translation: Update binary_search_insertion.md and binary_search_edge.md (#1554)

* translation: Update binary_search_insertion.md

* fix: Clarify binary search insertion and multiple targets handling

* fix: Update binary search insertion title and nav link

* translation: update binary_search _edge.md

* Fix typos and clarify binary search edge cases

* fix: Revert binary search insertion title

* revert the chapter 10.2 title

* Update binary_search_insertion.md

* Update binary search insertion and binary search edge

---------

Co-authored-by: Yudong Jin <krahets@163.com>

en/docs/chapter_searching/binary_search_edge.md

@@ -6,9 +6,9 @@
 
     Given a sorted array `nums` of length $n$, which may contain duplicate elements, return the index of the leftmost element `target`. If the element is not present in the array, return $-1$.
 
-Recall the method of binary search for an insertion point, after the search is completed, $i$ points to the leftmost `target`, **thus searching for the insertion point is essentially searching for the index of the leftmost `target`**.
+Recalling the method of binary search for an insertion point, after the search is completed, the index $i$ will point to the leftmost occurrence of `target`. Therefore, **searching for the insertion point is essentially the same as finding the index of the leftmost `target`**.
 
-Consider implementing the search for the left boundary using the function for finding an insertion point. Note that the array might not contain `target`, which could lead to the following two results:
+We can use the function for finding an insertion point to find the left boundary of `target`. Note that the array might not contain `target`, which could lead to the following two results:
 
 - The index $i$ of the insertion point is out of bounds.
 - The element `nums[i]` is not equal to `target`.
@@ -21,27 +21,27 @@ In these cases, simply return $-1$. The code is as follows:
 
 ## Find the right boundary
 
-So how do we find the rightmost `target`? The most straightforward way is to modify the code, replacing the pointer contraction operation in the case of `nums[m] == target`. The code is omitted here, but interested readers can implement it on their own.
+How do we find the rightmost occurrence of `target`? The most straightforward way is to modify the traditional binary search logic by changing how we adjust the search boundaries in the case of `nums[m] == target`. The code is omitted here. If you are interested, try to implement the code on your own.
 
-Below we introduce two more cunning methods.
+Below we are going to introduce two more ingenious methods.
 
-### Reusing the search for the left boundary
+### Reuse the left boundary search
 
-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`**.
+To find the rightmost occurrence of `target`, we can reuse the function used for locating the leftmost `target`. Specifically, we transform the search for the rightmost target into a search for the leftmost target + 1.
 
-As shown in the figure below, 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 the figure below, after the search is complete, pointer $i$ will point to the leftmost `target + 1` (if exists), while pointer $j$ will point to the rightmost occurrence of `target`. Therefore, returning $j$ will give us the right boundary.
 
 ![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)
 
-Please note, the insertion point returned is $i$, therefore, it should be subtracted by $1$ to obtain $j$:
+Note that the insertion point returned is $i$, therefore, it should be subtracted by $1$ to obtain $j$:
 
 ```src
 [file]{binary_search_edge}-[class]{}-[func]{binary_search_right_edge}
 ```
 
-### Transforming into an element search
+### Transform into an element search
 
-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.
+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 below, we can construct an element that does not exist in the array, to search for the left and right boundaries.
 
@@ -50,7 +50,7 @@ Thus, as shown in the figure below, we can construct an element that does not ex
 
 ![Transforming the search for boundaries into the search for an element](binary_search_edge.assets/binary_search_edge_by_element.png)
 
-The code is omitted here, but two points are worth noting.
+The code is omitted here, but here are two important points to note about this approach.
 
-- The given array does not contain decimals, meaning we do not need to worry about how to handle equal situations.
-- Since this method introduces decimals, the variable `target` in the function needs to be changed to a floating point type (no change needed in Python).
+- The given array `nums` does not contain decimal, so handling equal cases is not a concern.
+- However, introducing decimals in this approach requires modifying the `target` variable to a floating-point type (no change needed in Python).

en/docs/chapter_searching/binary_search_insertion.md

@@ -6,21 +6,21 @@ 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 below.
+    Given a sorted array `nums` of length $n$ with unique elements and an element `target`, insert `target` into `nums` while maintaining its sorted order. If `target` already exists in the array, insert it to the left of the existing element. Return the index of `target` in the array after insertion. See the example shown in the figure below.
 
 ![Example data for binary search insertion point](binary_search_insertion.assets/binary_search_insertion_example.png)
 
 If you want to reuse the binary search code from the previous section, you need to answer the following two questions.
 
-**Question one**: When the array contains `target`, is the insertion point index the index of that element?
+**Question one**: If the array already contains `target`, would the insertion point be the index of existing element?
 
-The requirement to insert `target` to the left of equal elements means that the newly inserted `target` replaces the original `target` position. Thus, **when the array contains `target`, the insertion point index is the index of that `target`**.
+The requirement to insert `target` to the left of equal elements means that the newly inserted `target` will replace the original `target` position. In other words, **when the array contains `target`, the insertion point is indeed the index of that `target`**.
 
-**Question two**: When the array does not contain `target`, what is the index of the insertion point?
+**Question two**: When the array does not contain `target`, at which index would it be inserted?
 
-Further consider the binary search process: when `nums[m] < target`, pointer $i$ moves, meaning that pointer $i$ is approaching an element greater than or equal to `target`. Similarly, pointer $j$ is always approaching an element less than or equal to `target`.
+Let's further consider the binary search process: when `nums[m] < target`, pointer $i$ moves, meaning that pointer $i$ is approaching an element greater than or equal to `target`. Similarly, pointer $j$ is always approaching an element less than or equal to `target`.
 
-Therefore, at the end of the binary, it is certain that: $i$ points to the first element greater than `target`, and $j$ points to the first element less than `target`. **It is easy to see that when the array does not contain `target`, the insertion index is $i$**. The code is as follows:
+Therefore, at the end of the binary, it is certain that: $i$ points to the first element greater than `target`, and $j$ points to the first element less than `target`. **It is easy to see that when the array does not contain `target`, the insertion point is $i$**. The code is as follows:
 
 ```src
 [file]{binary_search_insertion}-[class]{}-[func]{binary_search_insertion_simple}
@@ -32,21 +32,21 @@ Therefore, at the end of the binary, it is certain that: $i$ points to the first
 
     Based on the previous question, assume the array may contain duplicate elements, all else remains the same.
 
-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**.
+When there are multiple occurrences of `target` in the array, a regular binary search can only return the index of one occurrence of `target`, **and it cannot determine how many occurrences of `target` are to the left and right of that position**.
 
-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 below.
+The problem requires inserting the target element at the leftmost position, **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 below.
 
-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.
+1. Perform a binary search to find any index of `target`, say $k$.
+2. Starting from index $k$, conduct a linear search to the left until the leftmost occurrence of `target` is found, then return this index.
 
 ![Linear search for the insertion point of duplicate elements](binary_search_insertion.assets/binary_search_insertion_naive.png)
 
 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 below, 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 the figure below, the overall process remains the same. In each round, we first calculate the middle index $m$, then compare the value of `target` with `nums[m]`, leading to 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`**.
+- When `nums[m] < target` or `nums[m] > target`, it means `target` has not been found yet, thus use the normal binary search to narrow the search range, **bringing pointers $i$ and $j$ closer to `target`**.
+- When `nums[m] == target`, it indicates that the elements less than `target` are in the range $[i, m - 1]$, therefore use $j = m - 1$ to narrow the range, **thus bringing pointer $j$ closer to the elements less than `target`**.
 
 After the loop, $i$ points to the leftmost `target`, and $j$ points to the first element less than `target`, **therefore index $i$ is the insertion point**.
 
@@ -74,9 +74,9 @@ After the loop, $i$ points to the leftmost `target`, and $j$ points to the first
 === "<8>"
     ![binary_search_insertion_step8](binary_search_insertion.assets/binary_search_insertion_step8.png)
 
-Observe the code, the operations of the branch `nums[m] > target` and `nums[m] == target` are the same, so the two can be combined.
+Observe the following code. The operations in the branches `nums[m] > target` and `nums[m] == target` are the same, so these two branches can be merged.
 
-Even so, we can still keep the conditions expanded, as their logic is clearer and more readable.
+Even so, we can still keep the conditions expanded, as it makes the logic clearer and improves readability.
 
 ```src
 [file]{binary_search_insertion}-[class]{}-[func]{binary_search_insertion}
@@ -84,8 +84,8 @@ Even so, we can still keep the conditions expanded, as their logic is clearer an
 
 !!! tip
 
-    The code in this section uses "closed intervals". Readers interested can implement the "left-closed right-open" method themselves.
+    The code in this section uses "closed interval". If you are interested in "left-closed, right-open", try to implement the code on your own.
 
-In summary, binary search is merely about setting search targets for pointers $i$ and $j$, which might be a specific element (like `target`) or a range of elements (like elements less than `target`).
+In summary, binary search essentially involves setting search targets for pointers $i$ and $j$. These targets could be a specific element (like `target`) or a range of elements (such as those smaller than `target`).
 
 In the continuous loop of binary search, pointers $i$ and $j$ gradually approach the predefined target. Ultimately, they either find the answer or stop after crossing the boundary.