0A%20%20%20%20num_matrix%20%3D%20%5B%5B0%5D%20*%20n%20for%20_%20in%20range%28n%29%5D%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%205%0A%20%20%20%20print%28%22%E8%BE%93%E5%85%A5%E6%95%B0%E6%8D%AE%E5%A4%A7%E5%B0%8F%20n%20%3D%22,%20n%29%0A%0A%20%20%20%20%23%20%E5%B9%B3%E6%96%B9%E9%98%B6%0A%20%20%20%20quadratic%28n%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=16&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
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-As shown below, the recursive depth of this function is $n$, and in each recursive call, an array is initialized with lengths $n$, $n-1$, $\dots$, $2$, $1$, averaging $n/2$, thus overall occupying $O(n^2)$ space:
+As shown in Figure 2-18, the recursive depth of this function is $n$, and in each recursive call, an array is initialized with lengths $n$, $n-1$, $\dots$, $2$, $1$, averaging $n/2$, thus overall occupying $O(n^2)$ space:
 
 === "Python"
 
@@ -2068,7 +2068,7 @@ As shown below, the recursive depth of this function is $n$, and in each recursi
 
 ### 4. &nbsp; Exponential order $O(2^n)$ {data-toc-label="4. &nbsp; Exponential order"}
 
-Exponential order is common in binary trees. Observe the below image, a "full binary tree" with $n$ levels has $2^n - 1$ nodes, occupying $O(2^n)$ space:
+Exponential order is common in binary trees. Observe Figure 2-19, a "full binary tree" with $n$ levels has $2^n - 1$ nodes, occupying $O(2^n)$ space:
 
 === "Python"
 

en/docs/chapter_computational_complexity/time_complexity.md

@@ -468,7 +468,7 @@ Let's understand this concept of "time growth trend" with an example. Assume the
     }
     ```
 
-The following figure shows the time complexities of these three algorithms.
+Figure 2-7 shows the time complexities of these three algorithms.
 
 - Algorithm `A` has just one print operation, and its run time does not grow with $n$. Its time complexity is considered "constant order."
 - Algorithm `B` involves a print operation looping $n$ times, and its run time grows linearly with $n$. Its time complexity is "linear order."
@@ -1794,7 +1794,7 @@ Quadratic order means the number of operations grows quadratically with the inpu
     <div style="height: 477px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20quadratic%28n%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E5%B9%B3%E6%96%B9%E9%98%B6%22%22%22%0A%20%20%20%20count%20%3D%200%0A%20%20%20%20%23%20%E5%BE%AA%E7%8E%AF%E6%AC%A1%E6%95%B0%E4%B8%8E%E6%95%B0%E6%8D%AE%E5%A4%A7%E5%B0%8F%20n%20%E6%88%90%E5%B9%B3%E6%96%B9%E5%85%B3%E7%B3%BB%0A%20%20%20%20for%20i%20in%20range%28n%29%3A%0A%20%20%20%20%20%20%20%20for%20j%20in%20range%28n%29%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20count%20%2B%3D%201%0A%20%20%20%20return%20count%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%208%0A%20%20%20%20print%28%22%E8%BE%93%E5%85%A5%E6%95%B0%E6%8D%AE%E5%A4%A7%E5%B0%8F%20n%20%3D%22,%20n%29%0A%0A%20%20%20%20count%20%3D%20quadratic%28n%29%0A%20%20%20%20print%28%22%E5%B9%B3%E6%96%B9%E9%98%B6%E7%9A%84%E6%93%8D%E4%BD%9C%E6%95%B0%E9%87%8F%20%3D%22,%20count%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=3&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
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-The following image compares constant order, linear order, and quadratic order time complexities.
+Figure 2-10 compares constant order, linear order, and quadratic order time complexities.
 
 ![Constant, linear, and quadratic order time complexities](time_complexity.assets/time_complexity_constant_linear_quadratic.png){ class="animation-figure" }
 
@@ -2132,7 +2132,7 @@ For instance, in bubble sort, the outer loop runs $n - 1$ times, and the inner l
 
 Biological "cell division" is a classic example of exponential order growth: starting with one cell, it becomes two after one division, four after two divisions, and so on, resulting in $2^n$ cells after $n$ divisions.
 
-The following image and code simulate the cell division process, with a time complexity of $O(2^n)$:
+Figure 2-11 and code simulate the cell division process, with a time complexity of $O(2^n)$:
 
 === "Python"
 
@@ -2566,7 +2566,7 @@ Exponential order growth is extremely rapid and is commonly seen in exhaustive s
 
 In contrast to exponential order, logarithmic order reflects situations where "the size is halved each round." Given an input data size $n$, since the size is halved each round, the number of iterations is $\log_2 n$, the inverse function of $2^n$.
 
-The following image and code simulate the "halving each round" process, with a time complexity of $O(\log_2 n)$, commonly abbreviated as $O(\log n)$:
+Figure 2-12 and code simulate the "halving each round" process, with a time complexity of $O(\log_2 n)$, commonly abbreviated as $O(\log n)$:
 
 === "Python"
 
@@ -3161,7 +3161,7 @@ Linear-logarithmic order often appears in nested loops, with the complexities of
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-The image below demonstrates how linear-logarithmic order is generated. Each level of a binary tree has $n$ operations, and the tree has $\log_2 n + 1$ levels, resulting in a time complexity of $O(n \log n)$.
+Figure 2-13 demonstrates how linear-logarithmic order is generated. Each level of a binary tree has $n$ operations, and the tree has $\log_2 n + 1$ levels, resulting in a time complexity of $O(n \log n)$.
 
 ![Linear-logarithmic order time complexity](time_complexity.assets/time_complexity_logarithmic_linear.png){ class="animation-figure" }
 
@@ -3177,7 +3177,7 @@ $$
 n! = n \times (n - 1) \times (n - 2) \times \dots \times 2 \times 1
 $$
 
-Factorials are typically implemented using recursion. As shown in the image and code below, the first level splits into $n$ branches, the second level into $n - 1$ branches, and so on, stopping after the $n$th level:
+Factorials are typically implemented using recursion. As shown in the code and Figure 2-14, the first level splits into $n$ branches, the second level into $n - 1$ branches, and so on, stopping after the $n$th level:
 
 === "Python"
 

en/docs/chapter_data_structure/character_encoding.md

@@ -8,7 +8,7 @@ In the computer system, all data is stored in binary form, and characters (repre
 
 ## 3.4.1   ASCII character set
 
-The "ASCII code" is one of the earliest character sets, officially known as the American Standard Code for Information Interchange. It uses 7 binary digits (the lower 7 bits of a byte) to represent a character, allowing for a maximum of 128 different characters. As shown in the Figure 3-6 , ASCII includes uppercase and lowercase English letters, numbers 0 ~ 9, various punctuation marks, and certain control characters (such as newline and tab).
+The "ASCII code" is one of the earliest character sets, officially known as the American Standard Code for Information Interchange. It uses 7 binary digits (the lower 7 bits of a byte) to represent a character, allowing for a maximum of 128 different characters. As shown in Figure 3-6, ASCII includes uppercase and lowercase English letters, numbers 0 ~ 9, various punctuation marks, and certain control characters (such as newline and tab).
 
 ![ASCII code](character_encoding.assets/ascii_table.png){ class="animation-figure" }
 
@@ -36,7 +36,7 @@ Since its release in 1991, Unicode has continually expanded to include new langu
 
 Unicode is a universal character set that assigns a number (called a "code point") to each character, **but it does not specify how these character code points should be stored in a computer system**. One might ask: How does a system interpret Unicode code points of varying lengths within a text? For example, given a 2-byte code, how does the system determine if it represents a single 2-byte character or two 1-byte characters?
 
-A straightforward solution to this problem is to store all characters as equal-length encodings. As shown in the Figure 3-7 , each character in "Hello" occupies 1 byte, while each character in "算法" (algorithm) occupies 2 bytes. We could encode all characters in "Hello 算法" as 2 bytes by padding the higher bits with zeros. This method would enable the system to interpret a character every 2 bytes, recovering the content of the phrase.
+A straightforward solution to this problem is to store all characters as equal-length encodings. As shown in Figure 3-7, each character in "Hello" occupies 1 byte, while each character in "算法" (algorithm) occupies 2 bytes. We could encode all characters in "Hello 算法" as 2 bytes by padding the higher bits with zeros. This method would enable the system to interpret a character every 2 bytes, recovering the content of the phrase.
 
 ![Unicode encoding example](character_encoding.assets/unicode_hello_algo.png){ class="animation-figure" }
 
@@ -53,7 +53,7 @@ The encoding rules for UTF-8 are not complex and can be divided into two cases:
 - For 1-byte characters, set the highest bit to $0$, and the remaining 7 bits to the Unicode code point. Notably, ASCII characters occupy the first 128 code points in the Unicode set. This means that **UTF-8 encoding is backward compatible with ASCII**. This implies that UTF-8 can be used to parse ancient ASCII text.
 - For characters of length $n$ bytes (where $n > 1$), set the highest $n$ bits of the first byte to $1$, and the $(n + 1)^{\text{th}}$ bit to $0$; starting from the second byte, set the highest 2 bits of each byte to $10$; the rest of the bits are used to fill the Unicode code point.
 
-The Figure 3-8  shows the UTF-8 encoding for "Hello算法". It can be observed that since the highest $n$ bits are set to $1$, the system can determine the length of the character as $n$ by counting the number of highest bits set to $1$.
+Figure 3-8 shows the UTF-8 encoding for "Hello算法". It can be observed that since the highest $n$ bits are set to $1$, the system can determine the length of the character as $n$ by counting the number of highest bits set to $1$.
 
 But why set the highest 2 bits of the remaining bytes to $10$? Actually, this $10$ serves as a kind of checksum. If the system starts parsing text from an incorrect byte, the $10$ at the beginning of the byte can help the system quickly detect anomalies.
 

en/docs/chapter_data_structure/classification_of_data_structure.md

@@ -10,7 +10,7 @@ Common data structures include arrays, linked lists, stacks, queues, hash tables
 
 **The logical structures reveal the logical relationships between data elements**. In arrays and linked lists, data are arranged in a specific sequence, demonstrating the linear relationship between data; while in trees, data are arranged hierarchically from the top down, showing the derived relationship between "ancestors" and "descendants"; and graphs are composed of nodes and edges, reflecting the intricate network relationship.
 
-As shown in the Figure 3-1 , logical structures can be divided into two major categories: "linear" and "non-linear". Linear structures are more intuitive, indicating data is arranged linearly in logical relationships; non-linear structures, conversely, are arranged non-linearly.
+As shown in Figure 3-1, logical structures can be divided into two major categories: "linear" and "non-linear". Linear structures are more intuitive, indicating data is arranged linearly in logical relationships; non-linear structures, conversely, are arranged non-linearly.
 
 - **Linear data structures**: Arrays, Linked Lists, Stacks, Queues, Hash Tables.
 - **Non-linear data structures**: Trees, Heaps, Graphs, Hash Tables.
@@ -27,9 +27,9 @@ Non-linear data structures can be further divided into tree structures and netwo
 
 ## 3.1.2   Physical structure: contiguous and dispersed
 
-**During the execution of an algorithm, the data being processed is stored in memory**. The Figure 3-2  shows a computer memory stick where each black square is a physical memory space. We can think of memory as a vast Excel spreadsheet, with each cell capable of storing a certain amount of data.
+**During the execution of an algorithm, the data being processed is stored in memory**. Figure 3-2 shows a computer memory stick where each black square is a physical memory space. We can think of memory as a vast Excel spreadsheet, with each cell capable of storing a certain amount of data.
 
-**The system accesses the data at the target location by means of a memory address**. As shown in the Figure 3-2 , the computer assigns a unique identifier to each cell in the table according to specific rules, ensuring that each memory space has a unique memory address. With these addresses, the program can access the data stored in memory.
+**The system accesses the data at the target location by means of a memory address**. As shown in Figure 3-2, the computer assigns a unique identifier to each cell in the table according to specific rules, ensuring that each memory space has a unique memory address. With these addresses, the program can access the data stored in memory.
 
 ![Memory stick, memory spaces, memory addresses](classification_of_data_structure.assets/computer_memory_location.png){ class="animation-figure" }
 
@@ -41,7 +41,7 @@ Non-linear data structures can be further divided into tree structures and netwo
 
 Memory is a shared resource for all programs. When a block of memory is occupied by one program, it cannot be simultaneously used by other programs. **Therefore, considering memory resources is crucial in designing data structures and algorithms**. For instance, the algorithm's peak memory usage should not exceed the remaining free memory of the system; if there is a lack of contiguous memory blocks, then the data structure chosen must be able to be stored in non-contiguous memory blocks.
 
-As illustrated in the Figure 3-3 , **the physical structure reflects the way data is stored in computer memory** and it can be divided into contiguous space storage (arrays) and non-contiguous space storage (linked lists). The two types of physical structures exhibit complementary characteristics in terms of time efficiency and space efficiency.
+As illustrated in Figure 3-3, **the physical structure reflects the way data is stored in computer memory** and it can be divided into contiguous space storage (arrays) and non-contiguous space storage (linked lists). The two types of physical structures exhibit complementary characteristics in terms of time efficiency and space efficiency.
 
 ![Contiguous space storage and dispersed space storage](classification_of_data_structure.assets/classification_phisical_structure.png){ class="animation-figure" }
 

en/docs/chapter_data_structure/number_encoding.md

@@ -143,7 +143,7 @@ Now we can answer the initial question: **The representation of `float` includes
 
 **However, the trade-off for `float`'s expanded range is a sacrifice in precision**. The integer type `int` uses all 32 bits to represent the number, with values evenly distributed; but due to the exponent bit, the larger the value of a `float`, the greater the difference between adjacent numbers.
 
-As shown in the Table 3-2 , exponent bits $\mathrm{E} = 0$ and $\mathrm{E} = 255$ have special meanings, **used to represent zero, infinity, $\mathrm{NaN}$, etc.**
+As shown in Table 3-2, exponent bits $\mathrm{E} = 0$ and $\mathrm{E} = 255$ have special meanings, **used to represent zero, infinity, $\mathrm{NaN}$, etc.**
 
 <p align="center"> Table 3-2 &nbsp; Meaning of exponent bits </p>
 

en/docs/chapter_divide_and_conquer/build_binary_tree_problem.md

@@ -47,7 +47,7 @@ Based on the above division method, **we have now obtained the index intervals o
 - Let the index of the current tree's root node in `inorder` be denoted as $m$.
 - Let the index interval of the current tree in `inorder` be denoted as $[l, r]$.
 
-As shown in the Table 12-1 , the above variables can represent the index of the root node in `preorder` as well as the index intervals of the subtrees in `inorder`.
+As shown in Table 12-1, the above variables can represent the index of the root node in `preorder` as well as the index intervals of the subtrees in `inorder`.
 
 <p align="center"> Table 12-1 &nbsp; Indexes of the root node and subtrees in preorder and inorder traversals </p>
 

en/docs/chapter_divide_and_conquer/divide_and_conquer.md

@@ -9,7 +9,7 @@ comments: true
 1. **Divide (partition phase)**: Recursively decompose the original problem into two or more sub-problems until the smallest sub-problem is reached and the process terminates.
 2. **Conquer (merge phase)**: Starting from the smallest sub-problem with a known solution, merge the solutions of the sub-problems from bottom to top to construct the solution to the original problem.
 
-As shown in the Figure 12-1 , "merge sort" is one of the typical applications of the divide and conquer strategy.
+As shown in Figure 12-1, "merge sort" is one of the typical applications of the divide and conquer strategy.
 
 1. **Divide**: Recursively divide the original array (original problem) into two sub-arrays (sub-problems), until the sub-array has only one element (smallest sub-problem).
 2. **Conquer**: Merge the ordered sub-arrays (solutions to the sub-problems) from bottom to top to obtain an ordered original array (solution to the original problem).
@@ -40,7 +40,7 @@ Then, we may ask: **Why can divide and conquer improve algorithm efficiency, and
 
 ### 1.   Optimization of operation count
 
-Taking "bubble sort" as an example, it requires $O(n^2)$ time to process an array of length $n$. Suppose we divide the array from the midpoint into two sub-arrays as shown in the Figure 12-2 , then the division requires $O(n)$ time, sorting each sub-array requires $O((n / 2)^2)$ time, and merging the two sub-arrays requires $O(n)$ time, with the total time complexity being:
+Taking "bubble sort" as an example, it requires $O(n^2)$ time to process an array of length $n$. Suppose we divide the array from the midpoint into two sub-arrays as shown in Figure 12-2, then the division requires $O(n)$ time, sorting each sub-array requires $O((n / 2)^2)$ time, and merging the two sub-arrays requires $O(n)$ time, with the total time complexity being:
 
 $$
 O(n + (\frac{n}{2})^2 \times 2 + n) = O(\frac{n^2}{2} + 2n)
@@ -72,7 +72,7 @@ We know that the sub-problems generated by divide and conquer are independent of
 
 Parallel optimization is especially effective in environments with multiple cores or processors, as the system can process multiple sub-problems simultaneously, making fuller use of computing resources and significantly reducing the overall runtime.
 
-For example, in the "bucket sort" shown in the Figure 12-3 , we distribute massive data evenly across various buckets, then the sorting tasks of all buckets can be distributed to different computing units, and the results are merged after completion.
+For example, in the "bucket sort" shown in Figure 12-3, we distribute massive data evenly across various buckets, then the sorting tasks of all buckets can be distributed to different computing units, and the results are merged after completion.
 
 ![Bucket sort's parallel computation](divide_and_conquer.assets/divide_and_conquer_parallel_computing.png){ class="animation-figure" }
 

en/docs/chapter_divide_and_conquer/hanota_problem.md

@@ -8,7 +8,7 @@ In both merge sorting and building binary trees, we decompose the original probl
 
 !!! question
 
-    Given three pillars, denoted as `A`, `B`, and `C`. Initially, pillar `A` is stacked with $n$ discs, arranged in order from top to bottom from smallest to largest. Our task is to move these $n$ discs to pillar `C`, maintaining their original order (as shown below). The following rules must be followed during the disc movement process:
+    Given three pillars, denoted as `A`, `B`, and `C`. Initially, pillar `A` is stacked with $n$ discs, arranged in order from top to bottom from smallest to largest. Our task is to move these $n$ discs to pillar `C`, maintaining their original order (as shown in Figure 12-10). The following rules must be followed during the disc movement process:
     
     1. A disc can only be picked up from the top of a pillar and placed on top of another pillar.
     2. Only one disc can be moved at a time.
@@ -22,7 +22,7 @@ In both merge sorting and building binary trees, we decompose the original probl
 
 ### 1.   Consider the base case
 
-As shown below, for the problem $f(1)$, i.e., when there is only one disc, we can directly move it from `A` to `C`.
+As shown in Figure 12-11, for the problem $f(1)$, i.e., when there is only one disc, we can directly move it from `A` to `C`.
 
 === "<1>"
     ![Solution for a problem of size 1](hanota_problem.assets/hanota_f1_step1.png){ class="animation-figure" }
@@ -32,7 +32,7 @@ As shown below, for the problem $f(1)$, i.e., when there is only one disc, we ca
 
 <p align="center"> Figure 12-11 &nbsp; Solution for a problem of size 1 </p>
 
-As shown below, for the problem $f(2)$, i.e., when there are two discs, **since the smaller disc must always be above the larger disc, `B` is needed to assist in the movement**.
+As shown in Figure 12-12, for the problem $f(2)$, i.e., when there are two discs, **since the smaller disc must always be above the larger disc, `B` is needed to assist in the movement**.
 
 1. First, move the smaller disc from `A` to `B`.
 2. Then move the larger disc from `A` to `C`.
@@ -58,7 +58,7 @@ The process of solving the problem $f(2)$ can be summarized as: **moving two dis
 
 For the problem $f(3)$, i.e., when there are three discs, the situation becomes slightly more complicated.
 
-Since we already know the solutions to $f(1)$ and $f(2)$, we can think from a divide-and-conquer perspective and **consider the two top discs on `A` as a unit**, performing the steps shown below. This way, the three discs are successfully moved from `A` to `C`.
+Since we already know the solutions to $f(1)$ and $f(2)$, we can think from a divide-and-conquer perspective and **consider the two top discs on `A` as a unit**, performing the steps shown in Figure 12-13. This way, the three discs are successfully moved from `A` to `C`.
 
 1. Let `B` be the target pillar and `C` the buffer pillar, and move the two discs from `A` to `B`.
 2. Move the remaining disc from `A` directly to `C`.
@@ -80,7 +80,7 @@ Since we already know the solutions to $f(1)$ and $f(2)$, we can think from a di
 
 Essentially, **we divide the problem $f(3)$ into two subproblems $f(2)$ and one subproblem $f(1)$**. By solving these three subproblems in order, the original problem is resolved. This indicates that the subproblems are independent, and their solutions can be merged.
 
-From this, we can summarize the divide-and-conquer strategy for solving the Tower of Hanoi shown in the following image: divide the original problem $f(n)$ into two subproblems $f(n-1)$ and one subproblem $f(1)$, and solve these three subproblems in the following order.
+From this, we can summarize the divide-and-conquer strategy for solving the Tower of Hanoi shown in Figure 12-14: divide the original problem $f(n)$ into two subproblems $f(n-1)$ and one subproblem $f(1)$, and solve these three subproblems in the following order.
 
 1. Move $n-1$ discs with the help of `C` from `A` to `B`.
 2. Move the remaining one disc directly from `A` to `C`.
@@ -532,7 +532,7 @@ In the code, we declare a recursive function `dfs(i, src, buf, tar)` whose role
     <div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20move%28src%3A%20list%5Bint%5D,%20tar%3A%20list%5Bint%5D%29%3A%0A%20%20%20%20%22%22%22%E7%A7%BB%E5%8A%A8%E4%B8%80%E4%B8%AA%E5%9C%86%E7%9B%98%22%22%22%0A%20%20%20%20%23%20%E4%BB%8E%20src%20%E9%A1%B6%E9%83%A8%E6%8B%BF%E5%87%BA%E4%B8%80%E4%B8%AA%E5%9C%86%E7%9B%98%0A%20%20%20%20pan%20%3D%20src.pop%28%29%0A%20%20%20%20%23%20%E5%B0%86%E5%9C%86%E7%9B%98%E6%94%BE%E5%85%A5%20tar%20%E9%A1%B6%E9%83%A8%0A%20%20%20%20tar.append%28pan%29%0A%0A%0Adef%20dfs%28i%3A%20int,%20src%3A%20list%5Bint%5D,%20buf%3A%20list%5Bint%5D,%20tar%3A%20list%5Bint%5D%29%3A%0A%20%20%20%20%22%22%22%E6%B1%82%E8%A7%A3%E6%B1%89%E8%AF%BA%E5%A1%94%E9%97%AE%E9%A2%98%20f%28i%29%22%22%22%0A%20%20%20%20%23%20%E8%8B%A5%20src%20%E5%8F%AA%E5%89%A9%E4%B8%8B%E4%B8%80%E4%B8%AA%E5%9C%86%E7%9B%98%EF%BC%8C%E5%88%99%E7%9B%B4%E6%8E%A5%E5%B0%86%E5%85%B6%E7%A7%BB%E5%88%B0%20tar%0A%20%20%20%20if%20i%20%3D%3D%201%3A%0A%20%20%20%20%20%20%20%20move%28src,%20tar%29%0A%20%20%20%20%20%20%20%20return%0A%20%20%20%20%23%20%E5%AD%90%E9%97%AE%E9%A2%98%20f%28i-1%29%20%EF%BC%9A%E5%B0%86%20src%20%E9%A1%B6%E9%83%A8%20i-1%20%E4%B8%AA%E5%9C%86%E7%9B%98%E5%80%9F%E5%8A%A9%20tar%20%E7%A7%BB%E5%88%B0%20buf%0A%20%20%20%20dfs%28i%20-%201,%20src,%20tar,%20buf%29%0A%20%20%20%20%23%20%E5%AD%90%E9%97%AE%E9%A2%98%20f%281%29%20%EF%BC%9A%E5%B0%86%20src%20%E5%89%A9%E4%BD%99%E4%B8%80%E4%B8%AA%E5%9C%86%E7%9B%98%E7%A7%BB%E5%88%B0%20tar%0A%20%20%20%20move%28src,%20tar%29%0A%20%20%20%20%23%20%E5%AD%90%E9%97%AE%E9%A2%98%20f%28i-1%29%20%EF%BC%9A%E5%B0%86%20buf%20%E9%A1%B6%E9%83%A8%20i-1%20%E4%B8%AA%E5%9C%86%E7%9B%98%E5%80%9F%E5%8A%A9%20src%20%E7%A7%BB%E5%88%B0%20tar%0A%20%20%20%20dfs%28i%20-%201,%20buf,%20src,%20tar%29%0A%0A%0Adef%20solve_hanota%28A%3A%20list%5Bint%5D,%20B%3A%20list%5Bint%5D,%20C%3A%20list%5Bint%5D%29%3A%0A%20%20%20%20%22%22%22%E6%B1%82%E8%A7%A3%E6%B1%89%E8%AF%BA%E5%A1%94%E9%97%AE%E9%A2%98%22%22%22%0A%20%20%20%20n%20%3D%20len%28A%29%0A%20%20%20%20%23%20%E5%B0%86%20A%20%E9%A1%B6%E9%83%A8%20n%20%E4%B8%AA%E5%9C%86%E7%9B%98%E5%80%9F%E5%8A%A9%20B%20%E7%A7%BB%E5%88%B0%20C%0A%20%20%20%20dfs%28n,%20A,%20B,%20C%29%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20%23%20%E5%88%97%E8%A1%A8%E5%B0%BE%E9%83%A8%E6%98%AF%E6%9F%B1%E5%AD%90%E9%A1%B6%E9%83%A8%0A%20%20%20%20A%20%3D%20%5B5,%204,%203,%202,%201%5D%0A%20%20%20%20B%20%3D%20%5B%5D%0A%20%20%20%20C%20%3D%20%5B%5D%0A%20%20%20%20print%28%22%E5%88%9D%E5%A7%8B%E7%8A%B6%E6%80%81%E4%B8%8B%EF%BC%9A%22%29%0A%20%20%20%20print%28f%22A%20%3D%20%7BA%7D%22%29%0A%20%20%20%20print%28f%22B%20%3D%20%7BB%7D%22%29%0A%20%20%20%20print%28f%22C%20%3D%20%7BC%7D%22%29%0A%0A%20%20%20%20solve_hanota%28A,%20B,%20C%29%0A%0A%20%20%20%20print%28%22%E5%9C%86%E7%9B%98%E7%A7%BB%E5%8A%A8%E5%AE%8C%E6%88%90%E5%90%8E%EF%BC%9A%22%29%0A%20%20%20%20print%28f%22A%20%3D%20%7BA%7D%22%29%0A%20%20%20%20print%28f%22B%20%3D%20%7BB%7D%22%29%0A%20%20%20%20print%28f%22C%20%3D%20%7BC%7D%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=12&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
     <div style="margin-top: 5px;"><a href="https://pythontutor.com/iframe-embed.html#code=def%20move%28src%3A%20list%5Bint%5D,%20tar%3A%20list%5Bint%5D%29%3A%0A%20%20%20%20%22%22%22%E7%A7%BB%E5%8A%A8%E4%B8%80%E4%B8%AA%E5%9C%86%E7%9B%98%22%22%22%0A%20%20%20%20%23%20%E4%BB%8E%20src%20%E9%A1%B6%E9%83%A8%E6%8B%BF%E5%87%BA%E4%B8%80%E4%B8%AA%E5%9C%86%E7%9B%98%0A%20%20%20%20pan%20%3D%20src.pop%28%29%0A%20%20%20%20%23%20%E5%B0%86%E5%9C%86%E7%9B%98%E6%94%BE%E5%85%A5%20tar%20%E9%A1%B6%E9%83%A8%0A%20%20%20%20tar.append%28pan%29%0A%0A%0Adef%20dfs%28i%3A%20int,%20src%3A%20list%5Bint%5D,%20buf%3A%20list%5Bint%5D,%20tar%3A%20list%5Bint%5D%29%3A%0A%20%20%20%20%22%22%22%E6%B1%82%E8%A7%A3%E6%B1%89%E8%AF%BA%E5%A1%94%E9%97%AE%E9%A2%98%20f%28i%29%22%22%22%0A%20%20%20%20%23%20%E8%8B%A5%20src%20%E5%8F%AA%E5%89%A9%E4%B8%8B%E4%B8%80%E4%B8%AA%E5%9C%86%E7%9B%98%EF%BC%8C%E5%88%99%E7%9B%B4%E6%8E%A5%E5%B0%86%E5%85%B6%E7%A7%BB%E5%88%B0%20tar%0A%20%20%20%20if%20i%20%3D%3D%201%3A%0A%20%20%20%20%20%20%20%20move%28src,%20tar%29%0A%20%20%20%20%20%20%20%20return%0A%20%20%20%20%23%20%E5%AD%90%E9%97%AE%E9%A2%98%20f%28i-1%29%20%EF%BC%9A%E5%B0%86%20src%20%E9%A1%B6%E9%83%A8%20i-1%20%E4%B8%AA%E5%9C%86%E7%9B%98%E5%80%9F%E5%8A%A9%20tar%20%E7%A7%BB%E5%88%B0%20buf%0A%20%20%20%20dfs%28i%20-%201,%20src,%20tar,%20buf%29%0A%20%20%20%20%23%20%E5%AD%90%E9%97%AE%E9%A2%98%20f%281%29%20%EF%BC%9A%E5%B0%86%20src%20%E5%89%A9%E4%BD%99%E4%B8%80%E4%B8%AA%E5%9C%86%E7%9B%98%E7%A7%BB%E5%88%B0%20tar%0A%20%20%20%20move%28src,%20tar%29%0A%20%20%20%20%23%20%E5%AD%90%E9%97%AE%E9%A2%98%20f%28i-1%29%20%EF%BC%9A%E5%B0%86%20buf%20%E9%A1%B6%E9%83%A8%20i-1%20%E4%B8%AA%E5%9C%86%E7%9B%98%E5%80%9F%E5%8A%A9%20src%20%E7%A7%BB%E5%88%B0%20tar%0A%20%20%20%20dfs%28i%20-%201,%20buf,%20src,%20tar%29%0A%0A%0Adef%20solve_hanota%28A%3A%20list%5Bint%5D,%20B%3A%20list%5Bint%5D,%20C%3A%20list%5Bint%5D%29%3A%0A%20%20%20%20%22%22%22%E6%B1%82%E8%A7%A3%E6%B1%89%E8%AF%BA%E5%A1%94%E9%97%AE%E9%A2%98%22%22%22%0A%20%20%20%20n%20%3D%20len%28A%29%0A%20%20%20%20%23%20%E5%B0%86%20A%20%E9%A1%B6%E9%83%A8%20n%20%E4%B8%AA%E5%9C%86%E7%9B%98%E5%80%9F%E5%8A%A9%20B%20%E7%A7%BB%E5%88%B0%20C%0A%20%20%20%20dfs%28n,%20A,%20B,%20C%29%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20%23%20%E5%88%97%E8%A1%A8%E5%B0%BE%E9%83%A8%E6%98%AF%E6%9F%B1%E5%AD%90%E9%A1%B6%E9%83%A8%0A%20%20%20%20A%20%3D%20%5B5,%204,%203,%202,%201%5D%0A%20%20%20%20B%20%3D%20%5B%5D%0A%20%20%20%20C%20%3D%20%5B%5D%0A%20%20%20%20print%28%22%E5%88%9D%E5%A7%8B%E7%8A%B6%E6%80%81%E4%B8%8B%EF%BC%9A%22%29%0A%20%20%20%20print%28f%22A%20%3D%20%7BA%7D%22%29%0A%20%20%20%20print%28f%22B%20%3D%20%7BB%7D%22%29%0A%20%20%20%20print%28f%22C%20%3D%20%7BC%7D%22%29%0A%0A%20%20%20%20solve_hanota%28A,%20B,%20C%29%0A%0A%20%20%20%20print%28%22%E5%9C%86%E7%9B%98%E7%A7%BB%E5%8A%A8%E5%AE%8C%E6%88%90%E5%90%8E%EF%BC%9A%22%29%0A%20%20%20%20print%28f%22A%20%3D%20%7BA%7D%22%29%0A%20%20%20%20print%28f%22B%20%3D%20%7BB%7D%22%29%0A%20%20%20%20print%28f%22C%20%3D%20%7BC%7D%22%29&codeDivHeight=800&codeDivWidth=600&cumulative=false&curInstr=12&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">Full Screen ></a></div>
 
-As shown below, the Tower of Hanoi forms a recursive tree with a height of $n$, each node representing a subproblem, corresponding to an open `dfs()` function, **thus the time complexity is $O(2^n)$, and the space complexity is $O(n)$**.
+As shown in Figure 12-15, the Tower of Hanoi forms a recursive tree with a height of $n$, each node representing a subproblem, corresponding to an open `dfs()` function, **thus the time complexity is $O(2^n)$, and the space complexity is $O(n)$**.
 
 ![Recursive tree of the Tower of Hanoi](hanota_problem.assets/hanota_recursive_tree.png){ class="animation-figure" }
 

en/docs/chapter_dynamic_programming/dp_problem_features.md

@@ -20,7 +20,7 @@ We make a slight modification to the stair climbing problem to make it more suit
 
     Given a staircase, you can step up 1 or 2 steps at a time, and each step on the staircase has a non-negative integer representing the cost you need to pay at that step. Given a non-negative integer array $cost$, where $cost[i]$ represents the cost you need to pay at the $i$-th step, $cost[0]$ is the ground (starting point). What is the minimum cost required to reach the top?
 
-As shown in the Figure 14-6 , if the costs of the 1st, 2nd, and 3rd steps are $1$, $10$, and $1$ respectively, then the minimum cost to climb to the 3rd step from the ground is $2$.
+As shown in Figure 14-6, if the costs of the 1st, 2nd, and 3rd steps are $1$, $10$, and $1$ respectively, then the minimum cost to climb to the 3rd step from the ground is $2$.
 
 ![Minimum cost to climb to the 3rd step](dp_problem_features.assets/min_cost_cs_example.png){ class="animation-figure" }
 
@@ -334,7 +334,7 @@ According to the state transition equation, and the initial states $dp[1] = cost
     <div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20min_cost_climbing_stairs_dp%28cost%3A%20list%5Bint%5D%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E7%88%AC%E6%A5%BC%E6%A2%AF%E6%9C%80%E5%B0%8F%E4%BB%A3%E4%BB%B7%EF%BC%9A%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%22%22%22%0A%20%20%20%20n%20%3D%20len%28cost%29%20-%201%0A%20%20%20%20if%20n%20%3D%3D%201%20or%20n%20%3D%3D%202%3A%0A%20%20%20%20%20%20%20%20return%20cost%5Bn%5D%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%20dp%20%E8%A1%A8%EF%BC%8C%E7%94%A8%E4%BA%8E%E5%AD%98%E5%82%A8%E5%AD%90%E9%97%AE%E9%A2%98%E7%9A%84%E8%A7%A3%0A%20%20%20%20dp%20%3D%20%5B0%5D%20*%20%28n%20%2B%201%29%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E7%8A%B6%E6%80%81%EF%BC%9A%E9%A2%84%E8%AE%BE%E6%9C%80%E5%B0%8F%E5%AD%90%E9%97%AE%E9%A2%98%E7%9A%84%E8%A7%A3%0A%20%20%20%20dp%5B1%5D,%20dp%5B2%5D%20%3D%20cost%5B1%5D,%20cost%5B2%5D%0A%20%20%20%20%23%20%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%EF%BC%9A%E4%BB%8E%E8%BE%83%E5%B0%8F%E5%AD%90%E9%97%AE%E9%A2%98%E9%80%90%E6%AD%A5%E6%B1%82%E8%A7%A3%E8%BE%83%E5%A4%A7%E5%AD%90%E9%97%AE%E9%A2%98%0A%20%20%20%20for%20i%20in%20range%283,%20n%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20dp%5Bi%5D%20%3D%20min%28dp%5Bi%20-%201%5D,%20dp%5Bi%20-%202%5D%29%20%2B%20cost%5Bi%5D%0A%20%20%20%20return%20dp%5Bn%5D%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20cost%20%3D%20%5B0,%201,%2010,%201,%201,%201,%2010,%201,%201,%2010,%201%5D%0A%20%20%20%20print%28f%22%E8%BE%93%E5%85%A5%E6%A5%BC%E6%A2%AF%E7%9A%84%E4%BB%A3%E4%BB%B7%E5%88%97%E8%A1%A8%E4%B8%BA%20%7Bcost%7D%22%29%0A%0A%20%20%20%20res%20%3D%20min_cost_climbing_stairs_dp%28cost%29%0A%20%20%20%20print%28f%22%E7%88%AC%E5%AE%8C%E6%A5%BC%E6%A2%AF%E7%9A%84%E6%9C%80%E4%BD%8E%E4%BB%A3%E4%BB%B7%E4%B8%BA%20%7Bres%7D%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=4&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
     <div style="margin-top: 5px;"><a href="https://pythontutor.com/iframe-embed.html#code=def%20min_cost_climbing_stairs_dp%28cost%3A%20list%5Bint%5D%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E7%88%AC%E6%A5%BC%E6%A2%AF%E6%9C%80%E5%B0%8F%E4%BB%A3%E4%BB%B7%EF%BC%9A%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%22%22%22%0A%20%20%20%20n%20%3D%20len%28cost%29%20-%201%0A%20%20%20%20if%20n%20%3D%3D%201%20or%20n%20%3D%3D%202%3A%0A%20%20%20%20%20%20%20%20return%20cost%5Bn%5D%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%20dp%20%E8%A1%A8%EF%BC%8C%E7%94%A8%E4%BA%8E%E5%AD%98%E5%82%A8%E5%AD%90%E9%97%AE%E9%A2%98%E7%9A%84%E8%A7%A3%0A%20%20%20%20dp%20%3D%20%5B0%5D%20*%20%28n%20%2B%201%29%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E7%8A%B6%E6%80%81%EF%BC%9A%E9%A2%84%E8%AE%BE%E6%9C%80%E5%B0%8F%E5%AD%90%E9%97%AE%E9%A2%98%E7%9A%84%E8%A7%A3%0A%20%20%20%20dp%5B1%5D,%20dp%5B2%5D%20%3D%20cost%5B1%5D,%20cost%5B2%5D%0A%20%20%20%20%23%20%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%EF%BC%9A%E4%BB%8E%E8%BE%83%E5%B0%8F%E5%AD%90%E9%97%AE%E9%A2%98%E9%80%90%E6%AD%A5%E6%B1%82%E8%A7%A3%E8%BE%83%E5%A4%A7%E5%AD%90%E9%97%AE%E9%A2%98%0A%20%20%20%20for%20i%20in%20range%283,%20n%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20dp%5Bi%5D%20%3D%20min%28dp%5Bi%20-%201%5D,%20dp%5Bi%20-%202%5D%29%20%2B%20cost%5Bi%5D%0A%20%20%20%20return%20dp%5Bn%5D%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20cost%20%3D%20%5B0,%201,%2010,%201,%201,%201,%2010,%201,%201,%2010,%201%5D%0A%20%20%20%20print%28f%22%E8%BE%93%E5%85%A5%E6%A5%BC%E6%A2%AF%E7%9A%84%E4%BB%A3%E4%BB%B7%E5%88%97%E8%A1%A8%E4%B8%BA%20%7Bcost%7D%22%29%0A%0A%20%20%20%20res%20%3D%20min_cost_climbing_stairs_dp%28cost%29%0A%20%20%20%20print%28f%22%E7%88%AC%E5%AE%8C%E6%A5%BC%E6%A2%AF%E7%9A%84%E6%9C%80%E4%BD%8E%E4%BB%A3%E4%BB%B7%E4%B8%BA%20%7Bres%7D%22%29&codeDivHeight=800&codeDivWidth=600&cumulative=false&curInstr=4&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">Full Screen ></a></div>
 
-The Figure 14-7  shows the dynamic programming process for the above code.
+Figure 14-7 shows the dynamic programming process for the above code.
 
 ![Dynamic programming process for minimum cost of climbing stairs](dp_problem_features.assets/min_cost_cs_dp.png){ class="animation-figure" }
 
@@ -610,7 +610,7 @@ However, if we add a constraint to the stair climbing problem, the situation cha
 
     Given a staircase with $n$ steps, you can go up 1 or 2 steps each time, **but you cannot jump 1 step twice in a row**. How many ways are there to climb to the top?
 
-As shown in the Figure 14-8 , there are only 2 feasible options for climbing to the 3rd step, among which the option of jumping 1 step three times in a row does not meet the constraint condition and is therefore discarded.
+As shown in Figure 14-8, there are only 2 feasible options for climbing to the 3rd step, among which the option of jumping 1 step three times in a row does not meet the constraint condition and is therefore discarded.
 
 ![Number of feasible options for climbing to the 3rd step with constraints](dp_problem_features.assets/climbing_stairs_constraint_example.png){ class="animation-figure" }
 
@@ -625,7 +625,7 @@ For this, we need to expand the state definition: **State $[i, j]$ represents be
 - When the last round was a jump of 1 step, the round before last could only choose to jump 2 steps, that is, $dp[i, 1]$ can only be transferred from $dp[i-1, 2]$.
 - When the last round was a jump of 2 steps, the round before last could choose to jump 1 step or 2 steps, that is, $dp[i, 2]$ can be transferred from $dp[i-2, 1]$ or $dp[i-2, 2]$.
 
-As shown in the Figure 14-9 , $dp[i, j]$ represents the number of solutions for state $[i, j]$. At this point, the state transition equation is:
+As shown in Figure 14-9, $dp[i, j]$ represents the number of solutions for state $[i, j]$. At this point, the state transition equation is:
 
 $$
 \begin{cases}

en/docs/chapter_dynamic_programming/dp_solution_pipeline.md

@@ -39,7 +39,7 @@ To illustrate the problem-solving steps more vividly, we use a classic problem,
 
     Given an $n \times m$ two-dimensional grid `grid`, each cell in the grid contains a non-negative integer representing the cost of that cell. The robot starts from the top-left cell and can only move down or right at each step until it reaches the bottom-right cell. Return the minimum path sum from the top-left to the bottom-right.
 
-The following figure shows an example, where the given grid's minimum path sum is $13$.
+Figure 14-10 shows an example, where the given grid's minimum path sum is $13$.
 
 ![Minimum Path Sum Example Data](dp_solution_pipeline.assets/min_path_sum_example.png){ class="animation-figure" }
 
@@ -51,7 +51,7 @@ Each round of decisions in this problem is to move one step down or right from t
 
 The state $[i, j]$ corresponds to the subproblem: the minimum path sum from the starting point $[0, 0]$ to $[i, j]$, denoted as $dp[i, j]$.
 
-Thus, we obtain the two-dimensional $dp$ matrix shown below, whose size is the same as the input grid $grid$.
+Thus, we obtain the two-dimensional $dp$ matrix shown in Figure 14-11, whose size is the same as the input grid $grid$.
 
 ![State definition and DP table](dp_solution_pipeline.assets/min_path_sum_solution_state_definition.png){ class="animation-figure" }
 
@@ -67,7 +67,7 @@ Thus, we obtain the two-dimensional $dp$ matrix shown below, whose size is the s
 
 For the state $[i, j]$, it can only be derived from the cell above $[i-1, j]$ or the cell to the left $[i, j-1]$. Therefore, the optimal substructure is: the minimum path sum to reach $[i, j]$ is determined by the smaller of the minimum path sums of $[i, j-1]$ and $[i-1, j]$.
 
-Based on the above analysis, the state transition equation shown in the following figure can be derived:
+Based on the above analysis, the state transition equation shown in Figure 14-12 can be derived:
 
 $$
 dp[i, j] = \min(dp[i-1, j], dp[i, j-1]) + grid[i, j]
@@ -87,7 +87,7 @@ $$
 
 In this problem, the states in the first row can only come from the states to their left, and the states in the first column can only come from the states above them, so the first row $i = 0$ and the first column $j = 0$ are the boundary conditions.
 
-As shown in the Figure 14-13 , since each cell is derived from the cell to its left and the cell above it, we use loops to traverse the matrix, the outer loop iterating over the rows and the inner loop iterating over the columns.
+As shown in Figure 14-13, since each cell is derived from the cell to its left and the cell above it, we use loops to traverse the matrix, the outer loop iterating over the rows and the inner loop iterating over the columns.
 
 ![Boundary conditions and state transition order](dp_solution_pipeline.assets/min_path_sum_solution_initial_state.png){ class="animation-figure" }
 
@@ -398,7 +398,7 @@ Implementation code as follows:
     <div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=from%20math%20import%20inf%0A%0Adef%20min_path_sum_dfs%28grid%3A%20list%5Blist%5Bint%5D%5D,%20i%3A%20int,%20j%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E6%9C%80%E5%B0%8F%E8%B7%AF%E5%BE%84%E5%92%8C%EF%BC%9A%E6%9A%B4%E5%8A%9B%E6%90%9C%E7%B4%A2%22%22%22%0A%20%20%20%20%23%20%E8%8B%A5%E4%B8%BA%E5%B7%A6%E4%B8%8A%E8%A7%92%E5%8D%95%E5%85%83%E6%A0%BC%EF%BC%8C%E5%88%99%E7%BB%88%E6%AD%A2%E6%90%9C%E7%B4%A2%0A%20%20%20%20if%20i%20%3D%3D%200%20and%20j%20%3D%3D%200%3A%0A%20%20%20%20%20%20%20%20return%20grid%5B0%5D%5B0%5D%0A%20%20%20%20%23%20%E8%8B%A5%E8%A1%8C%E5%88%97%E7%B4%A2%E5%BC%95%E8%B6%8A%E7%95%8C%EF%BC%8C%E5%88%99%E8%BF%94%E5%9B%9E%20%2B%E2%88%9E%20%E4%BB%A3%E4%BB%B7%0A%20%20%20%20if%20i%20%3C%200%20or%20j%20%3C%200%3A%0A%20%20%20%20%20%20%20%20return%20inf%0A%20%20%20%20%23%20%E8%AE%A1%E7%AE%97%E4%BB%8E%E5%B7%A6%E4%B8%8A%E8%A7%92%E5%88%B0%20%28i-1,%20j%29%20%E5%92%8C%20%28i,%20j-1%29%20%E7%9A%84%E6%9C%80%E5%B0%8F%E8%B7%AF%E5%BE%84%E4%BB%A3%E4%BB%B7%0A%20%20%20%20up%20%3D%20min_path_sum_dfs%28grid,%20i%20-%201,%20j%29%0A%20%20%20%20left%20%3D%20min_path_sum_dfs%28grid,%20i,%20j%20-%201%29%0A%20%20%20%20%23%20%E8%BF%94%E5%9B%9E%E4%BB%8E%E5%B7%A6%E4%B8%8A%E8%A7%92%E5%88%B0%20%28i,%20j%29%20%E7%9A%84%E6%9C%80%E5%B0%8F%E8%B7%AF%E5%BE%84%E4%BB%A3%E4%BB%B7%0A%20%20%20%20return%20min%28left,%20up%29%20%2B%20grid%5Bi%5D%5Bj%5D%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20grid%20%3D%20%5B%5B1,%203,%201,%205%5D,%20%5B2,%202,%204,%202%5D,%20%5B5,%203,%202,%201%5D,%20%5B4,%203,%205,%202%5D%5D%0A%20%20%20%20n,%20m%20%3D%20len%28grid%29,%20len%28grid%5B0%5D%29%0A%0A%20%20%20%20%23%20%E6%9A%B4%E5%8A%9B%E6%90%9C%E7%B4%A2%0A%20%20%20%20res%20%3D%20min_path_sum_dfs%28grid,%20n%20-%201,%20m%20-%201%29%0A%20%20%20%20print%28f%22%E4%BB%8E%E5%B7%A6%E4%B8%8A%E8%A7%92%E5%88%B0%E5%8F%B3%E4%B8%8B%E8%A7%92%E7%9A%84%E5%81%9A%E5%B0%8F%E8%B7%AF%E5%BE%84%E5%92%8C%E4%B8%BA%20%7Bres%7D%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=6&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
     <div style="margin-top: 5px;"><a href="https://pythontutor.com/iframe-embed.html#code=from%20math%20import%20inf%0A%0Adef%20min_path_sum_dfs%28grid%3A%20list%5Blist%5Bint%5D%5D,%20i%3A%20int,%20j%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E6%9C%80%E5%B0%8F%E8%B7%AF%E5%BE%84%E5%92%8C%EF%BC%9A%E6%9A%B4%E5%8A%9B%E6%90%9C%E7%B4%A2%22%22%22%0A%20%20%20%20%23%20%E8%8B%A5%E4%B8%BA%E5%B7%A6%E4%B8%8A%E8%A7%92%E5%8D%95%E5%85%83%E6%A0%BC%EF%BC%8C%E5%88%99%E7%BB%88%E6%AD%A2%E6%90%9C%E7%B4%A2%0A%20%20%20%20if%20i%20%3D%3D%200%20and%20j%20%3D%3D%200%3A%0A%20%20%20%20%20%20%20%20return%20grid%5B0%5D%5B0%5D%0A%20%20%20%20%23%20%E8%8B%A5%E8%A1%8C%E5%88%97%E7%B4%A2%E5%BC%95%E8%B6%8A%E7%95%8C%EF%BC%8C%E5%88%99%E8%BF%94%E5%9B%9E%20%2B%E2%88%9E%20%E4%BB%A3%E4%BB%B7%0A%20%20%20%20if%20i%20%3C%200%20or%20j%20%3C%200%3A%0A%20%20%20%20%20%20%20%20return%20inf%0A%20%20%20%20%23%20%E8%AE%A1%E7%AE%97%E4%BB%8E%E5%B7%A6%E4%B8%8A%E8%A7%92%E5%88%B0%20%28i-1,%20j%29%20%E5%92%8C%20%28i,%20j-1%29%20%E7%9A%84%E6%9C%80%E5%B0%8F%E8%B7%AF%E5%BE%84%E4%BB%A3%E4%BB%B7%0A%20%20%20%20up%20%3D%20min_path_sum_dfs%28grid,%20i%20-%201,%20j%29%0A%20%20%20%20left%20%3D%20min_path_sum_dfs%28grid,%20i,%20j%20-%201%29%0A%20%20%20%20%23%20%E8%BF%94%E5%9B%9E%E4%BB%8E%E5%B7%A6%E4%B8%8A%E8%A7%92%E5%88%B0%20%28i,%20j%29%20%E7%9A%84%E6%9C%80%E5%B0%8F%E8%B7%AF%E5%BE%84%E4%BB%A3%E4%BB%B7%0A%20%20%20%20return%20min%28left,%20up%29%20%2B%20grid%5Bi%5D%5Bj%5D%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20grid%20%3D%20%5B%5B1,%203,%201,%205%5D,%20%5B2,%202,%204,%202%5D,%20%5B5,%203,%202,%201%5D,%20%5B4,%203,%205,%202%5D%5D%0A%20%20%20%20n,%20m%20%3D%20len%28grid%29,%20len%28grid%5B0%5D%29%0A%0A%20%20%20%20%23%20%E6%9A%B4%E5%8A%9B%E6%90%9C%E7%B4%A2%0A%20%20%20%20res%20%3D%20min_path_sum_dfs%28grid,%20n%20-%201,%20m%20-%201%29%0A%20%20%20%20print%28f%22%E4%BB%8E%E5%B7%A6%E4%B8%8A%E8%A7%92%E5%88%B0%E5%8F%B3%E4%B8%8B%E8%A7%92%E7%9A%84%E5%81%9A%E5%B0%8F%E8%B7%AF%E5%BE%84%E5%92%8C%E4%B8%BA%20%7Bres%7D%22%29&codeDivHeight=800&codeDivWidth=600&cumulative=false&curInstr=6&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">Full Screen ></a></div>
 
-The following figure shows the recursive tree rooted at $dp[2, 1]$, which includes some overlapping subproblems, the number of which increases sharply as the size of the grid `grid` increases.
+Figure 14-14 shows the recursive tree rooted at $dp[2, 1]$, which includes some overlapping subproblems, the number of which increases sharply as the size of the grid `grid` increases.
 
 Essentially, the reason for overlapping subproblems is: **there are multiple paths to reach a certain cell from the top-left corner**.
 
@@ -771,7 +771,7 @@ We introduce a memo list `mem` of the same size as the grid `grid`, used to reco
     <div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=from%20math%20import%20inf%0A%0Adef%20min_path_sum_dfs_mem%28%0A%20%20%20%20grid%3A%20list%5Blist%5Bint%5D%5D,%20mem%3A%20list%5Blist%5Bint%5D%5D,%20i%3A%20int,%20j%3A%20int%0A%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E6%9C%80%E5%B0%8F%E8%B7%AF%E5%BE%84%E5%92%8C%EF%BC%9A%E8%AE%B0%E5%BF%86%E5%8C%96%E6%90%9C%E7%B4%A2%22%22%22%0A%20%20%20%20%23%20%E8%8B%A5%E4%B8%BA%E5%B7%A6%E4%B8%8A%E8%A7%92%E5%8D%95%E5%85%83%E6%A0%BC%EF%BC%8C%E5%88%99%E7%BB%88%E6%AD%A2%E6%90%9C%E7%B4%A2%0A%20%20%20%20if%20i%20%3D%3D%200%20and%20j%20%3D%3D%200%3A%0A%20%20%20%20%20%20%20%20return%20grid%5B0%5D%5B0%5D%0A%20%20%20%20%23%20%E8%8B%A5%E8%A1%8C%E5%88%97%E7%B4%A2%E5%BC%95%E8%B6%8A%E7%95%8C%EF%BC%8C%E5%88%99%E8%BF%94%E5%9B%9E%20%2B%E2%88%9E%20%E4%BB%A3%E4%BB%B7%0A%20%20%20%20if%20i%20%3C%200%20or%20j%20%3C%200%3A%0A%20%20%20%20%20%20%20%20return%20inf%0A%20%20%20%20%23%20%E8%8B%A5%E5%B7%B2%E6%9C%89%E8%AE%B0%E5%BD%95%EF%BC%8C%E5%88%99%E7%9B%B4%E6%8E%A5%E8%BF%94%E5%9B%9E%0A%20%20%20%20if%20mem%5Bi%5D%5Bj%5D%20!%3D%20-1%3A%0A%20%20%20%20%20%20%20%20return%20mem%5Bi%5D%5Bj%5D%0A%20%20%20%20%23%20%E5%B7%A6%E8%BE%B9%E5%92%8C%E4%B8%8A%E8%BE%B9%E5%8D%95%E5%85%83%E6%A0%BC%E7%9A%84%E6%9C%80%E5%B0%8F%E8%B7%AF%E5%BE%84%E4%BB%A3%E4%BB%B7%0A%20%20%20%20up%20%3D%20min_path_sum_dfs_mem%28grid,%20mem,%20i%20-%201,%20j%29%0A%20%20%20%20left%20%3D%20min_path_sum_dfs_mem%28grid,%20mem,%20i,%20j%20-%201%29%0A%20%20%20%20%23%20%E8%AE%B0%E5%BD%95%E5%B9%B6%E8%BF%94%E5%9B%9E%E5%B7%A6%E4%B8%8A%E8%A7%92%E5%88%B0%20%28i,%20j%29%20%E7%9A%84%E6%9C%80%E5%B0%8F%E8%B7%AF%E5%BE%84%E4%BB%A3%E4%BB%B7%0A%20%20%20%20mem%5Bi%5D%5Bj%5D%20%3D%20min%28left,%20up%29%20%2B%20grid%5Bi%5D%5Bj%5D%0A%20%20%20%20return%20mem%5Bi%5D%5Bj%5D%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20grid%20%3D%20%5B%5B1,%203,%201,%205%5D,%20%5B2,%202,%204,%202%5D,%20%5B5,%203,%202,%201%5D,%20%5B4,%203,%205,%202%5D%5D%0A%20%20%20%20n,%20m%20%3D%20len%28grid%29,%20len%28grid%5B0%5D%29%0A%0A%20%20%20%23%20%E8%AE%B0%E5%BF%86%E5%8C%96%E6%90%9C%E7%B4%A2%0A%20%20%20%20mem%20%3D%20%5B%5B-1%5D%20*%20m%20for%20_%20in%20range%28n%29%5D%0A%20%20%20%20res%20%3D%20min_path_sum_dfs_mem%28grid,%20mem,%20n%20-%201,%20m%20-%201%29%0A%20%20%20%20print%28f%22%E4%BB%8E%E5%B7%A6%E4%B8%8A%E8%A7%92%E5%88%B0%E5%8F%B3%E4%B8%8B%E8%A7%92%E7%9A%84%E5%81%9A%E5%B0%8F%E8%B7%AF%E5%BE%84%E5%92%8C%E4%B8%BA%20%7Bres%7D%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=16&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
     <div style="margin-top: 5px;"><a href="https://pythontutor.com/iframe-embed.html#code=from%20math%20import%20inf%0A%0Adef%20min_path_sum_dfs_mem%28%0A%20%20%20%20grid%3A%20list%5Blist%5Bint%5D%5D,%20mem%3A%20list%5Blist%5Bint%5D%5D,%20i%3A%20int,%20j%3A%20int%0A%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E6%9C%80%E5%B0%8F%E8%B7%AF%E5%BE%84%E5%92%8C%EF%BC%9A%E8%AE%B0%E5%BF%86%E5%8C%96%E6%90%9C%E7%B4%A2%22%22%22%0A%20%20%20%20%23%20%E8%8B%A5%E4%B8%BA%E5%B7%A6%E4%B8%8A%E8%A7%92%E5%8D%95%E5%85%83%E6%A0%BC%EF%BC%8C%E5%88%99%E7%BB%88%E6%AD%A2%E6%90%9C%E7%B4%A2%0A%20%20%20%20if%20i%20%3D%3D%200%20and%20j%20%3D%3D%200%3A%0A%20%20%20%20%20%20%20%20return%20grid%5B0%5D%5B0%5D%0A%20%20%20%20%23%20%E8%8B%A5%E8%A1%8C%E5%88%97%E7%B4%A2%E5%BC%95%E8%B6%8A%E7%95%8C%EF%BC%8C%E5%88%99%E8%BF%94%E5%9B%9E%20%2B%E2%88%9E%20%E4%BB%A3%E4%BB%B7%0A%20%20%20%20if%20i%20%3C%200%20or%20j%20%3C%200%3A%0A%20%20%20%20%20%20%20%20return%20inf%0A%20%20%20%20%23%20%E8%8B%A5%E5%B7%B2%E6%9C%89%E8%AE%B0%E5%BD%95%EF%BC%8C%E5%88%99%E7%9B%B4%E6%8E%A5%E8%BF%94%E5%9B%9E%0A%20%20%20%20if%20mem%5Bi%5D%5Bj%5D%20!%3D%20-1%3A%0A%20%20%20%20%20%20%20%20return%20mem%5Bi%5D%5Bj%5D%0A%20%20%20%20%23%20%E5%B7%A6%E8%BE%B9%E5%92%8C%E4%B8%8A%E8%BE%B9%E5%8D%95%E5%85%83%E6%A0%BC%E7%9A%84%E6%9C%80%E5%B0%8F%E8%B7%AF%E5%BE%84%E4%BB%A3%E4%BB%B7%0A%20%20%20%20up%20%3D%20min_path_sum_dfs_mem%28grid,%20mem,%20i%20-%201,%20j%29%0A%20%20%20%20left%20%3D%20min_path_sum_dfs_mem%28grid,%20mem,%20i,%20j%20-%201%29%0A%20%20%20%20%23%20%E8%AE%B0%E5%BD%95%E5%B9%B6%E8%BF%94%E5%9B%9E%E5%B7%A6%E4%B8%8A%E8%A7%92%E5%88%B0%20%28i,%20j%29%20%E7%9A%84%E6%9C%80%E5%B0%8F%E8%B7%AF%E5%BE%84%E4%BB%A3%E4%BB%B7%0A%20%20%20%20mem%5Bi%5D%5Bj%5D%20%3D%20min%28left,%20up%29%20%2B%20grid%5Bi%5D%5Bj%5D%0A%20%20%20%20return%20mem%5Bi%5D%5Bj%5D%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20grid%20%3D%20%5B%5B1,%203,%201,%205%5D,%20%5B2,%202,%204,%202%5D,%20%5B5,%203,%202,%201%5D,%20%5B4,%203,%205,%202%5D%5D%0A%20%20%20%20n,%20m%20%3D%20len%28grid%29,%20len%28grid%5B0%5D%29%0A%0A%20%20%20%23%20%E8%AE%B0%E5%BF%86%E5%8C%96%E6%90%9C%E7%B4%A2%0A%20%20%20%20mem%20%3D%20%5B%5B-1%5D%20*%20m%20for%20_%20in%20range%28n%29%5D%0A%20%20%20%20res%20%3D%20min_path_sum_dfs_mem%28grid,%20mem,%20n%20-%201,%20m%20-%201%29%0A%20%20%20%20print%28f%22%E4%BB%8E%E5%B7%A6%E4%B8%8A%E8%A7%92%E5%88%B0%E5%8F%B3%E4%B8%8B%E8%A7%92%E7%9A%84%E5%81%9A%E5%B0%8F%E8%B7%AF%E5%BE%84%E5%92%8C%E4%B8%BA%20%7Bres%7D%22%29&codeDivHeight=800&codeDivWidth=600&cumulative=false&curInstr=16&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">Full Screen ></a></div>
 
-As shown in the Figure 14-15 , after introducing memoization, all subproblem solutions only need to be calculated once, so the time complexity depends on the total number of states, i.e., the grid size $O(nm)$.
+As shown in Figure 14-15, after introducing memoization, all subproblem solutions only need to be calculated once, so the time complexity depends on the total number of states, i.e., the grid size $O(nm)$.
 
 ![Memoized search recursive tree](dp_solution_pipeline.assets/min_path_sum_dfs_mem.png){ class="animation-figure" }
 
@@ -1157,7 +1157,7 @@ Implement the dynamic programming solution iteratively, code as shown below:
     <div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=from%20math%20import%20inf%0A%0Adef%20min_path_sum_dp%28grid%3A%20list%5Blist%5Bint%5D%5D%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E6%9C%80%E5%B0%8F%E8%B7%AF%E5%BE%84%E5%92%8C%EF%BC%9A%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%22%22%22%0A%20%20%20%20n,%20m%20%3D%20len%28grid%29,%20len%28grid%5B0%5D%29%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%20dp%20%E8%A1%A8%0A%20%20%20%20dp%20%3D%20%5B%5B0%5D%20*%20m%20for%20_%20in%20range%28n%29%5D%0A%20%20%20%20dp%5B0%5D%5B0%5D%20%3D%20grid%5B0%5D%5B0%5D%0A%20%20%20%20%23%20%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%EF%BC%9A%E9%A6%96%E8%A1%8C%0A%20%20%20%20for%20j%20in%20range%281,%20m%29%3A%0A%20%20%20%20%20%20%20%20dp%5B0%5D%5Bj%5D%20%3D%20dp%5B0%5D%5Bj%20-%201%5D%20%2B%20grid%5B0%5D%5Bj%5D%0A%20%20%20%20%23%20%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%EF%BC%9A%E9%A6%96%E5%88%97%0A%20%20%20%20for%20i%20in%20range%281,%20n%29%3A%0A%20%20%20%20%20%20%20%20dp%5Bi%5D%5B0%5D%20%3D%20dp%5Bi%20-%201%5D%5B0%5D%20%2B%20grid%5Bi%5D%5B0%5D%0A%20%20%20%20%23%20%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%EF%BC%9A%E5%85%B6%E4%BD%99%E8%A1%8C%E5%92%8C%E5%88%97%0A%20%20%20%20for%20i%20in%20range%281,%20n%29%3A%0A%20%20%20%20%20%20%20%20for%20j%20in%20range%281,%20m%29%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20dp%5Bi%5D%5Bj%5D%20%3D%20min%28dp%5Bi%5D%5Bj%20-%201%5D,%20dp%5Bi%20-%201%5D%5Bj%5D%29%20%2B%20grid%5Bi%5D%5Bj%5D%0A%20%20%20%20return%20dp%5Bn%20-%201%5D%5Bm%20-%201%5D%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20grid%20%3D%20%5B%5B1,%203,%201,%205%5D,%20%5B2,%202,%204,%202%5D,%20%5B5,%203,%202,%201%5D,%20%5B4,%203,%205,%202%5D%5D%0A%20%20%20%20n,%20m%20%3D%20len%28grid%29,%20len%28grid%5B0%5D%29%0A%0A%20%20%20%20%23%20%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%0A%20%20%20%20res%20%3D%20min_path_sum_dp%28grid%29%0A%20%20%20%20print%28f%22%E4%BB%8E%E5%B7%A6%E4%B8%8A%E8%A7%92%E5%88%B0%E5%8F%B3%E4%B8%8B%E8%A7%92%E7%9A%84%E5%81%9A%E5%B0%8F%E8%B7%AF%E5%BE%84%E5%92%8C%E4%B8%BA%20%7Bres%7D%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=6&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
     <div style="margin-top: 5px;"><a href="https://pythontutor.com/iframe-embed.html#code=from%20math%20import%20inf%0A%0Adef%20min_path_sum_dp%28grid%3A%20list%5Blist%5Bint%5D%5D%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E6%9C%80%E5%B0%8F%E8%B7%AF%E5%BE%84%E5%92%8C%EF%BC%9A%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%22%22%22%0A%20%20%20%20n,%20m%20%3D%20len%28grid%29,%20len%28grid%5B0%5D%29%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%20dp%20%E8%A1%A8%0A%20%20%20%20dp%20%3D%20%5B%5B0%5D%20*%20m%20for%20_%20in%20range%28n%29%5D%0A%20%20%20%20dp%5B0%5D%5B0%5D%20%3D%20grid%5B0%5D%5B0%5D%0A%20%20%20%20%23%20%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%EF%BC%9A%E9%A6%96%E8%A1%8C%0A%20%20%20%20for%20j%20in%20range%281,%20m%29%3A%0A%20%20%20%20%20%20%20%20dp%5B0%5D%5Bj%5D%20%3D%20dp%5B0%5D%5Bj%20-%201%5D%20%2B%20grid%5B0%5D%5Bj%5D%0A%20%20%20%20%23%20%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%EF%BC%9A%E9%A6%96%E5%88%97%0A%20%20%20%20for%20i%20in%20range%281,%20n%29%3A%0A%20%20%20%20%20%20%20%20dp%5Bi%5D%5B0%5D%20%3D%20dp%5Bi%20-%201%5D%5B0%5D%20%2B%20grid%5Bi%5D%5B0%5D%0A%20%20%20%20%23%20%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%EF%BC%9A%E5%85%B6%E4%BD%99%E8%A1%8C%E5%92%8C%E5%88%97%0A%20%20%20%20for%20i%20in%20range%281,%20n%29%3A%0A%20%20%20%20%20%20%20%20for%20j%20in%20range%281,%20m%29%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20dp%5Bi%5D%5Bj%5D%20%3D%20min%28dp%5Bi%5D%5Bj%20-%201%5D,%20dp%5Bi%20-%201%5D%5Bj%5D%29%20%2B%20grid%5Bi%5D%5Bj%5D%0A%20%20%20%20return%20dp%5Bn%20-%201%5D%5Bm%20-%201%5D%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20grid%20%3D%20%5B%5B1,%203,%201,%205%5D,%20%5B2,%202,%204,%202%5D,%20%5B5,%203,%202,%201%5D,%20%5B4,%203,%205,%202%5D%5D%0A%20%20%20%20n,%20m%20%3D%20len%28grid%29,%20len%28grid%5B0%5D%29%0A%0A%20%20%20%20%23%20%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%0A%20%20%20%20res%20%3D%20min_path_sum_dp%28grid%29%0A%20%20%20%20print%28f%22%E4%BB%8E%E5%B7%A6%E4%B8%8A%E8%A7%92%E5%88%B0%E5%8F%B3%E4%B8%8B%E8%A7%92%E7%9A%84%E5%81%9A%E5%B0%8F%E8%B7%AF%E5%BE%84%E5%92%8C%E4%B8%BA%20%7Bres%7D%22%29&codeDivHeight=800&codeDivWidth=600&cumulative=false&curInstr=6&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">Full Screen ></a></div>
 
-The following figures show the state transition process of the minimum path sum, traversing the entire grid, **thus the time complexity is $O(nm)$**.
+Figure 14-16 show the state transition process of the minimum path sum, traversing the entire grid, **thus the time complexity is $O(nm)$**.
 
 The array `dp` is of size $n \times m$, **therefore the space complexity is $O(nm)$**.
 

en/docs/chapter_dynamic_programming/edit_distance_problem.md

@@ -12,7 +12,7 @@ Edit distance, also known as Levenshtein distance, refers to the minimum number
 
     You can perform three types of edits on a string: insert a character, delete a character, or replace a character with any other character.
 
-As shown in the Figure 14-27 , transforming `kitten` into `sitting` requires 3 edits, including 2 replacements and 1 insertion; transforming `hello` into `algo` requires 3 steps, including 2 replacements and 1 deletion.
+As shown in Figure 14-27, transforming `kitten` into `sitting` requires 3 edits, including 2 replacements and 1 insertion; transforming `hello` into `algo` requires 3 steps, including 2 replacements and 1 deletion.
 
 ![Example data of edit distance](edit_distance_problem.assets/edit_distance_example.png){ class="animation-figure" }
 
@@ -20,7 +20,7 @@ As shown in the Figure 14-27 , transforming `kitten` into `sitting` requires 3 e
 
 **The edit distance problem can naturally be explained with a decision tree model**. Strings correspond to tree nodes, and a round of decision (an edit operation) corresponds to an edge of the tree.
 
-As shown in the Figure 14-28 , with unrestricted operations, each node can derive many edges, each corresponding to one operation, meaning there are many possible paths to transform `hello` into `algo`.
+As shown in Figure 14-28, with unrestricted operations, each node can derive many edges, each corresponding to one operation, meaning there are many possible paths to transform `hello` into `algo`.
 
 From the perspective of the decision tree, the goal of this problem is to find the shortest path between the node `hello` and the node `algo`.
 
@@ -47,7 +47,7 @@ From this, we obtain a two-dimensional $dp$ table of size $(i+1) \times (j+1)$.
 
 **Step two: Identify the optimal substructure and then derive the state transition equation**
 
-Consider the subproblem $dp[i, j]$, whose corresponding tail characters of the two strings are $s[i-1]$ and $t[j-1]$, which can be divided into three scenarios as shown below.
+Consider the subproblem $dp[i, j]$, whose corresponding tail characters of the two strings are $s[i-1]$ and $t[j-1]$, which can be divided into three scenarios as shown in Figure 14-29.
 
 1. Add $t[j-1]$ after $s[i-1]$, then the remaining subproblem is $dp[i, j-1]$.
 2. Delete $s[i-1]$, then the remaining subproblem is $dp[i-1, j]$.
@@ -493,7 +493,7 @@ Observing the state transition equation, solving $dp[i, j]$ depends on the solut
     <div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20edit_distance_dp%28s%3A%20str,%20t%3A%20str%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E7%BC%96%E8%BE%91%E8%B7%9D%E7%A6%BB%EF%BC%9A%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%22%22%22%0A%20%20%20%20n,%20m%20%3D%20len%28s%29,%20len%28t%29%0A%20%20%20%20dp%20%3D%20%5B%5B0%5D%20*%20%28m%20%2B%201%29%20for%20_%20in%20range%28n%20%2B%201%29%5D%0A%20%20%20%20%23%20%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%EF%BC%9A%E9%A6%96%E8%A1%8C%E9%A6%96%E5%88%97%0A%20%20%20%20for%20i%20in%20range%281,%20n%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20dp%5Bi%5D%5B0%5D%20%3D%20i%0A%20%20%20%20for%20j%20in%20range%281,%20m%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20dp%5B0%5D%5Bj%5D%20%3D%20j%0A%20%20%20%20%23%20%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%EF%BC%9A%E5%85%B6%E4%BD%99%E8%A1%8C%E5%92%8C%E5%88%97%0A%20%20%20%20for%20i%20in%20range%281,%20n%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20for%20j%20in%20range%281,%20m%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20if%20s%5Bi%20-%201%5D%20%3D%3D%20t%5Bj%20-%201%5D%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E8%8B%A5%E4%B8%A4%E5%AD%97%E7%AC%A6%E7%9B%B8%E7%AD%89%EF%BC%8C%E5%88%99%E7%9B%B4%E6%8E%A5%E8%B7%B3%E8%BF%87%E6%AD%A4%E4%B8%A4%E5%AD%97%E7%AC%A6%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20dp%5Bi%5D%5Bj%5D%20%3D%20dp%5Bi%20-%201%5D%5Bj%20-%201%5D%0A%20%20%20%20%20%20%20%20%20%20%20%20else%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E6%9C%80%E5%B0%91%E7%BC%96%E8%BE%91%E6%AD%A5%E6%95%B0%20%3D%20%E6%8F%92%E5%85%A5%E3%80%81%E5%88%A0%E9%99%A4%E3%80%81%E6%9B%BF%E6%8D%A2%E8%BF%99%E4%B8%89%E7%A7%8D%E6%93%8D%E4%BD%9C%E7%9A%84%E6%9C%80%E5%B0%91%E7%BC%96%E8%BE%91%E6%AD%A5%E6%95%B0%20%2B%201%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20dp%5Bi%5D%5Bj%5D%20%3D%20min%28dp%5Bi%5D%5Bj%20-%201%5D,%20dp%5Bi%20-%201%5D%5Bj%5D,%20dp%5Bi%20-%201%5D%5Bj%20-%201%5D%29%20%2B%201%0A%20%20%20%20return%20dp%5Bn%5D%5Bm%5D%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20s%20%3D%20%22bag%22%0A%20%20%20%20t%20%3D%20%22pack%22%0A%20%20%20%20n,%20m%20%3D%20len%28s%29,%20len%28t%29%0A%0A%20%20%20%20%23%20%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%0A%20%20%20%20res%20%3D%20edit_distance_dp%28s,%20t%29%0A%20%20%20%20print%28f%22%E5%B0%86%20%7Bs%7D%20%E6%9B%B4%E6%94%B9%E4%B8%BA%20%7Bt%7D%20%E6%9C%80%E5%B0%91%E9%9C%80%E8%A6%81%E7%BC%96%E8%BE%91%20%7Bres%7D%20%E6%AD%A5%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=6&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
     <div style="margin-top: 5px;"><a href="https://pythontutor.com/iframe-embed.html#code=def%20edit_distance_dp%28s%3A%20str,%20t%3A%20str%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E7%BC%96%E8%BE%91%E8%B7%9D%E7%A6%BB%EF%BC%9A%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%22%22%22%0A%20%20%20%20n,%20m%20%3D%20len%28s%29,%20len%28t%29%0A%20%20%20%20dp%20%3D%20%5B%5B0%5D%20*%20%28m%20%2B%201%29%20for%20_%20in%20range%28n%20%2B%201%29%5D%0A%20%20%20%20%23%20%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%EF%BC%9A%E9%A6%96%E8%A1%8C%E9%A6%96%E5%88%97%0A%20%20%20%20for%20i%20in%20range%281,%20n%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20dp%5Bi%5D%5B0%5D%20%3D%20i%0A%20%20%20%20for%20j%20in%20range%281,%20m%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20dp%5B0%5D%5Bj%5D%20%3D%20j%0A%20%20%20%20%23%20%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%EF%BC%9A%E5%85%B6%E4%BD%99%E8%A1%8C%E5%92%8C%E5%88%97%0A%20%20%20%20for%20i%20in%20range%281,%20n%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20for%20j%20in%20range%281,%20m%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20if%20s%5Bi%20-%201%5D%20%3D%3D%20t%5Bj%20-%201%5D%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E8%8B%A5%E4%B8%A4%E5%AD%97%E7%AC%A6%E7%9B%B8%E7%AD%89%EF%BC%8C%E5%88%99%E7%9B%B4%E6%8E%A5%E8%B7%B3%E8%BF%87%E6%AD%A4%E4%B8%A4%E5%AD%97%E7%AC%A6%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20dp%5Bi%5D%5Bj%5D%20%3D%20dp%5Bi%20-%201%5D%5Bj%20-%201%5D%0A%20%20%20%20%20%20%20%20%20%20%20%20else%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E6%9C%80%E5%B0%91%E7%BC%96%E8%BE%91%E6%AD%A5%E6%95%B0%20%3D%20%E6%8F%92%E5%85%A5%E3%80%81%E5%88%A0%E9%99%A4%E3%80%81%E6%9B%BF%E6%8D%A2%E8%BF%99%E4%B8%89%E7%A7%8D%E6%93%8D%E4%BD%9C%E7%9A%84%E6%9C%80%E5%B0%91%E7%BC%96%E8%BE%91%E6%AD%A5%E6%95%B0%20%2B%201%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20dp%5Bi%5D%5Bj%5D%20%3D%20min%28dp%5Bi%5D%5Bj%20-%201%5D,%20dp%5Bi%20-%201%5D%5Bj%5D,%20dp%5Bi%20-%201%5D%5Bj%20-%201%5D%29%20%2B%201%0A%20%20%20%20return%20dp%5Bn%5D%5Bm%5D%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20s%20%3D%20%22bag%22%0A%20%20%20%20t%20%3D%20%22pack%22%0A%20%20%20%20n,%20m%20%3D%20len%28s%29,%20len%28t%29%0A%0A%20%20%20%20%23%20%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%0A%20%20%20%20res%20%3D%20edit_distance_dp%28s,%20t%29%0A%20%20%20%20print%28f%22%E5%B0%86%20%7Bs%7D%20%E6%9B%B4%E6%94%B9%E4%B8%BA%20%7Bt%7D%20%E6%9C%80%E5%B0%91%E9%9C%80%E8%A6%81%E7%BC%96%E8%BE%91%20%7Bres%7D%20%E6%AD%A5%22%29&codeDivHeight=800&codeDivWidth=600&cumulative=false&curInstr=6&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">Full Screen ></a></div>
 
-As shown below, the process of state transition in the edit distance problem is very similar to that in the knapsack problem, which can be seen as filling a two-dimensional grid.
+As shown in Figure 14-30, the process of state transition in the edit distance problem is very similar to that in the knapsack problem, which can be seen as filling a two-dimensional grid.
 
 === "<1>"
     ![Dynamic programming process of edit distance](edit_distance_problem.assets/edit_distance_dp_step1.png){ class="animation-figure" }

en/docs/chapter_dynamic_programming/intro_to_dynamic_programming.md

@@ -12,7 +12,7 @@ In this section, we start with a classic problem, first presenting its brute for
 
     Given a staircase with $n$ steps, where you can climb $1$ or $2$ steps at a time, how many different ways are there to reach the top?
 
-As shown in the Figure 14-1 , there are $3$ ways to reach the top of a $3$-step staircase.
+As shown in Figure 14-1, there are $3$ ways to reach the top of a $3$-step staircase.
 
 ![Number of ways to reach the 3rd step](intro_to_dynamic_programming.assets/climbing_stairs_example.png){ class="animation-figure" }
 
@@ -456,7 +456,7 @@ $$
 dp[i] = dp[i-1] + dp[i-2]
 $$
 
-This means that in the stair climbing problem, there is a recursive relationship between the subproblems, **the solution to the original problem can be constructed from the solutions to the subproblems**. The following image shows this recursive relationship.
+This means that in the stair climbing problem, there is a recursive relationship between the subproblems, **the solution to the original problem can be constructed from the solutions to the subproblems**. Figure 14-2 shows this recursive relationship.
 
 ![Recursive relationship of solution counts](intro_to_dynamic_programming.assets/climbing_stairs_state_transfer.png){ class="animation-figure" }
 
@@ -724,13 +724,13 @@ Observe the following code, which, like standard backtracking code, belongs to d
     <div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20dfs%28i%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E6%90%9C%E7%B4%A2%22%22%22%0A%20%20%20%20%23%20%E5%B7%B2%E7%9F%A5%20dp%5B1%5D%20%E5%92%8C%20dp%5B2%5D%20%EF%BC%8C%E8%BF%94%E5%9B%9E%E4%B9%8B%0A%20%20%20%20if%20i%20%3D%3D%201%20or%20i%20%3D%3D%202%3A%0A%20%20%20%20%20%20%20%20return%20i%0A%20%20%20%20%23%20dp%5Bi%5D%20%3D%20dp%5Bi-1%5D%20%2B%20dp%5Bi-2%5D%0A%20%20%20%20count%20%3D%20dfs%28i%20-%201%29%20%2B%20dfs%28i%20-%202%29%0A%20%20%20%20return%20count%0A%0A%0Adef%20climbing_stairs_dfs%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E7%88%AC%E6%A5%BC%E6%A2%AF%EF%BC%9A%E6%90%9C%E7%B4%A2%22%22%22%0A%20%20%20%20return%20dfs%28n%29%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%209%0A%0A%20%20%20%20res%20%3D%20climbing_stairs_dfs%28n%29%0A%20%20%20%20print%28f%22%E7%88%AC%20%7Bn%7D%20%E9%98%B6%E6%A5%BC%E6%A2%AF%E5%85%B1%E6%9C%89%20%7Bres%7D%20%E7%A7%8D%E6%96%B9%E6%A1%88%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=5&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
     <div style="margin-top: 5px;"><a href="https://pythontutor.com/iframe-embed.html#code=def%20dfs%28i%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E6%90%9C%E7%B4%A2%22%22%22%0A%20%20%20%20%23%20%E5%B7%B2%E7%9F%A5%20dp%5B1%5D%20%E5%92%8C%20dp%5B2%5D%20%EF%BC%8C%E8%BF%94%E5%9B%9E%E4%B9%8B%0A%20%20%20%20if%20i%20%3D%3D%201%20or%20i%20%3D%3D%202%3A%0A%20%20%20%20%20%20%20%20return%20i%0A%20%20%20%20%23%20dp%5Bi%5D%20%3D%20dp%5Bi-1%5D%20%2B%20dp%5Bi-2%5D%0A%20%20%20%20count%20%3D%20dfs%28i%20-%201%29%20%2B%20dfs%28i%20-%202%29%0A%20%20%20%20return%20count%0A%0A%0Adef%20climbing_stairs_dfs%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E7%88%AC%E6%A5%BC%E6%A2%AF%EF%BC%9A%E6%90%9C%E7%B4%A2%22%22%22%0A%20%20%20%20return%20dfs%28n%29%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%209%0A%0A%20%20%20%20res%20%3D%20climbing_stairs_dfs%28n%29%0A%20%20%20%20print%28f%22%E7%88%AC%20%7Bn%7D%20%E9%98%B6%E6%A5%BC%E6%A2%AF%E5%85%B1%E6%9C%89%20%7Bres%7D%20%E7%A7%8D%E6%96%B9%E6%A1%88%22%29&codeDivHeight=800&codeDivWidth=600&cumulative=false&curInstr=5&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">Full Screen ></a></div>
 
-The following image shows the recursive tree formed by brute force search. For the problem $dp[n]$, the depth of its recursive tree is $n$, with a time complexity of $O(2^n)$. Exponential order represents explosive growth, and entering a long wait if a relatively large $n$ is input.
+Figure 14-3 shows the recursive tree formed by brute force search. For the problem $dp[n]$, the depth of its recursive tree is $n$, with a time complexity of $O(2^n)$. Exponential order represents explosive growth, and entering a long wait if a relatively large $n$ is input.
 
 ![Recursive tree for climbing stairs](intro_to_dynamic_programming.assets/climbing_stairs_dfs_tree.png){ class="animation-figure" }
 
 <p align="center"> Figure 14-3 &nbsp; Recursive tree for climbing stairs </p>
 
-Observing the above image, **the exponential time complexity is caused by 'overlapping subproblems'**. For example, $dp[9]$ is decomposed into $dp[8]$ and $dp[7]$, $dp[8]$ into $dp[7]$ and $dp[6]$, both containing the subproblem $dp[7]$.
+Observing Figure 14-3, **the exponential time complexity is caused by 'overlapping subproblems'**. For example, $dp[9]$ is decomposed into $dp[8]$ and $dp[7]$, $dp[8]$ into $dp[7]$ and $dp[6]$, both containing the subproblem $dp[7]$.
 
 Thus, subproblems include even smaller overlapping subproblems, endlessly. A vast majority of computational resources are wasted on these overlapping subproblems.
 
@@ -1103,7 +1103,7 @@ The code is as follows:
     <div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20dfs%28i%3A%20int,%20mem%3A%20list%5Bint%5D%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E8%AE%B0%E5%BF%86%E5%8C%96%E6%90%9C%E7%B4%A2%22%22%22%0A%20%20%20%20%23%20%E5%B7%B2%E7%9F%A5%20dp%5B1%5D%20%E5%92%8C%20dp%5B2%5D%20%EF%BC%8C%E8%BF%94%E5%9B%9E%E4%B9%8B%0A%20%20%20%20if%20i%20%3D%3D%201%20or%20i%20%3D%3D%202%3A%0A%20%20%20%20%20%20%20%20return%20i%0A%20%20%20%20%23%20%E8%8B%A5%E5%AD%98%E5%9C%A8%E8%AE%B0%E5%BD%95%20dp%5Bi%5D%20%EF%BC%8C%E5%88%99%E7%9B%B4%E6%8E%A5%E8%BF%94%E5%9B%9E%E4%B9%8B%0A%20%20%20%20if%20mem%5Bi%5D%20!%3D%20-1%3A%0A%20%20%20%20%20%20%20%20return%20mem%5Bi%5D%0A%20%20%20%20%23%20dp%5Bi%5D%20%3D%20dp%5Bi-1%5D%20%2B%20dp%5Bi-2%5D%0A%20%20%20%20count%20%3D%20dfs%28i%20-%201,%20mem%29%20%2B%20dfs%28i%20-%202,%20mem%29%0A%20%20%20%20%23%20%E8%AE%B0%E5%BD%95%20dp%5Bi%5D%0A%20%20%20%20mem%5Bi%5D%20%3D%20count%0A%20%20%20%20return%20count%0A%0A%0Adef%20climbing_stairs_dfs_mem%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E7%88%AC%E6%A5%BC%E6%A2%AF%EF%BC%9A%E8%AE%B0%E5%BF%86%E5%8C%96%E6%90%9C%E7%B4%A2%22%22%22%0A%20%20%20%20%23%20mem%5Bi%5D%20%E8%AE%B0%E5%BD%95%E7%88%AC%E5%88%B0%E7%AC%AC%20i%20%E9%98%B6%E7%9A%84%E6%96%B9%E6%A1%88%E6%80%BB%E6%95%B0%EF%BC%8C-1%20%E4%BB%A3%E8%A1%A8%E6%97%A0%E8%AE%B0%E5%BD%95%0A%20%20%20%20mem%20%3D%20%5B-1%5D%20*%20%28n%20%2B%201%29%0A%20%20%20%20return%20dfs%28n,%20mem%29%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%209%0A%0A%20%20%20%20res%20%3D%20climbing_stairs_dfs_mem%28n%29%0A%20%20%20%20print%28f%22%E7%88%AC%20%7Bn%7D%20%E9%98%B6%E6%A5%BC%E6%A2%AF%E5%85%B1%E6%9C%89%20%7Bres%7D%20%E7%A7%8D%E6%96%B9%E6%A1%88%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=5&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
     <div style="margin-top: 5px;"><a href="https://pythontutor.com/iframe-embed.html#code=def%20dfs%28i%3A%20int,%20mem%3A%20list%5Bint%5D%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E8%AE%B0%E5%BF%86%E5%8C%96%E6%90%9C%E7%B4%A2%22%22%22%0A%20%20%20%20%23%20%E5%B7%B2%E7%9F%A5%20dp%5B1%5D%20%E5%92%8C%20dp%5B2%5D%20%EF%BC%8C%E8%BF%94%E5%9B%9E%E4%B9%8B%0A%20%20%20%20if%20i%20%3D%3D%201%20or%20i%20%3D%3D%202%3A%0A%20%20%20%20%20%20%20%20return%20i%0A%20%20%20%20%23%20%E8%8B%A5%E5%AD%98%E5%9C%A8%E8%AE%B0%E5%BD%95%20dp%5Bi%5D%20%EF%BC%8C%E5%88%99%E7%9B%B4%E6%8E%A5%E8%BF%94%E5%9B%9E%E4%B9%8B%0A%20%20%20%20if%20mem%5Bi%5D%20!%3D%20-1%3A%0A%20%20%20%20%20%20%20%20return%20mem%5Bi%5D%0A%20%20%20%20%23%20dp%5Bi%5D%20%3D%20dp%5Bi-1%5D%20%2B%20dp%5Bi-2%5D%0A%20%20%20%20count%20%3D%20dfs%28i%20-%201,%20mem%29%20%2B%20dfs%28i%20-%202,%20mem%29%0A%20%20%20%20%23%20%E8%AE%B0%E5%BD%95%20dp%5Bi%5D%0A%20%20%20%20mem%5Bi%5D%20%3D%20count%0A%20%20%20%20return%20count%0A%0A%0Adef%20climbing_stairs_dfs_mem%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E7%88%AC%E6%A5%BC%E6%A2%AF%EF%BC%9A%E8%AE%B0%E5%BF%86%E5%8C%96%E6%90%9C%E7%B4%A2%22%22%22%0A%20%20%20%20%23%20mem%5Bi%5D%20%E8%AE%B0%E5%BD%95%E7%88%AC%E5%88%B0%E7%AC%AC%20i%20%E9%98%B6%E7%9A%84%E6%96%B9%E6%A1%88%E6%80%BB%E6%95%B0%EF%BC%8C-1%20%E4%BB%A3%E8%A1%A8%E6%97%A0%E8%AE%B0%E5%BD%95%0A%20%20%20%20mem%20%3D%20%5B-1%5D%20*%20%28n%20%2B%201%29%0A%20%20%20%20return%20dfs%28n,%20mem%29%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%209%0A%0A%20%20%20%20res%20%3D%20climbing_stairs_dfs_mem%28n%29%0A%20%20%20%20print%28f%22%E7%88%AC%20%7Bn%7D%20%E9%98%B6%E6%A5%BC%E6%A2%AF%E5%85%B1%E6%9C%89%20%7Bres%7D%20%E7%A7%8D%E6%96%B9%E6%A1%88%22%29&codeDivHeight=800&codeDivWidth=600&cumulative=false&curInstr=5&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">Full Screen ></a></div>
 
-Observe the following image, **after memoization, all overlapping subproblems need to be calculated only once, optimizing the time complexity to $O(n)$**, which is a significant leap.
+Observe Figure 14-4, **after memoization, all overlapping subproblems need to be calculated only once, optimizing the time complexity to $O(n)$**, which is a significant leap.
 
 ![Recursive tree with memoized search](intro_to_dynamic_programming.assets/climbing_stairs_dfs_memo_tree.png){ class="animation-figure" }
 
@@ -1389,7 +1389,7 @@ Since dynamic programming does not include a backtracking process, it only requi
     <div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20climbing_stairs_dp%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E7%88%AC%E6%A5%BC%E6%A2%AF%EF%BC%9A%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%22%22%22%0A%20%20%20%20if%20n%20%3D%3D%201%20or%20n%20%3D%3D%202%3A%0A%20%20%20%20%20%20%20%20return%20n%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%20dp%20%E8%A1%A8%EF%BC%8C%E7%94%A8%E4%BA%8E%E5%AD%98%E5%82%A8%E5%AD%90%E9%97%AE%E9%A2%98%E7%9A%84%E8%A7%A3%0A%20%20%20%20dp%20%3D%20%5B0%5D%20*%20%28n%20%2B%201%29%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E7%8A%B6%E6%80%81%EF%BC%9A%E9%A2%84%E8%AE%BE%E6%9C%80%E5%B0%8F%E5%AD%90%E9%97%AE%E9%A2%98%E7%9A%84%E8%A7%A3%0A%20%20%20%20dp%5B1%5D,%20dp%5B2%5D%20%3D%201,%202%0A%20%20%20%20%23%20%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%EF%BC%9A%E4%BB%8E%E8%BE%83%E5%B0%8F%E5%AD%90%E9%97%AE%E9%A2%98%E9%80%90%E6%AD%A5%E6%B1%82%E8%A7%A3%E8%BE%83%E5%A4%A7%E5%AD%90%E9%97%AE%E9%A2%98%0A%20%20%20%20for%20i%20in%20range%283,%20n%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20dp%5Bi%5D%20%3D%20dp%5Bi%20-%201%5D%20%2B%20dp%5Bi%20-%202%5D%0A%20%20%20%20return%20dp%5Bn%5D%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%209%0A%0A%20%20%20%20res%20%3D%20climbing_stairs_dp%28n%29%0A%20%20%20%20print%28f%22%E7%88%AC%20%7Bn%7D%20%E9%98%B6%E6%A5%BC%E6%A2%AF%E5%85%B1%E6%9C%89%20%7Bres%7D%20%E7%A7%8D%E6%96%B9%E6%A1%88%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=4&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
     <div style="margin-top: 5px;"><a href="https://pythontutor.com/iframe-embed.html#code=def%20climbing_stairs_dp%28n%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E7%88%AC%E6%A5%BC%E6%A2%AF%EF%BC%9A%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%22%22%22%0A%20%20%20%20if%20n%20%3D%3D%201%20or%20n%20%3D%3D%202%3A%0A%20%20%20%20%20%20%20%20return%20n%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%20dp%20%E8%A1%A8%EF%BC%8C%E7%94%A8%E4%BA%8E%E5%AD%98%E5%82%A8%E5%AD%90%E9%97%AE%E9%A2%98%E7%9A%84%E8%A7%A3%0A%20%20%20%20dp%20%3D%20%5B0%5D%20*%20%28n%20%2B%201%29%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E7%8A%B6%E6%80%81%EF%BC%9A%E9%A2%84%E8%AE%BE%E6%9C%80%E5%B0%8F%E5%AD%90%E9%97%AE%E9%A2%98%E7%9A%84%E8%A7%A3%0A%20%20%20%20dp%5B1%5D,%20dp%5B2%5D%20%3D%201,%202%0A%20%20%20%20%23%20%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%EF%BC%9A%E4%BB%8E%E8%BE%83%E5%B0%8F%E5%AD%90%E9%97%AE%E9%A2%98%E9%80%90%E6%AD%A5%E6%B1%82%E8%A7%A3%E8%BE%83%E5%A4%A7%E5%AD%90%E9%97%AE%E9%A2%98%0A%20%20%20%20for%20i%20in%20range%283,%20n%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20dp%5Bi%5D%20%3D%20dp%5Bi%20-%201%5D%20%2B%20dp%5Bi%20-%202%5D%0A%20%20%20%20return%20dp%5Bn%5D%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20n%20%3D%209%0A%0A%20%20%20%20res%20%3D%20climbing_stairs_dp%28n%29%0A%20%20%20%20print%28f%22%E7%88%AC%20%7Bn%7D%20%E9%98%B6%E6%A5%BC%E6%A2%AF%E5%85%B1%E6%9C%89%20%7Bres%7D%20%E7%A7%8D%E6%96%B9%E6%A1%88%22%29&codeDivHeight=800&codeDivWidth=600&cumulative=false&curInstr=4&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">Full Screen ></a></div>
 
-The image below simulates the execution process of the above code.
+Figure 14-5 simulates the execution process of the above code.
 
 ![Dynamic programming process for climbing stairs](intro_to_dynamic_programming.assets/climbing_stairs_dp.png){ class="animation-figure" }
 

en/docs/chapter_dynamic_programming/knapsack_problem.md

@@ -12,7 +12,7 @@ In this section, we will first solve the most common 0-1 knapsack problem.
 
     Given $n$ items, the weight of the $i$-th item is $wgt[i-1]$ and its value is $val[i-1]$, and a knapsack with a capacity of $cap$. Each item can be chosen only once. What is the maximum value of items that can be placed in the knapsack under the capacity limit?
 
-Observe the following figure, since the item number $i$ starts counting from 1, and the array index starts from 0, thus the weight of item $i$ corresponds to $wgt[i-1]$ and the value corresponds to $val[i-1]$.
+Observe Figure 14-17, since the item number $i$ starts counting from 1, and the array index starts from 0, thus the weight of item $i$ corresponds to $wgt[i-1]$ and the value corresponds to $val[i-1]$.
 
 ![Example data of the 0-1 knapsack](knapsack_problem.assets/knapsack_example.png){ class="animation-figure" }
 
@@ -353,7 +353,7 @@ The search code includes the following elements.
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-As shown in the Figure 14-18 , since each item generates two search branches of not selecting and selecting, the time complexity is $O(2^n)$.
+As shown in Figure 14-18, since each item generates two search branches of not selecting and selecting, the time complexity is $O(2^n)$.
 
 Observing the recursive tree, it is easy to see that there are overlapping sub-problems, such as $dp[1, 10]$, etc. When there are many items and the knapsack capacity is large, especially when there are many items of the same weight, the number of overlapping sub-problems will increase significantly.
 
@@ -734,7 +734,7 @@ After introducing memoization, **the time complexity depends on the number of su
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-The following figure shows the search branches that are pruned in memoized search.
+Figure 14-19 shows the search branches that are pruned in memoized search.
 
 ![The memoized search recursive tree of the 0-1 knapsack problem](knapsack_problem.assets/knapsack_dfs_mem.png){ class="animation-figure" }
 
@@ -742,7 +742,7 @@ The following figure shows the search branches that are pruned in memoized searc
 
 ### 3. &nbsp; Method three: Dynamic programming
 
-Dynamic programming essentially involves filling the $dp$ table during the state transition, the code is shown below:
+Dynamic programming essentially involves filling the $dp$ table during the state transition, the code is shown in Figure 14-20:
 
 === "Python"
 

en/docs/chapter_dynamic_programming/unbounded_knapsack_problem.md

@@ -386,7 +386,7 @@ Comparing the code for the two problems, the state transition changes from $i-1$
 
 Since the current state comes from the state to the left and above, **the space-optimized solution should perform a forward traversal for each row in the $dp$ table**.
 
-This traversal order is the opposite of that for the 0-1 knapsack. Please refer to the following figures to understand the difference.
+This traversal order is the opposite of that for the 0-1 knapsack. Please refer to Figure 14-23 to understand the difference.
 
 === "<1>"
     ![Dynamic programming process for the unbounded knapsack problem after space optimization](unbounded_knapsack_problem.assets/unbounded_knapsack_dp_comp_step1.png){ class="animation-figure" }
@@ -1200,7 +1200,7 @@ For this reason, we use the number $amt + 1$ to represent an invalid solution, b
     <div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20coin_change_dp%28coins%3A%20list%5Bint%5D,%20amt%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E9%9B%B6%E9%92%B1%E5%85%91%E6%8D%A2%EF%BC%9A%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%22%22%22%0A%20%20%20%20n%20%3D%20len%28coins%29%0A%20%20%20%20MAX%20%3D%20amt%20%2B%201%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%20dp%20%E8%A1%A8%0A%20%20%20%20dp%20%3D%20%5B%5B0%5D%20*%20%28amt%20%2B%201%29%20for%20_%20in%20range%28n%20%2B%201%29%5D%0A%20%20%20%20%23%20%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%EF%BC%9A%E9%A6%96%E8%A1%8C%E9%A6%96%E5%88%97%0A%20%20%20%20for%20a%20in%20range%281,%20amt%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20dp%5B0%5D%5Ba%5D%20%3D%20MAX%0A%20%20%20%20%23%20%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%EF%BC%9A%E5%85%B6%E4%BD%99%E8%A1%8C%E5%92%8C%E5%88%97%0A%20%20%20%20for%20i%20in%20range%281,%20n%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20for%20a%20in%20range%281,%20amt%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20if%20coins%5Bi%20-%201%5D%20%3E%20a%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E8%8B%A5%E8%B6%85%E8%BF%87%E7%9B%AE%E6%A0%87%E9%87%91%E9%A2%9D%EF%BC%8C%E5%88%99%E4%B8%8D%E9%80%89%E7%A1%AC%E5%B8%81%20i%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20dp%5Bi%5D%5Ba%5D%20%3D%20dp%5Bi%20-%201%5D%5Ba%5D%0A%20%20%20%20%20%20%20%20%20%20%20%20else%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E4%B8%8D%E9%80%89%E5%92%8C%E9%80%89%E7%A1%AC%E5%B8%81%20i%20%E8%BF%99%E4%B8%A4%E7%A7%8D%E6%96%B9%E6%A1%88%E7%9A%84%E8%BE%83%E5%B0%8F%E5%80%BC%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20dp%5Bi%5D%5Ba%5D%20%3D%20min%28dp%5Bi%20-%201%5D%5Ba%5D,%20dp%5Bi%5D%5Ba%20-%20coins%5Bi%20-%201%5D%5D%20%2B%201%29%0A%20%20%20%20return%20dp%5Bn%5D%5Bamt%5D%20if%20dp%5Bn%5D%5Bamt%5D%20!%3D%20MAX%20else%20-1%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20coins%20%3D%20%5B1,%202,%205%5D%0A%20%20%20%20amt%20%3D%204%0A%0A%20%20%20%20%23%20%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%0A%20%20%20%20res%20%3D%20coin_change_dp%28coins,%20amt%29%0A%20%20%20%20print%28f%22%E5%87%91%E5%88%B0%E7%9B%AE%E6%A0%87%E9%87%91%E9%A2%9D%E6%89%80%E9%9C%80%E7%9A%84%E6%9C%80%E5%B0%91%E7%A1%AC%E5%B8%81%E6%95%B0%E9%87%8F%E4%B8%BA%20%7Bres%7D%22%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=5&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
     <div style="margin-top: 5px;"><a href="https://pythontutor.com/iframe-embed.html#code=def%20coin_change_dp%28coins%3A%20list%5Bint%5D,%20amt%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E9%9B%B6%E9%92%B1%E5%85%91%E6%8D%A2%EF%BC%9A%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%22%22%22%0A%20%20%20%20n%20%3D%20len%28coins%29%0A%20%20%20%20MAX%20%3D%20amt%20%2B%201%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%20dp%20%E8%A1%A8%0A%20%20%20%20dp%20%3D%20%5B%5B0%5D%20*%20%28amt%20%2B%201%29%20for%20_%20in%20range%28n%20%2B%201%29%5D%0A%20%20%20%20%23%20%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%EF%BC%9A%E9%A6%96%E8%A1%8C%E9%A6%96%E5%88%97%0A%20%20%20%20for%20a%20in%20range%281,%20amt%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20dp%5B0%5D%5Ba%5D%20%3D%20MAX%0A%20%20%20%20%23%20%E7%8A%B6%E6%80%81%E8%BD%AC%E7%A7%BB%EF%BC%9A%E5%85%B6%E4%BD%99%E8%A1%8C%E5%92%8C%E5%88%97%0A%20%20%20%20for%20i%20in%20range%281,%20n%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20for%20a%20in%20range%281,%20amt%20%2B%201%29%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20if%20coins%5Bi%20-%201%5D%20%3E%20a%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E8%8B%A5%E8%B6%85%E8%BF%87%E7%9B%AE%E6%A0%87%E9%87%91%E9%A2%9D%EF%BC%8C%E5%88%99%E4%B8%8D%E9%80%89%E7%A1%AC%E5%B8%81%20i%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20dp%5Bi%5D%5Ba%5D%20%3D%20dp%5Bi%20-%201%5D%5Ba%5D%0A%20%20%20%20%20%20%20%20%20%20%20%20else%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%23%20%E4%B8%8D%E9%80%89%E5%92%8C%E9%80%89%E7%A1%AC%E5%B8%81%20i%20%E8%BF%99%E4%B8%A4%E7%A7%8D%E6%96%B9%E6%A1%88%E7%9A%84%E8%BE%83%E5%B0%8F%E5%80%BC%0A%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20%20dp%5Bi%5D%5Ba%5D%20%3D%20min%28dp%5Bi%20-%201%5D%5Ba%5D,%20dp%5Bi%5D%5Ba%20-%20coins%5Bi%20-%201%5D%5D%20%2B%201%29%0A%20%20%20%20return%20dp%5Bn%5D%5Bamt%5D%20if%20dp%5Bn%5D%5Bamt%5D%20!%3D%20MAX%20else%20-1%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20coins%20%3D%20%5B1,%202,%205%5D%0A%20%20%20%20amt%20%3D%204%0A%0A%20%20%20%20%23%20%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%0A%20%20%20%20res%20%3D%20coin_change_dp%28coins,%20amt%29%0A%20%20%20%20print%28f%22%E5%87%91%E5%88%B0%E7%9B%AE%E6%A0%87%E9%87%91%E9%A2%9D%E6%89%80%E9%9C%80%E7%9A%84%E6%9C%80%E5%B0%91%E7%A1%AC%E5%B8%81%E6%95%B0%E9%87%8F%E4%B8%BA%20%7Bres%7D%22%29&codeDivHeight=800&codeDivWidth=600&cumulative=false&curInstr=5&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">Full Screen ></a></div>
 
-The following images show the dynamic programming process for the coin change problem, which is very similar to the unbounded knapsack problem.
+Figure 14-25 show the dynamic programming process for the coin change problem, which is very similar to the unbounded knapsack problem.
 
 === "<1>"
     ![Dynamic programming process for the coin change problem](unbounded_knapsack_problem.assets/coin_change_dp_step1.png){ class="animation-figure" }

en/docs/chapter_graph/graph.md

@@ -14,7 +14,7 @@ G & = \{ V, E \} \newline
 \end{aligned}
 $$
 
-If vertices are viewed as nodes and edges as references (pointers) connecting the nodes, graphs can be seen as a data structure that extends from linked lists. As shown below, **compared to linear relationships (linked lists) and divide-and-conquer relationships (trees), network relationships (graphs) are more complex due to their higher degree of freedom**.
+If vertices are viewed as nodes and edges as references (pointers) connecting the nodes, graphs can be seen as a data structure that extends from linked lists. As shown in Figure 9-1, **compared to linear relationships (linked lists) and divide-and-conquer relationships (trees), network relationships (graphs) are more complex due to their higher degree of freedom**.
 
 ![Relationship between linked lists, trees, and graphs](graph.assets/linkedlist_tree_graph.png){ class="animation-figure" }
 
@@ -22,7 +22,7 @@ If vertices are viewed as nodes and edges as references (pointers) connecting th
 
 ## 9.1.1   Common types of graphs
 
-Based on whether edges have direction, graphs can be divided into "undirected graphs" and "directed graphs", as shown below.
+Based on whether edges have direction, graphs can be divided into "undirected graphs" and "directed graphs", as shown in Figure 9-2.
 
 - In undirected graphs, edges represent a "bidirectional" connection between two vertices, for example, the "friendship" in WeChat or QQ.
 - In directed graphs, edges have directionality, that is, the edges $A \rightarrow B$ and $A \leftarrow B$ are independent of each other, for example, the "follow" and "be followed" relationship on Weibo or TikTok.
@@ -31,7 +31,7 @@ Based on whether edges have direction, graphs can be divided into "undirected gr
 
 <p align="center"> Figure 9-2 &nbsp; Directed and undirected graphs </p>
 
-Based on whether all vertices are connected, graphs can be divided into "connected graphs" and "disconnected graphs", as shown below.
+Based on whether all vertices are connected, graphs can be divided into "connected graphs" and "disconnected graphs", as shown in Figure 9-3.
 
 - For connected graphs, it is possible to reach any other vertex starting from a certain vertex.
 - For disconnected graphs, there is at least one vertex that cannot be reached from a certain starting vertex.
@@ -40,7 +40,7 @@ Based on whether all vertices are connected, graphs can be divided into "connect
 
 <p align="center"> Figure 9-3 &nbsp; Connected and disconnected graphs </p>
 
-We can also add a "weight" variable to edges, resulting in "weighted graphs" as shown below. For example, in mobile games like "Honor of Kings", the system calculates the "closeness" between players based on shared gaming time, and this closeness network can be represented with a weighted graph.
+We can also add a "weight" variable to edges, resulting in "weighted graphs" as shown in Figure 9-4. For example, in mobile games like "Honor of Kings", the system calculates the "closeness" between players based on shared gaming time, and this closeness network can be represented with a weighted graph.
 
 ![Weighted and unweighted graphs](graph.assets/weighted_graph.png){ class="animation-figure" }
 
@@ -48,8 +48,8 @@ We can also add a "weight" variable to edges, resulting in "weighted graphs" as
 
 Graph data structures include the following commonly used terms.
 
-- "Adjacency": When there is an edge connecting two vertices, these two vertices are said to be "adjacent". In the above figure, the adjacent vertices of vertex 1 are vertices 2, 3, and 5.
-- "Path": The sequence of edges passed from vertex A to vertex B is called a "path" from A to B. In the above figure, the edge sequence 1-5-2-4 is a path from vertex 1 to vertex 4.
+- "Adjacency": When there is an edge connecting two vertices, these two vertices are said to be "adjacent". In Figure 9-4, the adjacent vertices of vertex 1 are vertices 2, 3, and 5.
+- "Path": The sequence of edges passed from vertex A to vertex B is called a "path" from A to B. In Figure 9-4, the edge sequence 1-5-2-4 is a path from vertex 1 to vertex 4.
 - "Degree": The number of edges a vertex has. For directed graphs, "in-degree" refers to how many edges point to the vertex, and "out-degree" refers to how many edges point out from the vertex.
 
 ## 9.1.2 &nbsp; Representation of graphs
@@ -60,7 +60,7 @@ Common representations of graphs include "adjacency matrices" and "adjacency lis
 
 Let the number of vertices in the graph be $n$, the "adjacency matrix" uses an $n \times n$ matrix to represent the graph, where each row (column) represents a vertex, and the matrix elements represent edges, with $1$ or $0$ indicating whether there is an edge between two vertices.
 
-As shown below, let the adjacency matrix be $M$, and the list of vertices be $V$, then the matrix element $M[i, j] = 1$ indicates there is an edge between vertex $V[i]$ and vertex $V[j]$, conversely $M[i, j] = 0$ indicates there is no edge between the two vertices.
+As shown in Figure 9-5, let the adjacency matrix be $M$, and the list of vertices be $V$, then the matrix element $M[i, j] = 1$ indicates there is an edge between vertex $V[i]$ and vertex $V[j]$, conversely $M[i, j] = 0$ indicates there is no edge between the two vertices.
 
 ![Representation of a graph with an adjacency matrix](graph.assets/adjacency_matrix.png){ class="animation-figure" }
 
@@ -76,7 +76,7 @@ When representing graphs with adjacency matrices, it is possible to directly acc
 
 ### 2. &nbsp; Adjacency list
 
-The "adjacency list" uses $n$ linked lists to represent the graph, with each linked list node representing a vertex. The $i$-th linked list corresponds to vertex $i$ and contains all adjacent vertices (vertices connected to that vertex). The Figure 9-6  shows an example of a graph stored using an adjacency list.
+The "adjacency list" uses $n$ linked lists to represent the graph, with each linked list node representing a vertex. The $i$-th linked list corresponds to vertex $i$ and contains all adjacent vertices (vertices connected to that vertex). Figure 9-6 shows an example of a graph stored using an adjacency list.
 
 ![Representation of a graph with an adjacency list](graph.assets/adjacency_list.png){ class="animation-figure" }
 
@@ -84,11 +84,11 @@ The "adjacency list" uses $n$ linked lists to represent the graph, with each lin
 
 The adjacency list only stores actual edges, and the total number of edges is often much less than $n^2$, making it more space-efficient. However, finding edges in the adjacency list requires traversing the linked list, so its time efficiency is not as good as that of the adjacency matrix.
 
-Observing the above figure, **the structure of the adjacency list is very similar to the "chaining" in hash tables, hence we can use similar methods to optimize efficiency**. For example, when the linked list is long, it can be transformed into an AVL tree or red-black tree, thus optimizing the time efficiency from $O(n)$ to $O(\log n)$; the linked list can also be transformed into a hash table, thus reducing the time complexity to $O(1)$.
+Observing Figure 9-6, **the structure of the adjacency list is very similar to the "chaining" in hash tables, hence we can use similar methods to optimize efficiency**. For example, when the linked list is long, it can be transformed into an AVL tree or red-black tree, thus optimizing the time efficiency from $O(n)$ to $O(\log n)$; the linked list can also be transformed into a hash table, thus reducing the time complexity to $O(1)$.
 
 ## 9.1.3 &nbsp; Common applications of graphs
 
-As shown in the Table 9-1 , many real-world systems can be modeled with graphs, and corresponding problems can be reduced to graph computing problems.
+As shown in Table 9-1, many real-world systems can be modeled with graphs, and corresponding problems can be reduced to graph computing problems.
 
 <p align="center"> Table 9-1 &nbsp; Common graphs in real life </p>
 

en/docs/chapter_graph/graph_operations.md

@@ -8,7 +8,7 @@ The basic operations on graphs can be divided into operations on "edges" and ope
 
 ## 9.2.1   Implementation based on adjacency matrix
 
-Given an undirected graph with $n$ vertices, the various operations are implemented as shown in the Figure 9-7 .
+Given an undirected graph with $n$ vertices, the various operations are implemented as shown in Figure 9-7.
 
 - **Adding or removing an edge**: Directly modify the specified edge in the adjacency matrix, using $O(1)$ time. Since it is an undirected graph, it is necessary to update the edges in both directions simultaneously.
 - **Adding a vertex**: Add a row and a column at the end of the adjacency matrix and fill them all with $0$s, using $O(n)$ time.
@@ -1219,7 +1219,7 @@ Below is the implementation code for graphs represented using an adjacency matri
 
 ## 9.2.2   Implementation based on adjacency list
 
-Given an undirected graph with a total of $n$ vertices and $m$ edges, the various operations can be implemented as shown in the Figure 9-8 .
+Given an undirected graph with a total of $n$ vertices and $m$ edges, the various operations can be implemented as shown in Figure 9-8.
 
 - **Adding an edge**: Simply add the edge at the end of the corresponding vertex's linked list, using $O(1)$ time. Because it is an undirected graph, it is necessary to add edges in both directions simultaneously.
 - **Removing an edge**: Find and remove the specified edge in the corresponding vertex's linked list, using $O(m)$ time. In an undirected graph, it is necessary to remove edges in both directions simultaneously.
@@ -2388,7 +2388,7 @@ Additionally, we use the `Vertex` class to represent vertices in the adjacency l
 
 ## 9.2.3   Efficiency comparison
 
-Assuming there are $n$ vertices and $m$ edges in the graph, the Table 9-2  compares the time efficiency and space efficiency of the adjacency matrix and adjacency list.
+Assuming there are $n$ vertices and $m$ edges in the graph, Table 9-2 compares the time efficiency and space efficiency of the adjacency matrix and adjacency list.
 
 <p align="center"> Table 9-2 &nbsp; Comparison of adjacency matrix and adjacency list </p>
 
@@ -2405,4 +2405,4 @@ Assuming there are $n$ vertices and $m$ edges in the graph, the Table 9-2  compa
 
 </div>
 
-Observing the Table 9-2 , it seems that the adjacency list (hash table) has the best time efficiency and space efficiency. However, in practice, operating on edges in the adjacency matrix is more efficient, requiring only a single array access or assignment operation. Overall, the adjacency matrix exemplifies the principle of "space for time", while the adjacency list exemplifies "time for space".
+Observing Table 9-2, it seems that the adjacency list (hash table) has the best time efficiency and space efficiency. However, in practice, operating on edges in the adjacency matrix is more efficient, requiring only a single array access or assignment operation. Overall, the adjacency matrix exemplifies the principle of "space for time", while the adjacency list exemplifies "time for space".

en/docs/chapter_greedy/fractional_knapsack_problem.md

@@ -6,7 +6,7 @@ comments: true
 
 !!! question
 
-    Given $n$ items, the weight of the $i$-th item is $wgt[i-1]$ and its value is $val[i-1]$, and a knapsack with a capacity of $cap$. Each item can be chosen only once, **but a part of the item can be selected, with its value calculated based on the proportion of the weight chosen**, what is the maximum value of the items in the knapsack under the limited capacity? An example is shown below.
+    Given $n$ items, the weight of the $i$-th item is $wgt[i-1]$ and its value is $val[i-1]$, and a knapsack with a capacity of $cap$. Each item can be chosen only once, **but a part of the item can be selected, with its value calculated based on the proportion of the weight chosen**, what is the maximum value of the items in the knapsack under the limited capacity? An example is shown in Figure 15-3.
 
 ![Example data of the fractional knapsack problem](fractional_knapsack_problem.assets/fractional_knapsack_example.png){ class="animation-figure" }
 
@@ -14,7 +14,7 @@ comments: true
 
 The fractional knapsack problem is very similar overall to the 0-1 knapsack problem, involving the current item $i$ and capacity $c$, aiming to maximize the value within the limited capacity of the knapsack.
 
-The difference is that, in this problem, only a part of an item can be chosen. As shown in the Figure 15-4 , **we can arbitrarily split the items and calculate the corresponding value based on the weight proportion**.
+The difference is that, in this problem, only a part of an item can be chosen. As shown in Figure 15-4, **we can arbitrarily split the items and calculate the corresponding value based on the weight proportion**.
 
 1. For item $i$, its value per unit weight is $val[i-1] / wgt[i-1]$, referred to as the unit value.
 2. Suppose we put a part of item $i$ with weight $w$ into the knapsack, then the value added to the knapsack is $w \times val[i-1] / wgt[i-1]$.
@@ -25,7 +25,7 @@ The difference is that, in this problem, only a part of an item can be chosen. A
 
 ### 1.   Greedy strategy determination
 
-Maximizing the total value of the items in the knapsack essentially means maximizing the value per unit weight. From this, the greedy strategy shown below can be deduced.
+Maximizing the total value of the items in the knapsack essentially means maximizing the value per unit weight. From this, the greedy strategy shown in Figure 15-5 can be deduced.
 
 1. Sort the items by their unit value from high to low.
 2. Iterate over all items, **greedily choosing the item with the highest unit value in each round**.
@@ -526,7 +526,7 @@ Now remove a unit weight of any item from the knapsack and replace it with a uni
 
 For other items in this solution, we can also construct the above contradiction. Overall, **items with greater unit value are always better choices**, proving that the greedy strategy is effective.
 
-As shown in the Figure 15-6 , if the item weight and unit value are viewed as the horizontal and vertical axes of a two-dimensional chart respectively, the fractional knapsack problem can be transformed into "seeking the largest area enclosed within a limited horizontal axis range". This analogy can help us understand the effectiveness of the greedy strategy from a geometric perspective.
+As shown in Figure 15-6, if the item weight and unit value are viewed as the horizontal and vertical axes of a two-dimensional chart respectively, the fractional knapsack problem can be transformed into "seeking the largest area enclosed within a limited horizontal axis range". This analogy can help us understand the effectiveness of the greedy strategy from a geometric perspective.
 
 ![Geometric representation of the fractional knapsack problem](fractional_knapsack_problem.assets/fractional_knapsack_area_chart.png){ class="animation-figure" }
 

en/docs/chapter_greedy/greedy_algorithm.md

@@ -17,7 +17,7 @@ Let's first understand the working principle of the greedy algorithm through the
 
     Given $n$ types of coins, where the denomination of the $i$th type of coin is $coins[i - 1]$, and the target amount is $amt$, with each type of coin available indefinitely, what is the minimum number of coins needed to make up the target amount? If it is not possible to make up the target amount, return $-1$.
 
-The greedy strategy adopted in this problem is shown in the following figure. Given the target amount, **we greedily choose the coin that is closest to and not greater than it**, repeatedly following this step until the target amount is met.
+The greedy strategy adopted in this problem is shown in Figure 15-1. Given the target amount, **we greedily choose the coin that is closest to and not greater than it**, repeatedly following this step until the target amount is met.
 
 ![Greedy strategy for coin change](greedy_algorithm.assets/coin_change_greedy_strategy.png){ class="animation-figure" }
 
@@ -330,7 +330,7 @@ You might exclaim: So clean! The greedy algorithm solves the coin change problem
 
 **Greedy algorithms are not only straightforward and simple to implement, but they are also usually very efficient**. In the code above, if the smallest coin denomination is $\min(coins)$, the greedy choice loops at most $amt / \min(coins)$ times, giving a time complexity of $O(amt / \min(coins))$. This is an order of magnitude smaller than the time complexity of the dynamic programming solution, which is $O(n \times amt)$.
 
-However, **for some combinations of coin denominations, greedy algorithms cannot find the optimal solution**. The following figure provides two examples.
+However, **for some combinations of coin denominations, greedy algorithms cannot find the optimal solution**. Figure 15-2 provides two examples.
 
 - **Positive example $coins = [1, 5, 10, 20, 50, 100]$**: In this coin combination, given any $amt$, the greedy algorithm can find the optimal solution.
 - **Negative example $coins = [1, 20, 50]$**: Suppose $amt = 60$, the greedy algorithm can only find the combination $50 + 1 \times 10$, totaling 11 coins, but dynamic programming can find the optimal solution of $20 + 20 + 20$, needing only 3 coins.

en/docs/chapter_greedy/max_capacity_problem.md

@@ -10,7 +10,7 @@ comments: true
     
     The capacity of the container is the product of the height and the width (area), where the height is determined by the shorter partition, and the width is the difference in array indices between the two partitions.
     
-    Please select two partitions in the array that maximize the container's capacity and return this maximum capacity. An example is shown in the following figure.
+    Please select two partitions in the array that maximize the container's capacity and return this maximum capacity. An example is shown in Figure 15-7.
 
 ![Example data for the maximum capacity problem](max_capacity_problem.assets/max_capacity_example.png){ class="animation-figure" }
 
@@ -28,13 +28,13 @@ Assuming the length of the array is $n$, the number of combinations of two parti
 
 ### 1.   Determination of a greedy strategy
 
-There is a more efficient solution to this problem. As shown in the following figure, we select a state $[i, j]$ where the indices $i < j$ and the height $ht[i] < ht[j]$, meaning $i$ is the shorter partition, and $j$ is the taller one.
+There is a more efficient solution to this problem. As shown in Figure 15-8, we select a state $[i, j]$ where the indices $i < j$ and the height $ht[i] < ht[j]$, meaning $i$ is the shorter partition, and $j$ is the taller one.
 
 ![Initial state](max_capacity_problem.assets/max_capacity_initial_state.png){ class="animation-figure" }
 
 <p align="center"> Figure 15-8 &nbsp; Initial state </p>
 
-As shown in the following figure, **if we move the taller partition $j$ closer to the shorter partition $i$, the capacity will definitely decrease**.
+As shown in Figure 15-9, **if we move the taller partition $j$ closer to the shorter partition $i$, the capacity will definitely decrease**.
 
 This is because when moving the taller partition $j$, the width $j-i$ definitely decreases; and since the height is determined by the shorter partition, the height can only remain the same (if $i$ remains the shorter partition) or decrease (if the moved $j$ becomes the shorter partition).
 
@@ -42,7 +42,7 @@ This is because when moving the taller partition $j$, the width $j-i$ definitely
 
 <p align="center"> Figure 15-9 &nbsp; State after moving the taller partition inward </p>
 
-Conversely, **we can only possibly increase the capacity by moving the shorter partition $i$ inward**. Although the width will definitely decrease, **the height may increase** (if the moved shorter partition $i$ becomes taller). For example, in the Figure 15-10 , the area increases after moving the shorter partition.
+Conversely, **we can only possibly increase the capacity by moving the shorter partition $i$ inward**. Although the width will definitely decrease, **the height may increase** (if the moved shorter partition $i$ becomes taller). For example, in Figure 15-10, the area increases after moving the shorter partition.
 
 ![State after moving the shorter partition inward](max_capacity_problem.assets/max_capacity_moving_short_board.png){ class="animation-figure" }
 
@@ -50,7 +50,7 @@ Conversely, **we can only possibly increase the capacity by moving the shorter p
 
 This leads us to the greedy strategy for this problem: initialize two pointers at the ends of the container, and in each round, move the pointer corresponding to the shorter partition inward until the two pointers meet.
 
-The following figures illustrate the execution of the greedy strategy.
+Figure 15-11 illustrate the execution of the greedy strategy.
 
 1. Initially, the pointers $i$ and $j$ are positioned at the ends of the array.
 2. Calculate the current state's capacity $cap[i, j]$ and update the maximum capacity.
@@ -415,7 +415,7 @@ The variables $i$, $j$, and $res$ use a constant amount of extra space, **thus t
 
 The reason why the greedy method is faster than enumeration is that each round of greedy selection "skips" some states.
 
-For example, under the state $cap[i, j]$ where $i$ is the shorter partition and $j$ is the taller partition, greedily moving the shorter partition $i$ inward by one step leads to the "skipped" states shown below. **This means that these states' capacities cannot be verified later**.
+For example, under the state $cap[i, j]$ where $i$ is the shorter partition and $j$ is the taller partition, greedily moving the shorter partition $i$ inward by one step leads to the "skipped" states shown in Figure 15-12. **This means that these states' capacities cannot be verified later**.
 
 $$
 cap[i, i+1], cap[i, i+2], \dots, cap[i, j-2], cap[i, j-1]

en/docs/chapter_greedy/max_product_cutting_problem.md

@@ -38,7 +38,7 @@ n & \geq 4
 \end{aligned}
 $$
 
-As shown below, when $n \geq 4$, splitting out a $2$ increases the product, **which indicates that integers greater than or equal to $4$ should be split**.
+As shown in Figure 15-14, when $n \geq 4$, splitting out a $2$ increases the product, **which indicates that integers greater than or equal to $4$ should be split**.
 
 **Greedy strategy one**: If the splitting scheme includes factors $\geq 4$, they should be further split. The final split should only include factors $1$, $2$, and $3$.
 
@@ -48,7 +48,7 @@ As shown below, when $n \geq 4$, splitting out a $2$ increases the product, **wh
 
 Next, consider which factor is optimal. Among the factors $1$, $2$, and $3$, clearly $1$ is the worst, as $1 \times (n-1) < n$ always holds, meaning splitting out $1$ actually decreases the product.
 
-As shown below, when $n = 6$, $3 \times 3 > 2 \times 2 \times 2$. **This means splitting out $3$ is better than splitting out $2$**.
+As shown in Figure 15-15, when $n = 6$, $3 \times 3 > 2 \times 2 \times 2$. **This means splitting out $3$ is better than splitting out $2$**.
 
 **Greedy strategy two**: In the splitting scheme, there should be at most two $2$s. Because three $2$s can always be replaced by two $3$s to obtain a higher product.
 
@@ -65,7 +65,7 @@ From the above, the following greedy strategies can be derived.
 
 ### 2. &nbsp; Code implementation
 
-As shown below, we do not need to use loops to split the integer but can use the floor division operation to get the number of $3$s, $a$, and the modulo operation to get the remainder, $b$, thus:
+As shown in Figure 15-16, we do not need to use loops to split the integer but can use the floor division operation to get the number of $3$s, $a$, and the modulo operation to get the remainder, $b$, thus:
 
 $$
 n = 3a + b

en/docs/chapter_hashing/hash_algorithm.md

@@ -6,7 +6,7 @@ comments: true
 
 The previous two sections introduced the working principle of hash tables and the methods to handle hash collisions. However, both open addressing and chaining can **only ensure that the hash table functions normally when collisions occur, but cannot reduce the frequency of hash collisions**.
 
-If hash collisions occur too frequently, the performance of the hash table will deteriorate drastically. As shown in the Figure 6-8 , for a chaining hash table, in the ideal case, the key-value pairs are evenly distributed across the buckets, achieving optimal query efficiency; in the worst case, all key-value pairs are stored in the same bucket, degrading the time complexity to $O(n)$.
+If hash collisions occur too frequently, the performance of the hash table will deteriorate drastically. As shown in Figure 6-8, for a chaining hash table, in the ideal case, the key-value pairs are evenly distributed across the buckets, achieving optimal query efficiency; in the worst case, all key-value pairs are stored in the same bucket, degrading the time complexity to $O(n)$.
 
 ![Ideal and worst cases of hash collisions](hash_algorithm.assets/hash_collision_best_worst_condition.png){ class="animation-figure" }
 
@@ -693,7 +693,7 @@ It is not hard to see that the simple hash algorithms mentioned above are quite
 
 In practice, we usually use some standard hash algorithms, such as MD5, SHA-1, SHA-2, and SHA-3. They can map input data of any length to a fixed-length hash value.
 
-Over the past century, hash algorithms have been in a continuous process of upgrading and optimization. Some researchers strive to improve the performance of hash algorithms, while others, including hackers, are dedicated to finding security issues in hash algorithms. The Table 6-2  shows hash algorithms commonly used in practical applications.
+Over the past century, hash algorithms have been in a continuous process of upgrading and optimization. Some researchers strive to improve the performance of hash algorithms, while others, including hackers, are dedicated to finding security issues in hash algorithms. Table 6-2 shows hash algorithms commonly used in practical applications.
 
 - MD5 and SHA-1 have been successfully attacked multiple times and are thus abandoned in various security applications.
 - SHA-2 series, especially SHA-256, is one of the most secure hash algorithms to date, with no successful attacks reported, hence commonly used in various security applications and protocols.

en/docs/chapter_hashing/hash_collision.md

@@ -15,7 +15,7 @@ There are mainly two methods for improving the structure of hash tables: "Separa
 
 ## 6.2.1   Separate chaining
 
-In the original hash table, each bucket can store only one key-value pair. "Separate chaining" transforms individual elements into a linked list, with key-value pairs as list nodes, storing all colliding key-value pairs in the same list. The Figure 6-5  shows an example of a hash table with separate chaining.
+In the original hash table, each bucket can store only one key-value pair. "Separate chaining" transforms individual elements into a linked list, with key-value pairs as list nodes, storing all colliding key-value pairs in the same list. Figure 6-5 shows an example of a hash table with separate chaining.
 
 ![Separate chaining hash table](hash_collision.assets/hash_table_chaining.png){ class="animation-figure" }
 
@@ -1547,7 +1547,7 @@ Linear probing uses a fixed-step linear search for probing, differing from ordin
 - **Inserting elements**: Calculate the bucket index using the hash function. If the bucket already contains an element, linearly traverse forward from the conflict position (usually with a step size of $1$) until an empty bucket is found, then insert the element.
 - **Searching for elements**: If a hash collision is found, use the same step size to linearly traverse forward until the corresponding element is found and return `value`; if an empty bucket is encountered, it means the target element is not in the hash table, so return `None`.
 
-The Figure 6-6  shows the distribution of key-value pairs in an open addressing (linear probing) hash table. According to this hash function, keys with the same last two digits will be mapped to the same bucket. Through linear probing, they are stored consecutively in that bucket and the buckets below it.
+Figure 6-6 shows the distribution of key-value pairs in an open addressing (linear probing) hash table. According to this hash function, keys with the same last two digits will be mapped to the same bucket. Through linear probing, they are stored consecutively in that bucket and the buckets below it.
 
 ![Distribution of key-value pairs in open addressing (linear probing) hash table](hash_collision.assets/hash_table_linear_probing.png){ class="animation-figure" }
 
@@ -1555,7 +1555,7 @@ The Figure 6-6  shows the distribution of key-value pairs in an open addressing
 
 However, **linear probing tends to create "clustering"**. Specifically, the longer a continuous position in the array is occupied, the more likely these positions are to encounter hash collisions, further promoting the growth of these clusters and eventually leading to deterioration in the efficiency of operations.
 
-It's important to note that **we cannot directly delete elements in an open addressing hash table**. Deleting an element creates an empty bucket `None` in the array. When searching for elements, if linear probing encounters this empty bucket, it will return, making the elements below this bucket inaccessible. The program may incorrectly assume these elements do not exist, as shown in the Figure 6-7 .
+It's important to note that **we cannot directly delete elements in an open addressing hash table**. Deleting an element creates an empty bucket `None` in the array. When searching for elements, if linear probing encounters this empty bucket, it will return, making the elements below this bucket inaccessible. The program may incorrectly assume these elements do not exist, as shown in Figure 6-7.
 
 ![Query issues caused by deletion in open addressing](hash_collision.assets/hash_table_open_addressing_deletion.png){ class="animation-figure" }
 

en/docs/chapter_hashing/hash_map.md

@@ -6,13 +6,13 @@ comments: true
 
 A "hash table", also known as a "hash map", achieves efficient element querying by establishing a mapping between keys and values. Specifically, when we input a `key` into the hash table, we can retrieve the corresponding `value` in $O(1)$ time.
 
-As shown in the Figure 6-1 , given $n$ students, each with two pieces of data: "name" and "student number". If we want to implement a query feature that returns the corresponding name when given a student number, we can use the hash table shown in the Figure 6-1 .
+As shown in Figure 6-1, given $n$ students, each with two pieces of data: "name" and "student number". If we want to implement a query feature that returns the corresponding name when given a student number, we can use the hash table shown in Figure 6-1.
 
 ![Abstract representation of a hash table](hash_map.assets/hash_table_lookup.png){ class="animation-figure" }
 
 <p align="center"> Figure 6-1 &nbsp; Abstract representation of a hash table </p>
 
-Apart from hash tables, arrays and linked lists can also be used to implement querying functions. Their efficiency is compared in the Table 6-1 .
+Apart from hash tables, arrays and linked lists can also be used to implement querying functions. Their efficiency is compared in Table 6-1.
 
 - **Adding elements**: Simply add the element to the end of the array (or linked list), using $O(1)$ time.
 - **Querying elements**: Since the array (or linked list) is unordered, it requires traversing all the elements, using $O(n)$ time.
@@ -513,7 +513,7 @@ index = hash(key) % capacity
 
 Afterward, we can use `index` to access the corresponding bucket in the hash table and thereby retrieve the `value`.
 
-Assuming array length `capacity = 100` and hash algorithm `hash(key) = key`, the hash function is `key % 100`. The Figure 6-2  uses `key` as the student number and `value` as the name to demonstrate the working principle of the hash function.
+Assuming array length `capacity = 100` and hash algorithm `hash(key) = key`, the hash function is `key % 100`. Figure 6-2 uses `key` as the student number and `value` as the name to demonstrate the working principle of the hash function.
 
 ![Working principle of hash function](hash_map.assets/hash_function.png){ class="animation-figure" }
 
@@ -1824,7 +1824,7 @@ For the hash function in the above example, if the last two digits of the input
 20336 % 100 = 36
 ```
 
-As shown in the Figure 6-3 , both student numbers point to the same name, which is obviously incorrect. This situation where multiple inputs correspond to the same output is known as "hash collision".
+As shown in Figure 6-3, both student numbers point to the same name, which is obviously incorrect. This situation where multiple inputs correspond to the same output is known as "hash collision".
 
 ![Example of hash collision](hash_map.assets/hash_collision.png){ class="animation-figure" }
 
@@ -1832,7 +1832,7 @@ As shown in the Figure 6-3 , both student numbers point to the same name, which
 
 It is easy to understand that the larger the capacity $n$ of the hash table, the lower the probability of multiple keys being allocated to the same bucket, and the fewer the collisions. Therefore, **expanding the capacity of the hash table can reduce hash collisions**.
 
-As shown in the Figure 6-4 , before expansion, key-value pairs `(136, A)` and `(236, D)` collided; after expansion, the collision is resolved.
+As shown in Figure 6-4, before expansion, key-value pairs `(136, A)` and `(236, D)` collided; after expansion, the collision is resolved.
 
 ![Hash table expansion](hash_map.assets/hash_table_reshash.png){ class="animation-figure" }
 

en/docs/chapter_heap/build_heap.md

@@ -351,7 +351,7 @@ Let's perform a more accurate calculation. To simplify the calculation, assume a
 
 <p align="center"> Figure 8-5 &nbsp; Node counts at each level of a perfect binary tree </p>
 
-As shown in the Figure 8-5 , the maximum number of iterations for a node "to be heapified from top to bottom" is equal to the distance from that node to the leaf nodes, which is precisely "node height." Therefore, we can sum the "number of nodes $\times$ node height" at each level, **to get the total number of heapification iterations for all nodes**.
+As shown in Figure 8-5, the maximum number of iterations for a node "to be heapified from top to bottom" is equal to the distance from that node to the leaf nodes, which is precisely "node height." Therefore, we can sum the "number of nodes $\times$ node height" at each level, **to get the total number of heapification iterations for all nodes**.
 
 $$
 T(h) = 2^0h + 2^1(h-1) + 2^2(h-2) + \dots + 2^{(h-1)}\times1

en/docs/chapter_heap/heap.md

@@ -4,7 +4,7 @@ comments: true
 
 # 8.1   Heap
 
-A "heap" is a complete binary tree that satisfies specific conditions and can be mainly divided into two types, as shown in the Figure 8-1 .
+A "heap" is a complete binary tree that satisfies specific conditions and can be mainly divided into two types, as shown in Figure 8-1.
 
 - "Min heap": The value of any node $\leq$ the values of its child nodes.
 - "Max heap": The value of any node $\geq$ the values of its child nodes.
@@ -25,7 +25,7 @@ It should be noted that many programming languages provide a "priority queue," w
 
 In fact, **heaps are often used to implement priority queues, with max heaps equivalent to priority queues where elements are dequeued in descending order**. From a usage perspective, we can consider "priority queue" and "heap" as equivalent data structures. Therefore, this book does not make a special distinction between the two, uniformly referring to them as "heap."
 
-Common operations on heaps are shown in the Table 8-1 , and the method names depend on the programming language.
+Common operations on heaps are shown in Table 8-1, and the method names depend on the programming language.
 
 <p align="center"> Table 8-1 &nbsp; Efficiency of Heap Operations </p>
 
@@ -438,7 +438,7 @@ As mentioned in the "Binary Trees" section, complete binary trees are well-suite
 
 When using an array to represent a binary tree, elements represent node values, and indexes represent node positions in the binary tree. **Node pointers are implemented through an index mapping formula**.
 
-As shown in the Figure 8-2 , given an index $i$, the index of its left child is $2i + 1$, the index of its right child is $2i + 2$, and the index of its parent is $(i - 1) / 2$ (floor division). When the index is out of bounds, it signifies a null node or the node does not exist.
+As shown in Figure 8-2, given an index $i$, the index of its left child is $2i + 1$, the index of its right child is $2i + 2$, and the index of its parent is $(i - 1) / 2$ (floor division). When the index is out of bounds, it signifies a null node or the node does not exist.
 
 ![Representation and storage of heaps](heap.assets/representation_of_heap.png){ class="animation-figure" }
 
@@ -849,7 +849,7 @@ The top element of the heap is the root node of the binary tree, which is also t
 
 Given an element `val`, we first add it to the bottom of the heap. After addition, since `val` may be larger than other elements in the heap, the heap's integrity might be compromised, **thus it's necessary to repair the path from the inserted node to the root node**. This operation is called "heapifying".
 
-Considering starting from the node inserted, **perform heapify from bottom to top**. As shown in the Figure 8-3 , we compare the value of the inserted node with its parent node, and if the inserted node is larger, we swap them. Then continue this operation, repairing each node in the heap from bottom to top until passing the root node or encountering a node that does not need to be swapped.
+Considering starting from the node inserted, **perform heapify from bottom to top**. As shown in Figure 8-3, we compare the value of the inserted node with its parent node, and if the inserted node is larger, we swap them. Then continue this operation, repairing each node in the heap from bottom to top until passing the root node or encountering a node that does not need to be swapped.
 
 === "<1>"
     ![Steps of element insertion into the heap](heap.assets/heap_push_step1.png){ class="animation-figure" }
@@ -1285,7 +1285,7 @@ The top element of the heap is the root node of the binary tree, that is, the fi
 2. After swapping, remove the bottom of the heap from the list (note, since it has been swapped, what is actually being removed is the original top element).
 3. Starting from the root node, **perform heapify from top to bottom**.
 
-As shown in the Figure 8-4 , **the direction of "heapify from top to bottom" is opposite to "heapify from bottom to top"**. We compare the value of the root node with its two children and swap it with the largest child. Then repeat this operation until passing the leaf node or encountering a node that does not need to be swapped.
+As shown in Figure 8-4, **the direction of "heapify from top to bottom" is opposite to "heapify from bottom to top"**. We compare the value of the root node with its two children and swap it with the largest child. Then repeat this operation until passing the leaf node or encountering a node that does not need to be swapped.
 
 === "<1>"
     ![Steps of removing the top element from the heap](heap.assets/heap_pop_step1.png){ class="animation-figure" }

en/docs/chapter_heap/top_k.md

@@ -12,7 +12,7 @@ For this problem, we will first introduce two straightforward solutions, then ex
 
 ## 8.3.1   Method 1: Iterative selection
 
-We can perform $k$ rounds of iterations as shown in the Figure 8-6 , extracting the $1^{st}$, $2^{nd}$, $\dots$, $k^{th}$ largest elements in each round, with a time complexity of $O(nk)$.
+We can perform $k$ rounds of iterations as shown in Figure 8-6, extracting the $1^{st}$, $2^{nd}$, $\dots$, $k^{th}$ largest elements in each round, with a time complexity of $O(nk)$.
 
 This method is only suitable when $k \ll n$, as the time complexity approaches $O(n^2)$ when $k$ is close to $n$, which is very time-consuming.
 
@@ -26,7 +26,7 @@ This method is only suitable when $k \ll n$, as the time complexity approaches $
 
 ## 8.3.2   Method 2: Sorting
 
-As shown in the Figure 8-7 , we can first sort the array `nums` and then return the last $k$ elements, with a time complexity of $O(n \log n)$.
+As shown in Figure 8-7, we can first sort the array `nums` and then return the last $k$ elements, with a time complexity of $O(n \log n)$.
 
 Clearly, this method "overachieves" the task, as we only need to find the largest $k$ elements, without the need to sort the other elements.
 

en/docs/chapter_introduction/algorithms_are_everywhere.md

@@ -45,7 +45,7 @@ This essential skill for elementary students, looking up a dictionary, is actual
 
 The above method of organizing playing cards is essentially the "Insertion Sort" algorithm, which is very efficient for small datasets. Many programming languages' sorting functions include the insertion sort.
 
-**Example 3: Making Change**. Suppose we buy goods worth $69$ yuan at a supermarket and give the cashier $100$ yuan, then the cashier needs to give us $31$ yuan in change. They would naturally complete the thought process as shown below.
+**Example 3: Making Change**. Suppose we buy goods worth $69$ yuan at a supermarket and give the cashier $100$ yuan, then the cashier needs to give us $31$ yuan in change. They would naturally complete the thought process as shown in Figure 1-3.
 
 1. The options are currencies smaller than $31$, including $1$, $5$, $10$, and $20$.
 2. Take out the largest $20$ from the options, leaving $31 - 20 = 11$.

en/docs/chapter_introduction/what_is_dsa.md

@@ -27,7 +27,7 @@ A "data structure" is a way of organizing and storing data in a computer, with t
 
 ## 1.2.3   Relationship between data structures and algorithms
 
-As shown in the Figure 1-4 , data structures and algorithms are highly related and closely integrated, specifically in the following three aspects:
+As shown in Figure 1-4, data structures and algorithms are highly related and closely integrated, specifically in the following three aspects:
 
 - Data structures are the foundation of algorithms. They provide structured data storage and methods for manipulating data for algorithms.
 - Algorithms are the stage where data structures come into play. The data structure alone only stores data information; it is through the application of algorithms that specific problems can be solved.
@@ -37,13 +37,13 @@ As shown in the Figure 1-4 , data structures and algorithms are highly related a
 
 <p align="center"> Figure 1-4 &nbsp; Relationship between data structures and algorithms </p>
 
-Data structures and algorithms can be likened to a set of building blocks, as illustrated in the Figure 1-5 . A building block set includes numerous pieces, accompanied by detailed assembly instructions. Following these instructions step by step allows us to construct an intricate block model.
+Data structures and algorithms can be likened to a set of building blocks, as illustrated in Figure 1-5. A building block set includes numerous pieces, accompanied by detailed assembly instructions. Following these instructions step by step allows us to construct an intricate block model.
 
 ![Assembling blocks](what_is_dsa.assets/assembling_blocks.png){ class="animation-figure" }
 
 <p align="center"> Figure 1-5 &nbsp; Assembling blocks </p>
 
-The detailed correspondence between the two is shown in the Table 1-1 .
+The detailed correspondence between the two is shown in Table 1-1.
 
 <p align="center"> Table 1-1 &nbsp; Comparing data structures and algorithms to building blocks </p>
 

en/docs/chapter_preface/about_the_book.md

@@ -24,7 +24,7 @@ If you are an algorithm expert, we look forward to receiving your valuable sugge
 
 ## 0.1.2   Content structure
 
-The main content of the book is shown in the following figure.
+The main content of the book is shown in Figure 0-1.
 
 - **Complexity analysis**: explores aspects and methods for evaluating data structures and algorithms. Covers methods of deriving time complexity and space complexity, along with common types and examples.
 - **Data structures**: focuses on fundamental data types, classification methods, definitions, pros and cons, common operations, types, applications, and implementation methods of data structures such as array, linked list, stack, queue, hash table, tree, heap, graph, etc.

en/docs/chapter_preface/suggestions.md

@@ -188,7 +188,7 @@ comments: true
 
 Compared with text, videos and pictures have a higher density of information and are more structured, making them easier to understand. In this book, **key and difficult concepts are mainly presented through animations and illustrations**, with text serving as explanations and supplements.
 
-When encountering content with animations or illustrations as shown in the Figure 0-2 , **prioritize understanding the figure, with text as supplementary**, integrating both for a comprehensive understanding.
+When encountering content with animations or illustrations as shown in Figure 0-2, **prioritize understanding the figure, with text as supplementary**, integrating both for a comprehensive understanding.
 
 ![Animated illustration example](../index.assets/animation.gif){ class="animation-figure" }
 
@@ -196,7 +196,7 @@ When encountering content with animations or illustrations as shown in the Figur
 
 ## 0.2.3   Deepen understanding through coding practice
 
-The source code of this book is hosted on the [GitHub Repository](https://github.com/krahets/hello-algo). As shown in the Figure 0-3 , **the source code comes with test examples and can be executed with just a single click**.
+The source code of this book is hosted on the [GitHub Repository](https://github.com/krahets/hello-algo). As shown in Figure 0-3, **the source code comes with test examples and can be executed with just a single click**.
 
 If time permits, **it's recommended to type out the code yourself**. If pressed for time, at least read and run all the codes.
 
@@ -218,13 +218,13 @@ If [Git](https://git-scm.com/downloads) is installed, use the following command
 git clone https://github.com/krahets/hello-algo.git
 ```
 
-Alternatively, you can also click the "Download ZIP" button at the location shown in the Figure 0-4  to directly download the code as a compressed ZIP file. Then, you can simply extract it locally.
+Alternatively, you can also click the "Download ZIP" button at the location shown in Figure 0-4 to directly download the code as a compressed ZIP file. Then, you can simply extract it locally.
 
 ![Cloning repository and downloading code](suggestions.assets/download_code.png){ class="animation-figure" }
 
 <p align="center"> Figure 0-4 &nbsp; Cloning repository and downloading code </p>
 
-**Step 3: Run the source code**. As shown in the Figure 0-5 , for the code block labeled with the file name at the top, we can find the corresponding source code file in the `codes` folder of the repository. These files can be executed with a single click, which will help you save unnecessary debugging time and allow you to focus on learning.
+**Step 3: Run the source code**. As shown in Figure 0-5, for the code block labeled with the file name at the top, we can find the corresponding source code file in the `codes` folder of the repository. These files can be executed with a single click, which will help you save unnecessary debugging time and allow you to focus on learning.
 
 ![Code block and corresponding source code file](suggestions.assets/code_md_to_repo.png){ class="animation-figure" }
 
@@ -234,7 +234,7 @@ Alternatively, you can also click the "Download ZIP" button at the location show
 
 While reading this book, please don't skip over the points that you didn't learn. **Feel free to post your questions in the comment section**. We will be happy to answer them and can usually respond within two days.
 
-As illustrated in the Figure 0-6 , each chapter features a comment section at the bottom. I encourage you to pay attention to these comments. They not only expose you to others' encountered problems, aiding in identifying knowledge gaps and sparking deeper contemplation, but also invite you to generously contribute by answering fellow readers' inquiries, sharing insights, and fostering mutual improvement.
+As illustrated in Figure 0-6, each chapter features a comment section at the bottom. I encourage you to pay attention to these comments. They not only expose you to others' encountered problems, aiding in identifying knowledge gaps and sparking deeper contemplation, but also invite you to generously contribute by answering fellow readers' inquiries, sharing insights, and fostering mutual improvement.
 
 ![Comment section example](../index.assets/comment.gif){ class="animation-figure" }
 
@@ -248,7 +248,7 @@ Overall, the journey of mastering data structures and algorithms can be divided
 2. **Stage 2: Practicing algorithm problems**. It is recommended to start from popular problems, such as [Sword for Offer](https://leetcode.cn/studyplan/coding-interviews/) and [LeetCode Hot 100](https://leetcode.cn/studyplan/top-100- liked/), and accumulate at least 100 questions to familiarize yourself with mainstream algorithmic problems. Forgetfulness can be a challenge when you start practicing, but rest assured that this is normal. We can follow the "Ebbinghaus Forgetting Curve" to review the questions, and usually after 3~5 rounds of repetitions, we will be able to memorize them.
 3. **Stage 3: Building the knowledge system**. In terms of learning, we can read algorithm column articles, solution frameworks, and algorithm textbooks to continuously enrich the knowledge system. In terms of practicing, we can try advanced strategies, such as categorizing by topic, multiple solutions for a single problem,  and one solution for multiple problems, etc. Insights on these strategies can be found in various communities.
 
-As shown in the Figure 0-7 , this book mainly covers “Stage 1,” aiming to help you more efficiently embark on Stages 2 and 3.
+As shown in Figure 0-7, this book mainly covers “Stage 1,” aiming to help you more efficiently embark on Stages 2 and 3.
 
 ![Algorithm learning path](suggestions.assets/learning_route.png){ class="animation-figure" }
 

en/docs/chapter_searching/binary_search.md

@@ -8,13 +8,13 @@ comments: true
 
 !!! question
 
-    Given an array `nums` of length $n$, with elements arranged in ascending order and non-repeating. Please find and return the index of element `target` in this array. If the array does not contain the element, return $-1$. An example is shown below.
+    Given an array `nums` of length $n$, with elements arranged in ascending order and non-repeating. Please find and return the index of element `target` in this array. If the array does not contain the element, return $-1$. An example is shown in Figure 10-1.
 
 ![Binary search example data](binary_search.assets/binary_search_example.png){ class="animation-figure" }
 
 <p align="center"> Figure 10-1 &nbsp; Binary search example data </p>
 
-As shown in the Figure 10-2 , we first initialize pointers $i = 0$ and $j = n - 1$, pointing to the first and last elements of the array, representing the search interval $[0, n - 1]$. Please note that square brackets indicate a closed interval, which includes the boundary values themselves.
+As shown in Figure 10-2, we first initialize pointers $i = 0$ and $j = n - 1$, pointing to the first and last elements of the array, representing the search interval $[0, n - 1]$. Please note that square brackets indicate a closed interval, which includes the boundary values themselves.
 
 Next, perform the following two steps in a loop.
 
@@ -739,7 +739,7 @@ We can implement a binary search algorithm with the same functionality based on
     <div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20binary_search_lcro%28nums%3A%20list%5Bint%5D,%20target%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E4%BA%8C%E5%88%86%E6%9F%A5%E6%89%BE%EF%BC%88%E5%B7%A6%E9%97%AD%E5%8F%B3%E5%BC%80%E5%8C%BA%E9%97%B4%EF%BC%89%22%22%22%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%E5%B7%A6%E9%97%AD%E5%8F%B3%E5%BC%80%E5%8C%BA%E9%97%B4%20%5B0,%20n%29%20%EF%BC%8C%E5%8D%B3%20i,%20j%20%E5%88%86%E5%88%AB%E6%8C%87%E5%90%91%E6%95%B0%E7%BB%84%E9%A6%96%E5%85%83%E7%B4%A0%E3%80%81%E5%B0%BE%E5%85%83%E7%B4%A0%2B1%0A%20%20%20%20i,%20j%20%3D%200,%20len%28nums%29%0A%20%20%20%20%23%20%E5%BE%AA%E7%8E%AF%EF%BC%8C%E5%BD%93%E6%90%9C%E7%B4%A2%E5%8C%BA%E9%97%B4%E4%B8%BA%E7%A9%BA%E6%97%B6%E8%B7%B3%E5%87%BA%EF%BC%88%E5%BD%93%20i%20%3D%20j%20%E6%97%B6%E4%B8%BA%E7%A9%BA%EF%BC%89%0A%20%20%20%20while%20i%20%3C%20j%3A%0A%20%20%20%20%20%20%20%20m%20%3D%20%28i%20%2B%20j%29%20//%202%20%20%23%20%E8%AE%A1%E7%AE%97%E4%B8%AD%E7%82%B9%E7%B4%A2%E5%BC%95%20m%0A%20%20%20%20%20%20%20%20if%20nums%5Bm%5D%20%3C%20target%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20i%20%3D%20m%20%2B%201%20%20%23%20%E6%AD%A4%E6%83%85%E5%86%B5%E8%AF%B4%E6%98%8E%20target%20%E5%9C%A8%E5%8C%BA%E9%97%B4%20%5Bm%2B1,%20j%29%20%E4%B8%AD%0A%20%20%20%20%20%20%20%20elif%20nums%5Bm%5D%20%3E%20target%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20j%20%3D%20m%20%20%23%20%E6%AD%A4%E6%83%85%E5%86%B5%E8%AF%B4%E6%98%8E%20target%20%E5%9C%A8%E5%8C%BA%E9%97%B4%20%5Bi,%20m%29%20%E4%B8%AD%0A%20%20%20%20%20%20%20%20else%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20return%20m%20%20%23%20%E6%89%BE%E5%88%B0%E7%9B%AE%E6%A0%87%E5%85%83%E7%B4%A0%EF%BC%8C%E8%BF%94%E5%9B%9E%E5%85%B6%E7%B4%A2%E5%BC%95%0A%20%20%20%20return%20-1%20%20%23%20%E6%9C%AA%E6%89%BE%E5%88%B0%E7%9B%AE%E6%A0%87%E5%85%83%E7%B4%A0%EF%BC%8C%E8%BF%94%E5%9B%9E%20-1%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20target%20%3D%206%0A%20%20%20%20nums%20%3D%20%5B1,%203,%206,%208,%2012,%2015,%2023,%2026,%2031,%2035%5D%0A%0A%20%20%20%20%23%20%E4%BA%8C%E5%88%86%E6%9F%A5%E6%89%BE%EF%BC%88%E5%B7%A6%E9%97%AD%E5%8F%B3%E5%BC%80%E5%8C%BA%E9%97%B4%EF%BC%89%0A%20%20%20%20index%20%3D%20binary_search_lcro%28nums,%20target%29%0A%20%20%20%20print%28%22%E7%9B%AE%E6%A0%87%E5%85%83%E7%B4%A0%206%20%E7%9A%84%E7%B4%A2%E5%BC%95%20%3D%20%22,%20index%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=5&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
     <div style="margin-top: 5px;"><a href="https://pythontutor.com/iframe-embed.html#code=def%20binary_search_lcro%28nums%3A%20list%5Bint%5D,%20target%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E4%BA%8C%E5%88%86%E6%9F%A5%E6%89%BE%EF%BC%88%E5%B7%A6%E9%97%AD%E5%8F%B3%E5%BC%80%E5%8C%BA%E9%97%B4%EF%BC%89%22%22%22%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%E5%B7%A6%E9%97%AD%E5%8F%B3%E5%BC%80%E5%8C%BA%E9%97%B4%20%5B0,%20n%29%20%EF%BC%8C%E5%8D%B3%20i,%20j%20%E5%88%86%E5%88%AB%E6%8C%87%E5%90%91%E6%95%B0%E7%BB%84%E9%A6%96%E5%85%83%E7%B4%A0%E3%80%81%E5%B0%BE%E5%85%83%E7%B4%A0%2B1%0A%20%20%20%20i,%20j%20%3D%200,%20len%28nums%29%0A%20%20%20%20%23%20%E5%BE%AA%E7%8E%AF%EF%BC%8C%E5%BD%93%E6%90%9C%E7%B4%A2%E5%8C%BA%E9%97%B4%E4%B8%BA%E7%A9%BA%E6%97%B6%E8%B7%B3%E5%87%BA%EF%BC%88%E5%BD%93%20i%20%3D%20j%20%E6%97%B6%E4%B8%BA%E7%A9%BA%EF%BC%89%0A%20%20%20%20while%20i%20%3C%20j%3A%0A%20%20%20%20%20%20%20%20m%20%3D%20%28i%20%2B%20j%29%20//%202%20%20%23%20%E8%AE%A1%E7%AE%97%E4%B8%AD%E7%82%B9%E7%B4%A2%E5%BC%95%20m%0A%20%20%20%20%20%20%20%20if%20nums%5Bm%5D%20%3C%20target%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20i%20%3D%20m%20%2B%201%20%20%23%20%E6%AD%A4%E6%83%85%E5%86%B5%E8%AF%B4%E6%98%8E%20target%20%E5%9C%A8%E5%8C%BA%E9%97%B4%20%5Bm%2B1,%20j%29%20%E4%B8%AD%0A%20%20%20%20%20%20%20%20elif%20nums%5Bm%5D%20%3E%20target%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20j%20%3D%20m%20%20%23%20%E6%AD%A4%E6%83%85%E5%86%B5%E8%AF%B4%E6%98%8E%20target%20%E5%9C%A8%E5%8C%BA%E9%97%B4%20%5Bi,%20m%29%20%E4%B8%AD%0A%20%20%20%20%20%20%20%20else%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20return%20m%20%20%23%20%E6%89%BE%E5%88%B0%E7%9B%AE%E6%A0%87%E5%85%83%E7%B4%A0%EF%BC%8C%E8%BF%94%E5%9B%9E%E5%85%B6%E7%B4%A2%E5%BC%95%0A%20%20%20%20return%20-1%20%20%23%20%E6%9C%AA%E6%89%BE%E5%88%B0%E7%9B%AE%E6%A0%87%E5%85%83%E7%B4%A0%EF%BC%8C%E8%BF%94%E5%9B%9E%20-1%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20target%20%3D%206%0A%20%20%20%20nums%20%3D%20%5B1,%203,%206,%208,%2012,%2015,%2023,%2026,%2031,%2035%5D%0A%0A%20%20%20%20%23%20%E4%BA%8C%E5%88%86%E6%9F%A5%E6%89%BE%EF%BC%88%E5%B7%A6%E9%97%AD%E5%8F%B3%E5%BC%80%E5%8C%BA%E9%97%B4%EF%BC%89%0A%20%20%20%20index%20%3D%20binary_search_lcro%28nums,%20target%29%0A%20%20%20%20print%28%22%E7%9B%AE%E6%A0%87%E5%85%83%E7%B4%A0%206%20%E7%9A%84%E7%B4%A2%E5%BC%95%20%3D%20%22,%20index%29&codeDivHeight=800&codeDivWidth=600&cumulative=false&curInstr=5&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">Full Screen ></a></div>
 
-As shown in the Figure 10-3 , in the two types of interval representations, the initialization of the binary search algorithm, the loop condition, and the narrowing interval operation are different.
+As shown in Figure 10-3, in the two types of interval representations, the initialization of the binary search algorithm, the loop condition, and the narrowing interval operation are different.
 
 Since both boundaries in the "closed interval" representation are defined as closed, the operations to narrow the interval through pointers $i$ and $j$ are also symmetrical. This makes it less prone to errors, **therefore, it is generally recommended to use the "closed interval" approach**.
 

en/docs/chapter_searching/binary_search_edge.md

@@ -245,7 +245,7 @@ Below we introduce two more cunning methods.
 
 In fact, we can use the function for finding the leftmost element to find the rightmost element, specifically by **transforming the search for the rightmost `target` into a search for the leftmost `target + 1`**.
 
-As shown in the Figure 10-7 , after the search is completed, the pointer $i$ points to the leftmost `target + 1` (if it exists), while $j$ points to the rightmost `target`, **thus returning $j$ is sufficient**.
+As shown in Figure 10-7, after the search is completed, the pointer $i$ points to the leftmost `target + 1` (if it exists), while $j$ points to the rightmost `target`, **thus returning $j$ is sufficient**.
 
 ![Transforming the search for the right boundary into the search for the left boundary](binary_search_edge.assets/binary_search_right_edge_by_left_edge.png){ class="animation-figure" }
 
@@ -500,7 +500,7 @@ Please note, the insertion point returned is $i$, therefore, it should be subtra
 
 We know that when the array does not contain `target`, $i$ and $j$ will eventually point to the first element greater and smaller than `target` respectively.
 
-Thus, as shown in the Figure 10-8 , we can construct an element that does not exist in the array, to search for the left and right boundaries.
+Thus, as shown in Figure 10-8, we can construct an element that does not exist in the array, to search for the left and right boundaries.
 
 - To find the leftmost `target`: it can be transformed into searching for `target - 0.5`, and return the pointer $i$.
 - To find the rightmost `target`: it can be transformed into searching for `target + 0.5`, and return the pointer $j$.

en/docs/chapter_searching/binary_search_insertion.md

@@ -10,7 +10,7 @@ Binary search is not only used to search for target elements but also to solve m
 
 !!! question
 
-    Given an ordered array `nums` of length $n$ and an element `target`, where the array has no duplicate elements. Now insert `target` into the array `nums` while maintaining its order. If the element `target` already exists in the array, insert it to its left side. Please return the index of `target` in the array after insertion. See the example shown in the Figure 10-4 .
+    Given an ordered array `nums` of length $n$ and an element `target`, where the array has no duplicate elements. Now insert `target` into the array `nums` while maintaining its order. If the element `target` already exists in the array, insert it to its left side. Please return the index of `target` in the array after insertion. See the example shown in Figure 10-4.
 
 ![Example data for binary search insertion point](binary_search_insertion.assets/binary_search_insertion_example.png){ class="animation-figure" }
 
@@ -334,7 +334,7 @@ Therefore, at the end of the binary, it is certain that: $i$ points to the first
 
 Suppose there are multiple `target`s in the array, ordinary binary search can only return the index of one of the `target`s, **and it cannot determine how many `target`s are to the left and right of that element**.
 
-The task requires inserting the target element to the very left, **so we need to find the index of the leftmost `target` in the array**. Initially consider implementing this through the steps shown in the Figure 10-5 .
+The task requires inserting the target element to the very left, **so we need to find the index of the leftmost `target` in the array**. Initially consider implementing this through the steps shown in Figure 10-5.
 
 1. Perform a binary search, get an arbitrary index of `target`, denoted as $k$.
 2. Start from index $k$, and perform a linear search to the left until the leftmost `target` is found and return.
@@ -345,7 +345,7 @@ The task requires inserting the target element to the very left, **so we need to
 
 Although this method is feasible, it includes linear search, so its time complexity is $O(n)$. This method is inefficient when the array contains many duplicate `target`s.
 
-Now consider extending the binary search code. As shown in the Figure 10-6 , the overall process remains the same, each round first calculates the midpoint index $m$, then judges the size relationship between `target` and `nums[m]`, divided into the following cases.
+Now consider extending the binary search code. As shown in Figure 10-6, the overall process remains the same, each round first calculates the midpoint index $m$, then judges the size relationship between `target` and `nums[m]`, divided into the following cases.
 
 - When `nums[m] < target` or `nums[m] > target`, it means `target` has not been found yet, thus use the normal binary search interval reduction operation, **thus making pointers $i$ and $j$ approach `target`**.
 - When `nums[m] == target`, it indicates that the elements less than `target` are in the interval $[i, m - 1]$, therefore use $j = m - 1$ to narrow the interval, **thus making pointer $j$ approach elements less than `target`**.

en/docs/chapter_searching/replace_linear_by_hashing.md

@@ -12,7 +12,7 @@ In algorithm problems, **we often reduce the time complexity of algorithms by re
 
 ## 10.4.1   Linear search: trading time for space
 
-Consider traversing all possible combinations directly. As shown in the Figure 10-9 , we initiate a two-layer loop, and in each round, we determine whether the sum of the two integers equals `target`. If so, we return their indices.
+Consider traversing all possible combinations directly. As shown in Figure 10-9, we initiate a two-layer loop, and in each round, we determine whether the sum of the two integers equals `target`. If so, we return their indices.
 
 ![Linear search solution for two-sum problem](replace_linear_by_hashing.assets/two_sum_brute_force.png){ class="animation-figure" }
 

en/docs/chapter_searching/searching_algorithm_revisited.md

@@ -42,7 +42,7 @@ However, **using these algorithms often requires data preprocessing**. For examp
 
 ## 10.5.3   Choosing a search method
 
-Given a set of data of size $n$, we can use linear search, binary search, tree search, hash search, and other methods to search for the target element from it. The working principles of these methods are shown in the following figure.
+Given a set of data of size $n$, we can use linear search, binary search, tree search, hash search, and other methods to search for the target element from it. The working principles of these methods are shown in Figure 10-11.
 
 ![Various search strategies](searching_algorithm_revisited.assets/searching_algorithms.png){ class="animation-figure" }
 

en/docs/chapter_sorting/bubble_sort.md

@@ -6,7 +6,7 @@ comments: true
 
 <u>Bubble sort</u> achieves sorting by continuously comparing and swapping adjacent elements. This process resembles bubbles rising from the bottom to the top, hence the name bubble sort.
 
-As shown in the following figures, the bubbling process can be simulated using element swap operations: starting from the leftmost end of the array and moving right, sequentially compare the size of adjacent elements. If "left element > right element," then swap them. After the traversal, the largest element will be moved to the far right end of the array.
+As shown in Figure 11-4, the bubbling process can be simulated using element swap operations: starting from the leftmost end of the array and moving right, sequentially compare the size of adjacent elements. If "left element > right element," then swap them. After the traversal, the largest element will be moved to the far right end of the array.
 
 === "<1>"
     ![Simulating bubble process using element swap](bubble_sort.assets/bubble_operation_step1.png){ class="animation-figure" }
@@ -33,7 +33,7 @@ As shown in the following figures, the bubbling process can be simulated using e
 
 ## 11.3.1 &nbsp; Algorithm process
 
-Assuming the length of the array is $n$, the steps of bubble sort are shown below.
+Assuming the length of the array is $n$, the steps of bubble sort are shown in Figure 11-5.
 
 1. First, perform a "bubble" on $n$ elements, **swapping the largest element to its correct position**.
 2. Next, perform a "bubble" on the remaining $n - 1$ elements, **swapping the second largest element to its correct position**.

en/docs/chapter_sorting/bucket_sort.md

@@ -10,7 +10,7 @@ The previously mentioned sorting algorithms are all "comparison-based sorting al
 
 ## 11.8.1   Algorithm process
 
-Consider an array of length $n$, with elements in the range $[0, 1)$. The bucket sort process is illustrated in the Figure 11-13 .
+Consider an array of length $n$, with elements in the range $[0, 1)$. The bucket sort process is illustrated in Figure 11-13.
 
 1. Initialize $k$ buckets and distribute $n$ elements into these $k$ buckets.
 2. Sort each bucket individually (using the built-in sorting function of the programming language).
@@ -463,7 +463,7 @@ The theoretical time complexity of bucket sort can reach $O(n)$, **the key is to
 
 To achieve even distribution, we can initially set a rough dividing line, roughly dividing the data into 3 buckets. **After the distribution is complete, the buckets with more products can be further divided into 3 buckets, until the number of elements in all buckets is roughly equal**.
 
-As shown in the Figure 11-14 , this method essentially creates a recursive tree, aiming to make the leaf node values as even as possible. Of course, you don't have to divide the data into 3 buckets each round; the specific division method can be flexibly chosen based on data characteristics.
+As shown in Figure 11-14, this method essentially creates a recursive tree, aiming to make the leaf node values as even as possible. Of course, you don't have to divide the data into 3 buckets each round; the specific division method can be flexibly chosen based on data characteristics.
 
 ![Recursive division of buckets](bucket_sort.assets/scatter_in_buckets_recursively.png){ class="animation-figure" }
 
@@ -471,7 +471,7 @@ As shown in the Figure 11-14 , this method essentially creates a recursive tree,
 
 If we know the probability distribution of product prices in advance, **we can set the price dividing line for each bucket based on the data probability distribution**. It is worth noting that it is not necessarily required to specifically calculate the data distribution; it can also be approximated based on data characteristics using some probability model.
 
-As shown in the Figure 11-15 , we assume that product prices follow a normal distribution, allowing us to reasonably set the price intervals, thereby evenly distributing the products into the respective buckets.
+As shown in Figure 11-15, we assume that product prices follow a normal distribution, allowing us to reasonably set the price intervals, thereby evenly distributing the products into the respective buckets.
 
 ![Dividing buckets based on probability distribution](bucket_sort.assets/scatter_in_buckets_distribution.png){ class="animation-figure" }
 

en/docs/chapter_sorting/insertion_sort.md

@@ -8,7 +8,7 @@ comments: true
 
 Specifically, we select a pivot element from the unsorted interval, compare it with the elements in the sorted interval to its left, and insert the element into the correct position.
 
-The Figure 11-6  shows the process of inserting an element into an array. Assuming the pivot element is `base`, we need to move all elements between the target index and `base` one position to the right, then assign `base` to the target index.
+Figure 11-6 shows the process of inserting an element into an array. Assuming the pivot element is `base`, we need to move all elements between the target index and `base` one position to the right, then assign `base` to the target index.
 
 ![Single insertion operation](insertion_sort.assets/insertion_operation.png){ class="animation-figure" }
 
@@ -16,7 +16,7 @@ The Figure 11-6  shows the process of inserting an element into an array. Assumi
 
 ## 11.4.1   Algorithm process
 
-The overall process of insertion sort is shown in the following figure.
+The overall process of insertion sort is shown in Figure 11-7.
 
 1. Initially, the first element of the array is sorted.
 2. The second element of the array is taken as `base`, and after inserting it into the correct position, **the first two elements of the array are sorted**.

en/docs/chapter_sorting/merge_sort.md

@@ -4,7 +4,7 @@ comments: true
 
 # 11.6   Merge sort
 
-<u>Merge sort</u> is a sorting algorithm based on the divide-and-conquer strategy, involving the "divide" and "merge" phases shown in the following figure.
+<u>Merge sort</u> is a sorting algorithm based on the divide-and-conquer strategy, involving the "divide" and "merge" phases shown in Figure 11-10.
 
 1. **Divide phase**: Recursively split the array from the midpoint, transforming the sorting problem of a long array into that of shorter arrays.
 2. **Merge phase**: Stop dividing when the length of the sub-array is 1, start merging, and continuously combine two shorter ordered arrays into one longer ordered array until the process is complete.
@@ -15,7 +15,7 @@ comments: true
 
 ## 11.6.1 &nbsp; Algorithm workflow
 
-As shown in the Figure 11-11 , the "divide phase" recursively splits the array from the midpoint into two sub-arrays from top to bottom.
+As shown in Figure 11-11, the "divide phase" recursively splits the array from the midpoint into two sub-arrays from top to bottom.
 
 1. Calculate the midpoint `mid`, recursively divide the left sub-array (interval `[left, mid]`) and the right sub-array (interval `[mid + 1, right]`).
 2. Continue with step `1.` recursively until the sub-array interval length is 1 to stop.

en/docs/chapter_sorting/quick_sort.md

@@ -388,7 +388,7 @@ After the pivot partitioning, the original array is divided into three parts: le
 
 ## 11.5.1   Algorithm process
 
-The overall process of quick sort is shown in the following figure.
+The overall process of quick sort is shown in Figure 11-9.
 
 1. First, perform a "pivot partitioning" on the original array to obtain the unsorted left and right sub-arrays.
 2. Then, recursively perform "pivot partitioning" on both the left and right sub-arrays.

en/docs/chapter_sorting/selection_sort.md

@@ -6,7 +6,7 @@ comments: true
 
 <u>Selection sort</u> works on a very simple principle: it starts a loop where each iteration selects the smallest element from the unsorted interval and moves it to the end of the sorted interval.
 
-Suppose the length of the array is $n$, the algorithm flow of selection sort is as shown below.
+Suppose the length of the array is $n$, the algorithm flow of selection sort is as shown in Figure 11-2.
 
 1. Initially, all elements are unsorted, i.e., the unsorted (index) interval is $[0, n-1]$.
 2. Select the smallest element in the interval $[0, n-1]$ and swap it with the element at index $0$. After this, the first element of the array is sorted.
@@ -324,7 +324,7 @@ In the code, we use $k$ to record the smallest element within the unsorted inter
 
 - **Time complexity of $O(n^2)$, non-adaptive sort**: There are $n - 1$ rounds in the outer loop, with the unsorted interval length starting at $n$ in the first round and decreasing to $2$ in the last round, i.e., the outer loops contain $n$, $n - 1$, $\dots$, $3$, $2$ inner loops respectively, summing up to $\frac{(n - 1)(n + 2)}{2}$.
 - **Space complexity of $O(1)$, in-place sort**: Uses constant extra space with pointers $i$ and $j$.
-- **Non-stable sort**: As shown in the Figure 11-3 , an element `nums[i]` may be swapped to the right of an equal element, causing their relative order to change.
+- **Non-stable sort**: As shown in Figure 11-3, an element `nums[i]` may be swapped to the right of an equal element, causing their relative order to change.
 
 ![Selection sort instability example](selection_sort.assets/selection_sort_instability.png){ class="animation-figure" }
 

en/docs/chapter_sorting/sorting_algorithm.md

@@ -6,7 +6,7 @@ comments: true
 
 <u>Sorting algorithms (sorting algorithm)</u> are used to arrange a set of data in a specific order. Sorting algorithms have a wide range of applications because ordered data can usually be searched, analyzed, and processed more efficiently.
 
-As shown in the following figure, the data types in sorting algorithms can be integers, floating point numbers, characters, or strings, etc. Sorting rules can be set according to needs, such as numerical size, character ASCII order, or custom rules.
+As shown in Figure 11-1, the data types in sorting algorithms can be integers, floating point numbers, characters, or strings, etc. Sorting rules can be set according to needs, such as numerical size, character ASCII order, or custom rules.
 
 ![Data types and comparator examples](sorting_algorithm.assets/sorting_examples.png){ class="animation-figure" }
 

en/docs/chapter_sorting/summary.md

@@ -14,7 +14,7 @@ comments: true
 - Counting sort is a special case of bucket sort, which sorts by counting the occurrences of each data point. Counting sort is suitable for large datasets with a limited range of data and requires that data can be converted to positive integers.
 - Radix sort sorts data by sorting digit by digit, requiring data to be represented as fixed-length numbers.
 - Overall, we hope to find a sorting algorithm that has high efficiency, stability, in-place operation, and positive adaptability. However, like other data structures and algorithms, no sorting algorithm can meet all these conditions simultaneously. In practical applications, we need to choose the appropriate sorting algorithm based on the characteristics of the data.
-- The following figure compares mainstream sorting algorithms in terms of efficiency, stability, in-place nature, and adaptability.
+- Figure 11-19 compares mainstream sorting algorithms in terms of efficiency, stability, in-place nature, and adaptability.
 
 ![Sorting Algorithm Comparison](summary.assets/sorting_algorithms_comparison.png){ class="animation-figure" }
 

en/docs/chapter_stack_and_queue/deque.md

@@ -18,14 +18,14 @@ The common operations in a double-ended queue are listed below, and the names of
 
 <div class="center-table" markdown>
 
-| Method Name   | Description                 | Time Complexity |
-| ------------- | --------------------------- | --------------- |
+| Method Name   | Description                | Time Complexity |
+| ------------- | -------------------------- | --------------- |
 | `pushFirst()` | Add an element to the head | $O(1)$          |
-| `pushLast()`  | Add an element to the tail  | $O(1)$          |
-| `popFirst()`  | Remove the first element    | $O(1)$          |
-| `popLast()`   | Remove the last element     | $O(1)$          |
-| `peekFirst()` | Access the first element    | $O(1)$          |
-| `peekLast()`  | Access the last element     | $O(1)$          |
+| `pushLast()`  | Add an element to the tail | $O(1)$          |
+| `popFirst()`  | Remove the first element   | $O(1)$          |
+| `popLast()`   | Remove the last element    | $O(1)$          |
+| `peekFirst()` | Access the first element   | $O(1)$          |
+| `peekLast()`  | Access the last element    | $O(1)$          |
 
 </div>
 
@@ -360,7 +360,7 @@ Recall from the previous section that we used a regular singly linked list to im
 
 For a double-ended queue, both the head and the tail can perform enqueue and dequeue operations. In other words, a double-ended queue needs to implement operations in the opposite direction as well. For this, we use a "doubly linked list" as the underlying data structure of the double-ended queue.
 
-As shown in the Figure 5-8 , we treat the head and tail nodes of the doubly linked list as the front and rear of the double-ended queue, respectively, and implement the functionality to add and remove nodes at both ends.
+As shown in Figure 5-8, we treat the head and tail nodes of the doubly linked list as the front and rear of the double-ended queue, respectively, and implement the functionality to add and remove nodes at both ends.
 
 === "LinkedListDeque"
     ![Implementing Double-Ended Queue with Doubly Linked List for Enqueue and Dequeue Operations](deque.assets/linkedlist_deque_step1.png){ class="animation-figure" }
@@ -2264,7 +2264,7 @@ The implementation code is as follows:
 
 ### 2. &nbsp; Implementation based on array
 
-As shown in the Figure 5-9 , similar to implementing a queue with an array, we can also use a circular array to implement a double-ended queue.
+As shown in Figure 5-9, similar to implementing a queue with an array, we can also use a circular array to implement a double-ended queue.
 
 === "ArrayDeque"
     ![Implementing Double-Ended Queue with Array for Enqueue and Dequeue Operations](deque.assets/array_deque_step1.png){ class="animation-figure" }

en/docs/chapter_stack_and_queue/queue.md

@@ -6,7 +6,7 @@ comments: true
 
 "Queue" is a linear data structure that follows the First-In-First-Out (FIFO) rule. As the name suggests, a queue simulates the phenomenon of lining up, where newcomers join the queue at the rear, and the person at the front leaves the queue first.
 
-As shown in the Figure 5-4 , we call the front of the queue the "head" and the back the "tail." The operation of adding elements to the rear of the queue is termed "enqueue," and the operation of removing elements from the front is termed "dequeue."
+As shown in Figure 5-4, we call the front of the queue the "head" and the back the "tail." The operation of adding elements to the rear of the queue is termed "enqueue," and the operation of removing elements from the front is termed "dequeue."
 
 ![Queue's first-in-first-out rule](queue.assets/queue_operations.png){ class="animation-figure" }
 
@@ -14,7 +14,7 @@ As shown in the Figure 5-4 , we call the front of the queue the "head" and the b
 
 ## 5.2.1   Common operations on queue
 
-The common operations on a queue are shown in the Table 5-2 . Note that method names may vary across different programming languages. Here, we use the same naming convention as that used for stacks.
+The common operations on a queue are shown in Table 5-2. Note that method names may vary across different programming languages. Here, we use the same naming convention as that used for stacks.
 
 <p align="center"> Table 5-2 &nbsp; Efficiency of queue operations </p>
 
@@ -335,7 +335,7 @@ To implement a queue, we need a data structure that allows adding elements at on
 
 ### 1. &nbsp; Implementation based on a linked list
 
-As shown in the Figure 5-5 , we can consider the "head node" and "tail node" of a linked list as the "front" and "rear" of the queue, respectively. It is stipulated that nodes can only be added at the rear and removed at the front.
+As shown in Figure 5-5, we can consider the "head node" and "tail node" of a linked list as the "front" and "rear" of the queue, respectively. It is stipulated that nodes can only be added at the rear and removed at the front.
 
 === "LinkedListQueue"
     ![Implementing Queue with Linked List for Enqueue and Dequeue Operations](queue.assets/linkedlist_queue_step1.png){ class="animation-figure" }
@@ -1363,7 +1363,7 @@ Deleting the first element in an array has a time complexity of $O(n)$, which wo
 
 We use a variable `front` to indicate the index of the front element and maintain a variable `size` to record the queue's length. Define `rear = front + size`, which points to the position immediately following the tail element.
 
-With this design, **the effective interval of elements in the array is `[front, rear - 1]`**. The implementation methods for various operations are shown in the Figure 5-6 .
+With this design, **the effective interval of elements in the array is `[front, rear - 1]`**. The implementation methods for various operations are shown in Figure 5-6.
 
 - Enqueue operation: Assign the input element to the `rear` index and increase `size` by 1.
 - Dequeue operation: Simply increase `front` by 1 and decrease `size` by 1.

en/docs/chapter_stack_and_queue/stack.md

@@ -8,7 +8,7 @@ A "Stack" is a linear data structure that follows the principle of Last-In-First
 
 We can compare a stack to a pile of plates on a table. To access the bottom plate, one must first remove the plates on top. By replacing the plates with various types of elements (such as integers, characters, objects, etc.), we obtain the data structure known as a stack.
 
-As shown in the Figure 5-1 , we refer to the top of the pile of elements as the "top of the stack" and the bottom as the "bottom of the stack." The operation of adding elements to the top of the stack is called "push," and the operation of removing the top element is called "pop."
+As shown in Figure 5-1, we refer to the top of the pile of elements as the "top of the stack" and the bottom as the "bottom of the stack." The operation of adding elements to the top of the stack is called "push," and the operation of removing the top element is called "pop."
 
 ![Stack's last-in-first-out rule](stack.assets/stack_operations.png){ class="animation-figure" }
 
@@ -16,7 +16,7 @@ As shown in the Figure 5-1 , we refer to the top of the pile of elements as the
 
 ## 5.1.1   Common operations on stack
 
-The common operations on a stack are shown in the Table 5-1 . The specific method names depend on the programming language used. Here, we use `push()`, `pop()`, and `peek()` as examples.
+The common operations on a stack are shown in Table 5-1. The specific method names depend on the programming language used. Here, we use `push()`, `pop()`, and `peek()` as examples.
 
 <p align="center"> Table 5-1 &nbsp; Efficiency of stack operations </p>
 
@@ -333,7 +333,7 @@ A stack follows the principle of Last-In-First-Out, which means we can only add
 
 When implementing a stack using a linked list, we can consider the head node of the list as the top of the stack and the tail node as the bottom of the stack.
 
-As shown in the Figure 5-2 , for the push operation, we simply insert elements at the head of the linked list. This method of node insertion is known as "head insertion." For the pop operation, we just need to remove the head node from the list.
+As shown in Figure 5-2, for the push operation, we simply insert elements at the head of the linked list. This method of node insertion is known as "head insertion." For the pop operation, we just need to remove the head node from the list.
 
 === "LinkedListStack"
     ![Implementing Stack with Linked List for Push and Pop Operations](stack.assets/linkedlist_stack_step1.png){ class="animation-figure" }
@@ -1206,7 +1206,7 @@ Below is an example code for implementing a stack based on a linked list:
 
 ### 2. &nbsp; Implementation based on an array
 
-When implementing a stack using an array, we can consider the end of the array as the top of the stack. As shown in the Figure 5-3 , push and pop operations correspond to adding and removing elements at the end of the array, respectively, both with a time complexity of $O(1)$.
+When implementing a stack using an array, we can consider the end of the array as the top of the stack. As shown in Figure 5-3, push and pop operations correspond to adding and removing elements at the end of the array, respectively, both with a time complexity of $O(1)$.
 
 === "ArrayStack"
     ![Implementing Stack with Array for Push and Pop Operations](stack.assets/array_stack_step1.png){ class="animation-figure" }

en/docs/chapter_tree/array_representation_of_tree.md

@@ -12,7 +12,7 @@ So, can we use an array to represent a binary tree? The answer is yes.
 
 Let's analyze a simple case first. Given a perfect binary tree, we store all nodes in an array according to the order of level-order traversal, where each node corresponds to a unique array index.
 
-Based on the characteristics of level-order traversal, we can deduce a "mapping formula" between the index of a parent node and its children: **If a node's index is $i$, then the index of its left child is $2i + 1$ and the right child is $2i + 2$**. The Figure 7-12  shows the mapping relationship between the indices of various nodes.
+Based on the characteristics of level-order traversal, we can deduce a "mapping formula" between the index of a parent node and its children: **If a node's index is $i$, then the index of its left child is $2i + 1$ and the right child is $2i + 2$**. Figure 7-12 shows the mapping relationship between the indices of various nodes.
 
 ![Array representation of a perfect binary tree](array_representation_of_tree.assets/array_representation_binary_tree.png){ class="animation-figure" }
 
@@ -24,13 +24,13 @@ Based on the characteristics of level-order traversal, we can deduce a "mapping
 
 Perfect binary trees are a special case; there are often many `None` values in the middle levels of a binary tree. Since the sequence of level-order traversal does not include these `None` values, we cannot solely rely on this sequence to deduce the number and distribution of `None` values. **This means that multiple binary tree structures can match the same level-order traversal sequence**.
 
-As shown in the Figure 7-13 , given a non-perfect binary tree, the above method of array representation fails.
+As shown in Figure 7-13, given a non-perfect binary tree, the above method of array representation fails.
 
 ![Level-order traversal sequence corresponds to multiple binary tree possibilities](array_representation_of_tree.assets/array_representation_without_empty.png){ class="animation-figure" }
 
 <p align="center"> Figure 7-13 &nbsp; Level-order traversal sequence corresponds to multiple binary tree possibilities </p>
 
-To solve this problem, **we can consider explicitly writing out all `None` values in the level-order traversal sequence**. As shown in the following figure, after this treatment, the level-order traversal sequence can uniquely represent a binary tree. Example code is as follows:
+To solve this problem, **we can consider explicitly writing out all `None` values in the level-order traversal sequence**. As shown in Figure 7-14, after this treatment, the level-order traversal sequence can uniquely represent a binary tree. Example code is as follows:
 
 === "Python"
 
@@ -146,7 +146,7 @@ To solve this problem, **we can consider explicitly writing out all `None` value
 
 It's worth noting that **complete binary trees are very suitable for array representation**. Recalling the definition of a complete binary tree, `None` appears only at the bottom level and towards the right, **meaning all `None` values definitely appear at the end of the level-order traversal sequence**.
 
-This means that when using an array to represent a complete binary tree, it's possible to omit storing all `None` values, which is very convenient. The Figure 7-15  gives an example.
+This means that when using an array to represent a complete binary tree, it's possible to omit storing all `None` values, which is very convenient. Figure 7-15 gives an example.
 
 ![Array representation of a complete binary tree](array_representation_of_tree.assets/array_representation_complete_binary_tree.png){ class="animation-figure" }
 

en/docs/chapter_tree/avl_tree.md

@@ -6,13 +6,13 @@ comments: true
 
 In the "Binary Search Tree" section, we mentioned that after multiple insertions and removals, a binary search tree might degrade to a linked list. In such cases, the time complexity of all operations degrades from $O(\log n)$ to $O(n)$.
 
-As shown in the Figure 7-24 , after two node removal operations, this binary search tree will degrade into a linked list.
+As shown in Figure 7-24, after two node removal operations, this binary search tree will degrade into a linked list.
 
 ![Degradation of an AVL tree after removing nodes](avl_tree.assets/avltree_degradation_from_removing_node.png){ class="animation-figure" }
 
 <p align="center"> Figure 7-24 &nbsp; Degradation of an AVL tree after removing nodes </p>
 
-For example, in the perfect binary tree shown in the Figure 7-25 , after inserting two nodes, the tree will lean heavily to the left, and the time complexity of search operations will also degrade.
+For example, in the perfect binary tree shown in Figure 7-25, after inserting two nodes, the tree will lean heavily to the left, and the time complexity of search operations will also degrade.
 
 ![Degradation of an AVL tree after inserting nodes](avl_tree.assets/avltree_degradation_from_inserting_node.png){ class="animation-figure" }
 
@@ -682,7 +682,7 @@ We call nodes with an absolute balance factor $> 1$ "unbalanced nodes". Dependin
 
 ### 1. &nbsp; Right rotation
 
-As shown in the Figure 7-26 , the first unbalanced node from the bottom up in the binary tree is "node 3". Focusing on the subtree with this unbalanced node as the root, denoted as `node`, and its left child as `child`, perform a "right rotation". After the right rotation, the subtree is balanced again while still maintaining the properties of a binary search tree.
+As shown in Figure 7-26, the first unbalanced node from the bottom up in the binary tree is "node 3". Focusing on the subtree with this unbalanced node as the root, denoted as `node`, and its left child as `child`, perform a "right rotation". After the right rotation, the subtree is balanced again while still maintaining the properties of a binary search tree.
 
 === "<1>"
     ![Steps of right rotation](avl_tree.assets/avltree_right_rotate_step1.png){ class="animation-figure" }
@@ -698,7 +698,7 @@ As shown in the Figure 7-26 , the first unbalanced node from the bottom up in th
 
 <p align="center"> Figure 7-26 &nbsp; Steps of right rotation </p>
 
-As shown in the Figure 7-27 , when the `child` node has a right child (denoted as `grand_child`), a step needs to be added in the right rotation: set `grand_child` as the left child of `node`.
+As shown in Figure 7-27, when the `child` node has a right child (denoted as `grand_child`), a step needs to be added in the right rotation: set `grand_child` as the left child of `node`.
 
 ![Right rotation with grand_child](avl_tree.assets/avltree_right_rotate_with_grandchild.png){ class="animation-figure" }
 
@@ -965,13 +965,13 @@ As shown in the Figure 7-27 , when the `child` node has a right child (denoted a
 
 ### 2. &nbsp; Left rotation
 
-Correspondingly, if considering the "mirror" of the above unbalanced binary tree, the "left rotation" operation shown in the Figure 7-28  needs to be performed.
+Correspondingly, if considering the "mirror" of the above unbalanced binary tree, the "left rotation" operation shown in Figure 7-28 needs to be performed.
 
 ![Left rotation operation](avl_tree.assets/avltree_left_rotate.png){ class="animation-figure" }
 
 <p align="center"> Figure 7-28 &nbsp; Left rotation operation </p>
 
-Similarly, as shown in the Figure 7-29 , when the `child` node has a left child (denoted as `grand_child`), a step needs to be added in the left rotation: set `grand_child` as the right child of `node`.
+Similarly, as shown in Figure 7-29, when the `child` node has a left child (denoted as `grand_child`), a step needs to be added in the left rotation: set `grand_child` as the right child of `node`.
 
 ![Left rotation with grand_child](avl_tree.assets/avltree_left_rotate_with_grandchild.png){ class="animation-figure" }
 
@@ -1238,7 +1238,7 @@ It can be observed that **the right and left rotation operations are logically s
 
 ### 3. &nbsp; Right-left rotation
 
-For the unbalanced node 3 shown in the Figure 7-30 , using either left or right rotation alone cannot restore balance to the subtree. In this case, a "left rotation" needs to be performed on `child` first, followed by a "right rotation" on `node`.
+For the unbalanced node 3 shown in Figure 7-30, using either left or right rotation alone cannot restore balance to the subtree. In this case, a "left rotation" needs to be performed on `child` first, followed by a "right rotation" on `node`.
 
 ![Right-left rotation](avl_tree.assets/avltree_left_right_rotate.png){ class="animation-figure" }
 
@@ -1246,7 +1246,7 @@ For the unbalanced node 3 shown in the Figure 7-30 , using either left or right
 
 ### 4. &nbsp; Left-right rotation
 
-As shown in the Figure 7-31 , for the mirror case of the above unbalanced binary tree, a "right rotation" needs to be performed on `child` first, followed by a "left rotation" on `node`.
+As shown in Figure 7-31, for the mirror case of the above unbalanced binary tree, a "right rotation" needs to be performed on `child` first, followed by a "left rotation" on `node`.
 
 ![Left-right rotation](avl_tree.assets/avltree_right_left_rotate.png){ class="animation-figure" }
 
@@ -1254,13 +1254,13 @@ As shown in the Figure 7-31 , for the mirror case of the above unbalanced binary
 
 ### 5. &nbsp; Choice of rotation
 
-The four kinds of imbalances shown in the Figure 7-32  correspond to the cases described above, respectively requiring right rotation, left-right rotation, right-left rotation, and left rotation.
+The four kinds of imbalances shown in Figure 7-32 correspond to the cases described above, respectively requiring right rotation, left-right rotation, right-left rotation, and left rotation.
 
 ![The four rotation cases of AVL tree](avl_tree.assets/avltree_rotation_cases.png){ class="animation-figure" }
 
 <p align="center"> Figure 7-32 &nbsp; The four rotation cases of AVL tree </p>
 
-As shown in the Table 7-3 , we determine which of the above cases an unbalanced node belongs to by judging the sign of the balance factor of the unbalanced node and its higher-side child's balance factor.
+As shown in Table 7-3, we determine which of the above cases an unbalanced node belongs to by judging the sign of the balance factor of the unbalanced node and its higher-side child's balance factor.
 
 <p align="center"> Table 7-3 &nbsp; Conditions for Choosing Among the Four Rotation Cases </p>
 

en/docs/chapter_tree/binary_search_tree.md

@@ -4,7 +4,7 @@ comments: true
 
 # 7.4   Binary search tree
 
-As shown in the Figure 7-16 , a "binary search tree" satisfies the following conditions.
+As shown in Figure 7-16, a "binary search tree" satisfies the following conditions.
 
 1. For the root node, the value of all nodes in the left subtree < the value of the root node < the value of all nodes in the right subtree.
 2. The left and right subtrees of any node are also binary search trees, i.e., they satisfy condition `1.` as well.
@@ -19,7 +19,7 @@ We encapsulate the binary search tree as a class `BinarySearchTree` and declare
 
 ### 1. &nbsp; Searching for a node
 
-Given a target node value `num`, one can search according to the properties of the binary search tree. As shown in the Figure 7-17 , we declare a node `cur` and start from the binary tree's root node `root`, looping to compare the size relationship between the node value `cur.val` and `num`.
+Given a target node value `num`, one can search according to the properties of the binary search tree. As shown in Figure 7-17, we declare a node `cur` and start from the binary tree's root node `root`, looping to compare the size relationship between the node value `cur.val` and `num`.
 
 - If `cur.val < num`, it means the target node is in `cur`'s right subtree, thus execute `cur = cur.right`.
 - If `cur.val > num`, it means the target node is in `cur`'s left subtree, thus execute `cur = cur.left`.
@@ -369,7 +369,7 @@ The search operation in a binary search tree works on the same principle as the
 
 ### 2. &nbsp; Inserting a node
 
-Given an element `num` to be inserted, to maintain the property of the binary search tree "left subtree < root node < right subtree," the insertion operation proceeds as shown in the Figure 7-18 .
+Given an element `num` to be inserted, to maintain the property of the binary search tree "left subtree < root node < right subtree," the insertion operation proceeds as shown in Figure 7-18.
 
 1. **Finding the insertion position**: Similar to the search operation, start from the root node and loop downwards according to the size relationship between the current node value and `num` until passing through the leaf node (traversing to `None`) then exit the loop.
 2. **Insert the node at that position**: Initialize the node `num` and place it where `None` was.
@@ -873,13 +873,13 @@ Similar to searching for a node, inserting a node uses $O(\log n)$ time.
 
 First, find the target node in the binary tree, then remove it. Similar to inserting a node, we need to ensure that after the removal operation is completed, the property of the binary search tree "left subtree < root node < right subtree" is still satisfied. Therefore, based on the number of child nodes of the target node, we divide it into 0, 1, and 2 cases, performing the corresponding node removal operations.
 
-As shown in the Figure 7-19 , when the degree of the node to be removed is $0$, it means the node is a leaf node, and it can be directly removed.
+As shown in Figure 7-19, when the degree of the node to be removed is $0$, it means the node is a leaf node, and it can be directly removed.
 
 ![Removing a node in a binary search tree (degree 0)](binary_search_tree.assets/bst_remove_case1.png){ class="animation-figure" }
 
 <p align="center"> Figure 7-19 &nbsp; Removing a node in a binary search tree (degree 0) </p>
 
-As shown in the Figure 7-20 , when the degree of the node to be removed is $1$, replacing the node to be removed with its child node is sufficient.
+As shown in Figure 7-20, when the degree of the node to be removed is $1$, replacing the node to be removed with its child node is sufficient.
 
 ![Removing a node in a binary search tree (degree 1)](binary_search_tree.assets/bst_remove_case2.png){ class="animation-figure" }
 
@@ -887,7 +887,7 @@ As shown in the Figure 7-20 , when the degree of the node to be removed is $1$,
 
 When the degree of the node to be removed is $2$, we cannot remove it directly, but need to use a node to replace it. To maintain the property of the binary search tree "left subtree < root node < right subtree," **this node can be either the smallest node of the right subtree or the largest node of the left subtree**.
 
-Assuming we choose the smallest node of the right subtree (the next node in in-order traversal), then the removal operation proceeds as shown in the Figure 7-21 .
+Assuming we choose the smallest node of the right subtree (the next node in in-order traversal), then the removal operation proceeds as shown in Figure 7-21.
 
 1. Find the next node in the "in-order traversal sequence" of the node to be removed, denoted as `tmp`.
 2. Replace the value of the node to be removed with `tmp`'s value, and recursively remove the node `tmp` in the tree.
@@ -1705,7 +1705,7 @@ The operation of removing a node also uses $O(\log n)$ time, where finding the n
 
 ### 4. &nbsp; In-order traversal is ordered
 
-As shown in the Figure 7-22 , the in-order traversal of a binary tree follows the "left $\rightarrow$ root $\rightarrow$ right" traversal order, and a binary search tree satisfies the size relationship "left child node < root node < right child node".
+As shown in Figure 7-22, the in-order traversal of a binary tree follows the "left $\rightarrow$ root $\rightarrow$ right" traversal order, and a binary search tree satisfies the size relationship "left child node < root node < right child node".
 
 This means that in-order traversal in a binary search tree always traverses the next smallest node first, thus deriving an important property: **The in-order traversal sequence of a binary search tree is ascending**.
 
@@ -1717,7 +1717,7 @@ Using the ascending property of in-order traversal, obtaining ordered data in a
 
 ## 7.4.2 &nbsp; Efficiency of binary search trees
 
-Given a set of data, we consider using an array or a binary search tree for storage. Observing the Table 7-2 , the operations on a binary search tree all have logarithmic time complexity, which is stable and efficient. Only in scenarios of high-frequency addition and low-frequency search and removal, arrays are more efficient than binary search trees.
+Given a set of data, we consider using an array or a binary search tree for storage. Observing Table 7-2, the operations on a binary search tree all have logarithmic time complexity, which is stable and efficient. Only in scenarios of high-frequency addition and low-frequency search and removal, arrays are more efficient than binary search trees.
 
 <p align="center"> Table 7-2 &nbsp; Efficiency comparison between arrays and search trees </p>
 
@@ -1733,7 +1733,7 @@ Given a set of data, we consider using an array or a binary search tree for stor
 
 In ideal conditions, the binary search tree is "balanced," thus any node can be found within $\log n$ loops.
 
-However, continuously inserting and removing nodes in a binary search tree may lead to the binary tree degenerating into a chain list as shown in the Figure 7-23 , at which point the time complexity of various operations also degrades to $O(n)$.
+However, continuously inserting and removing nodes in a binary search tree may lead to the binary tree degenerating into a chain list as shown in Figure 7-23, at which point the time complexity of various operations also degrades to $O(n)$.
 
 ![Degradation of a binary search tree](binary_search_tree.assets/bst_degradation.png){ class="animation-figure" }
 

en/docs/chapter_tree/binary_tree.md

@@ -204,7 +204,7 @@ A "binary tree" is a non-linear data structure that represents the ancestral and
 
 Each node has two references (pointers), pointing to the "left-child node" and "right-child node," respectively. This node is called the "parent node" of these two child nodes. When given a node of a binary tree, we call the tree formed by this node's left child and all nodes under it the "left subtree" of this node. Similarly, the "right subtree" can be defined.
 
-**In a binary tree, except for leaf nodes, all other nodes contain child nodes and non-empty subtrees.** As shown in the Figure 7-1 , if "Node 2" is considered as the parent node, then its left and right child nodes are "Node 4" and "Node 5," respectively. The left subtree is "the tree formed by Node 4 and all nodes under it," and the right subtree is "the tree formed by Node 5 and all nodes under it."
+**In a binary tree, except for leaf nodes, all other nodes contain child nodes and non-empty subtrees.** As shown in Figure 7-1, if "Node 2" is considered as the parent node, then its left and right child nodes are "Node 4" and "Node 5," respectively. The left subtree is "the tree formed by Node 4 and all nodes under it," and the right subtree is "the tree formed by Node 5 and all nodes under it."
 
 ![Parent Node, child Node, subtree](binary_tree.assets/binary_tree_definition.png){ class="animation-figure" }
 
@@ -212,7 +212,7 @@ Each node has two references (pointers), pointing to the "left-child node" and "
 
 ## 7.1.1   Common terminology of binary trees
 
-The commonly used terminology of binary trees is shown in the following figure.
+The commonly used terminology of binary trees is shown in Figure 7-2.
 
 - "Root node": The node at the top level of the binary tree, which has no parent node.
 - "Leaf node": A node with no children, both of its pointers point to `None`.
@@ -455,7 +455,7 @@ Similar to a linked list, initialize nodes first, then construct references (poi
 
 ### 2.   Inserting and removing nodes
 
-Similar to a linked list, inserting and removing nodes in a binary tree can be achieved by modifying pointers. The Figure 7-3  provides an example.
+Similar to a linked list, inserting and removing nodes in a binary tree can be achieved by modifying pointers. Figure 7-3 provides an example.
 
 ![Inserting and removing nodes in a binary tree](binary_tree.assets/binary_tree_add_remove.png){ class="animation-figure" }
 
@@ -625,7 +625,7 @@ Similar to a linked list, inserting and removing nodes in a binary tree can be a
 
 ### 1.   Perfect binary tree
 
-As shown in the Figure 7-4 , in a "perfect binary tree," all levels of nodes are fully filled. In a perfect binary tree, the degree of leaf nodes is $0$, and the degree of all other nodes is $2$; if the tree's height is $h$, then the total number of nodes is $2^{h+1} - 1$, showing a standard exponential relationship, reflecting the common phenomenon of cell division in nature.
+As shown in Figure 7-4, in a "perfect binary tree," all levels of nodes are fully filled. In a perfect binary tree, the degree of leaf nodes is $0$, and the degree of all other nodes is $2$; if the tree's height is $h$, then the total number of nodes is $2^{h+1} - 1$, showing a standard exponential relationship, reflecting the common phenomenon of cell division in nature.
 
 !!! tip
 
@@ -637,7 +637,7 @@ As shown in the Figure 7-4 , in a "perfect binary tree," all levels of nodes are
 
 ### 2.   Complete binary tree
 
-As shown in the Figure 7-5 , a "complete binary tree" has only the bottom level nodes not fully filled, and the bottom level nodes are filled as far left as possible.
+As shown in Figure 7-5, a "complete binary tree" has only the bottom level nodes not fully filled, and the bottom level nodes are filled as far left as possible.
 
 ![Complete binary tree](binary_tree.assets/complete_binary_tree.png){ class="animation-figure" }
 
@@ -645,7 +645,7 @@ As shown in the Figure 7-5 , a "complete binary tree" has only the bottom level
 
 ### 3.   Full binary tree
 
-As shown in the Figure 7-6 , a "full binary tree" has all nodes except leaf nodes having two children.
+As shown in Figure 7-6, a "full binary tree" has all nodes except leaf nodes having two children.
 
 ![Full binary tree](binary_tree.assets/full_binary_tree.png){ class="animation-figure" }
 
@@ -653,7 +653,7 @@ As shown in the Figure 7-6 , a "full binary tree" has all nodes except leaf node
 
 ### 4.   Balanced binary tree
 
-As shown in the Figure 7-7 , in a "balanced binary tree," the absolute difference in height between the left and right subtrees of any node does not exceed 1.
+As shown in Figure 7-7, in a "balanced binary tree," the absolute difference in height between the left and right subtrees of any node does not exceed 1.
 
 ![Balanced binary tree](binary_tree.assets/balanced_binary_tree.png){ class="animation-figure" }
 
@@ -661,7 +661,7 @@ As shown in the Figure 7-7 , in a "balanced binary tree," the absolute differenc
 
 ## 7.1.4   Degeneration of binary trees
 
-The Figure 7-8  shows the ideal and degenerate structures of binary trees. When every level of a binary tree is filled, it reaches the "perfect binary tree"; when all nodes are biased towards one side, the binary tree degenerates into a "linked list".
+Figure 7-8 shows the ideal and degenerate structures of binary trees. When every level of a binary tree is filled, it reaches the "perfect binary tree"; when all nodes are biased towards one side, the binary tree degenerates into a "linked list".
 
 - The perfect binary tree is the ideal situation, fully leveraging the "divide and conquer" advantage of binary trees.
 - A linked list is another extreme, where operations become linear, degrading the time complexity to $O(n)$.
@@ -670,7 +670,7 @@ The Figure 7-8  shows the ideal and degenerate structures of binary trees. When
 
 <p align="center"> Figure 7-8 &nbsp; The Best and Worst Structures of Binary Trees </p>
 
-As shown in the Table 7-1 , in the best and worst structures, the number of leaf nodes, total number of nodes, and height of the binary tree reach their maximum or minimum values.
+As shown in Table 7-1, in the best and worst structures, the number of leaf nodes, total number of nodes, and height of the binary tree reach their maximum or minimum values.
 
 <p align="center"> Table 7-1 &nbsp; The Best and Worst Structures of Binary Trees </p>
 

en/docs/chapter_tree/binary_tree_traversal.md

@@ -10,7 +10,7 @@ Common traversal methods for binary trees include level-order traversal, preorde
 
 ## 7.2.1   Level-order traversal
 
-As shown in the Figure 7-9 , "level-order traversal" traverses the binary tree from top to bottom, layer by layer, and accesses nodes in each layer in a left-to-right order.
+As shown in Figure 7-9, "level-order traversal" traverses the binary tree from top to bottom, layer by layer, and accesses nodes in each layer in a left-to-right order.
 
 Level-order traversal essentially belongs to "breadth-first traversal", also known as "breadth-first search (BFS)", which embodies a "circumferentially outward expanding" layer-by-layer traversal method.
 
@@ -380,7 +380,7 @@ Breadth-first traversal is usually implemented with the help of a "queue". The q
 
 Correspondingly, preorder, inorder, and postorder traversal all belong to "depth-first traversal", also known as "depth-first search (DFS)", which embodies a "proceed to the end first, then backtrack and continue" traversal method.
 
-The Figure 7-10  shows the working principle of performing a depth-first traversal on a binary tree. **Depth-first traversal is like walking around the perimeter of the entire binary tree**, encountering three positions at each node, corresponding to preorder traversal, inorder traversal, and postorder traversal.
+Figure 7-10 shows the working principle of performing a depth-first traversal on a binary tree. **Depth-first traversal is like walking around the perimeter of the entire binary tree**, encountering three positions at each node, corresponding to preorder traversal, inorder traversal, and postorder traversal.
 
 ![Preorder, inorder, and postorder traversal of a binary search tree](binary_tree_traversal.assets/binary_tree_dfs.png){ class="animation-figure" }
 
@@ -875,7 +875,7 @@ Depth-first search is usually implemented based on recursion:
 
     Depth-first search can also be implemented based on iteration, interested readers can study this on their own.
 
-The Figure 7-11  shows the recursive process of preorder traversal of a binary tree, which can be divided into two opposite parts: "recursion" and "return".
+Figure 7-11 shows the recursive process of preorder traversal of a binary tree, which can be divided into two opposite parts: "recursion" and "return".
 
 1. "Recursion" means starting a new method, the program accesses the next node in this process.
 2. "Return" means the function returns, indicating the current node has been fully accessed.