diff --git a/codes/cpp/chapter_dynamic_programming/CMakeLists.txt b/codes/cpp/chapter_dynamic_programming/CMakeLists.txt
index 908e55db8..ed185458a 100644
--- a/codes/cpp/chapter_dynamic_programming/CMakeLists.txt
+++ b/codes/cpp/chapter_dynamic_programming/CMakeLists.txt
@@ -6,4 +6,5 @@ add_executable(min_cost_climbing_stairs_dp min_cost_climbing_stairs_dp.cpp)
 add_executable(min_path_sum min_path_sum.cpp)
 add_executable(unbounded_knapsack unbounded_knapsack.cpp)
 add_executable(coin_change coin_change.cpp)
-add_executable(coin_change_ii coin_change_ii.cpp)
\ No newline at end of file
+add_executable(coin_change_ii coin_change_ii.cpp)
+add_executable(edit_distance edit_distance.cpp)
\ No newline at end of file
diff --git a/codes/cpp/chapter_dynamic_programming/edit_distance.cpp b/codes/cpp/chapter_dynamic_programming/edit_distance.cpp
new file mode 100644
index 000000000..647d7de1e
--- /dev/null
+++ b/codes/cpp/chapter_dynamic_programming/edit_distance.cpp
@@ -0,0 +1,136 @@
+/**
+ * File: edit_distance.cpp
+ * Created Time: 2023-07-13
+ * Author: Krahets (krahets@163.com)
+ */
+
+#include "../utils/common.hpp"
+
+/* 编辑距离：暴力搜索 */
+int editDistanceDFS(string s, string t, int i, int j) {
+    // 若 s 和 t 都为空，则返回 0
+    if (i == 0 && j == 0)
+        return 0;
+    // 若 s 为空，则返回 t 长度
+    if (i == 0)
+        return j;
+    // 若 t 为空，则返回 s 长度
+    if (j == 0)
+        return i;
+    // 若两字符相等，则直接跳过此两字符
+    if (s[i - 1] == t[j - 1])
+        return editDistanceDFS(s, t, i - 1, j - 1);
+    // 最少编辑步数 = 插入、删除、替换这三种操作的最少编辑步数 + 1
+    int insert = editDistanceDFS(s, t, i, j - 1);
+    int del = editDistanceDFS(s, t, i - 1, j);
+    int replace = editDistanceDFS(s, t, i - 1, j - 1);
+    // 返回最少编辑步数
+    return min(min(insert, del), replace) + 1;
+}
+
+/* 编辑距离：记忆化搜索 */
+int editDistanceDFSMem(string s, string t, vector<vector<int>> &mem, int i, int j) {
+    // 若 s 和 t 都为空，则返回 0
+    if (i == 0 && j == 0)
+        return 0;
+    // 若 s 为空，则返回 t 长度
+    if (i == 0)
+        return j;
+    // 若 t 为空，则返回 s 长度
+    if (j == 0)
+        return i;
+    // 若已有记录，则直接返回之
+    if (mem[i][j] != -1)
+        return mem[i][j];
+    // 若两字符相等，则直接跳过此两字符
+    if (s[i - 1] == t[j - 1])
+        return editDistanceDFSMem(s, t, mem, i - 1, j - 1);
+    // 最少编辑步数 = 插入、删除、替换这三种操作的最少编辑步数 + 1
+    int insert = editDistanceDFSMem(s, t, mem, i, j - 1);
+    int del = editDistanceDFSMem(s, t, mem, i - 1, j);
+    int replace = editDistanceDFSMem(s, t, mem, i - 1, j - 1);
+    // 记录并返回最少编辑步数
+    mem[i][j] = min(min(insert, del), replace) + 1;
+    return mem[i][j];
+}
+
+/* 编辑距离：动态规划 */
+int editDistanceDP(string s, string t) {
+    int n = s.length(), m = t.length();
+    vector<vector<int>> dp(n + 1, vector<int>(m + 1, 0));
+    // 状态转移：首行首列
+    for (int i = 1; i <= n; i++) {
+        dp[i][0] = i;
+    }
+    for (int j = 1; j <= m; j++) {
+        dp[0][j] = j;
+    }
+    // 状态转移：其余行列
+    for (int i = 1; i <= n; i++) {
+        for (int j = 1; j <= m; j++) {
+            if (s[i - 1] == t[j - 1]) {
+                // 若两字符相等，则直接跳过此两字符
+                dp[i][j] = dp[i - 1][j - 1];
+            } else {
+                // 最少编辑步数 = 插入、删除、替换这三种操作的最少编辑步数 + 1
+                dp[i][j] = min(min(dp[i][j - 1], dp[i - 1][j]), dp[i - 1][j - 1]) + 1;
+            }
+        }
+    }
+    return dp[n][m];
+}
+
+/* 编辑距离：状态压缩后的动态规划 */
+int editDistanceDPComp(string s, string t) {
+    int n = s.length(), m = t.length();
+    vector<int> dp(m + 1, 0);
+    // 状态转移：首行
+    for (int j = 1; j <= m; j++) {
+        dp[j] = j;
+    }
+    // 状态转移：其余行
+    for (int i = 1; i <= n; i++) {
+        // 状态转移：首列
+        int leftup = dp[0]; // 暂存 dp[i-1, j-1]
+        dp[0] = i;
+        // 状态转移：其余列
+        for (int j = 1; j <= m; j++) {
+            int temp = dp[j];
+            if (s[i - 1] == t[j - 1]) {
+                // 若两字符相等，则直接跳过此两字符
+                dp[j] = leftup;
+            } else {
+                // 最少编辑步数 = 插入、删除、替换这三种操作的最少编辑步数 + 1
+                dp[j] = min(min(dp[j - 1], dp[j]), leftup) + 1;
+            }
+            leftup = temp; // 更新为下一轮的 dp[i-1, j-1]
+        }
+    }
+    return dp[m];
+}
+
+/* Driver Code */
+int main() {
+    string s = "bag";
+    string t = "pack";
+    int n = s.length(), m = t.length();
+
+    // 暴力搜索
+    int res = editDistanceDFS(s, t, n, m);
+    cout << "将 " << s << " 更改为 " << t << " 最少需要编辑 " << res << " 步\n";
+
+    // 记忆化搜索
+    vector<vector<int>> mem(n + 1, vector<int>(m + 1, -1));
+    res = editDistanceDFSMem(s, t, mem, n, m);
+    cout << "将 " << s << " 更改为 " << t << " 最少需要编辑 " << res << " 步\n";
+
+    // 动态规划
+    res = editDistanceDP(s, t);
+    cout << "将 " << s << " 更改为 " << t << " 最少需要编辑 " << res << " 步\n";
+
+    // 状态压缩后的动态规划
+    res = editDistanceDPComp(s, t);
+    cout << "将 " << s << " 更改为 " << t << " 最少需要编辑 " << res << " 步\n";
+
+    return 0;
+}
diff --git a/codes/java/chapter_dynamic_programming/edit_distance.java b/codes/java/chapter_dynamic_programming/edit_distance.java
new file mode 100644
index 000000000..89954d90b
--- /dev/null
+++ b/codes/java/chapter_dynamic_programming/edit_distance.java
@@ -0,0 +1,139 @@
+/**
+ * File: edit_distance.java
+ * Created Time: 2023-07-13
+ * Author: Krahets (krahets@163.com)
+ */
+
+package chapter_dynamic_programming;
+
+import java.util.Arrays;
+
+public class edit_distance {
+    /* 编辑距离：暴力搜索 */
+    static int editDistanceDFS(String s, String t, int i, int j) {
+        // 若 s 和 t 都为空，则返回 0
+        if (i == 0 && j == 0)
+            return 0;
+        // 若 s 为空，则返回 t 长度
+        if (i == 0)
+            return j;
+        // 若 t 为空，则返回 s 长度
+        if (j == 0)
+            return i;
+        // 若两字符相等，则直接跳过此两字符
+        if (s.charAt(i - 1) == t.charAt(j - 1))
+            return editDistanceDFS(s, t, i - 1, j - 1);
+        // 最少编辑步数 = 插入、删除、替换这三种操作的最少编辑步数 + 1
+        int insert = editDistanceDFS(s, t, i, j - 1);
+        int delete = editDistanceDFS(s, t, i - 1, j);
+        int replace = editDistanceDFS(s, t, i - 1, j - 1);
+        // 返回最少编辑步数
+        return Math.min(Math.min(insert, delete), replace) + 1;
+    }
+
+    /* 编辑距离：记忆化搜索 */
+    static int editDistanceDFSMem(String s, String t, int[][] mem, int i, int j) {
+        // 若 s 和 t 都为空，则返回 0
+        if (i == 0 && j == 0)
+            return 0;
+        // 若 s 为空，则返回 t 长度
+        if (i == 0)
+            return j;
+        // 若 t 为空，则返回 s 长度
+        if (j == 0)
+            return i;
+        // 若已有记录，则直接返回之
+        if (mem[i][j] != -1)
+            return mem[i][j];
+        // 若两字符相等，则直接跳过此两字符
+        if (s.charAt(i - 1) == t.charAt(j - 1))
+            return editDistanceDFSMem(s, t, mem, i - 1, j - 1);
+        // 最少编辑步数 = 插入、删除、替换这三种操作的最少编辑步数 + 1
+        int insert = editDistanceDFSMem(s, t, mem, i, j - 1);
+        int delete = editDistanceDFSMem(s, t, mem, i - 1, j);
+        int replace = editDistanceDFSMem(s, t, mem, i - 1, j - 1);
+        // 记录并返回最少编辑步数
+        mem[i][j] = Math.min(Math.min(insert, delete), replace) + 1;
+        return mem[i][j];
+    }
+
+    /* 编辑距离：动态规划 */
+    static int editDistanceDP(String s, String t) {
+        int n = s.length(), m = t.length();
+        int[][] dp = new int[n + 1][m + 1];
+        // 状态转移：首行首列
+        for (int i = 1; i <= n; i++) {
+            dp[i][0] = i;
+        }
+        for (int j = 1; j <= m; j++) {
+            dp[0][j] = j;
+        }
+        // 状态转移：其余行列
+        for (int i = 1; i <= n; i++) {
+            for (int j = 1; j <= m; j++) {
+                if (s.charAt(i - 1) == t.charAt(j - 1)) {
+                    // 若两字符相等，则直接跳过此两字符
+                    dp[i][j] = dp[i - 1][j - 1];
+                } else {
+                    // 最少编辑步数 = 插入、删除、替换这三种操作的最少编辑步数 + 1
+                    dp[i][j] = Math.min(Math.min(dp[i][j - 1], dp[i - 1][j]), dp[i - 1][j - 1]) + 1;
+                }
+            }
+        }
+        return dp[n][m];
+    }
+
+    /* 编辑距离：状态压缩后的动态规划 */
+    static int editDistanceDPComp(String s, String t) {
+        int n = s.length(), m = t.length();
+        int[] dp = new int[m + 1];
+        // 状态转移：首行
+        for (int j = 1; j <= m; j++) {
+            dp[j] = j;
+        }
+        // 状态转移：其余行
+        for (int i = 1; i <= n; i++) {
+            // 状态转移：首列
+            int leftup = dp[0]; // 暂存 dp[i-1, j-1]
+            dp[0] = i;
+            // 状态转移：其余列
+            for (int j = 1; j <= m; j++) {
+                int temp = dp[j];
+                if (s.charAt(i - 1) == t.charAt(j - 1)) {
+                    // 若两字符相等，则直接跳过此两字符
+                    dp[j] = leftup;
+                } else {
+                    // 最少编辑步数 = 插入、删除、替换这三种操作的最少编辑步数 + 1
+                    dp[j] = Math.min(Math.min(dp[j - 1], dp[j]), leftup) + 1;
+                }
+                leftup = temp; // 更新为下一轮的 dp[i-1, j-1]
+            }
+        }
+        return dp[m];
+    }
+
+    public static void main(String[] args) {
+        String s = "bag";
+        String t = "pack";
+        int n = s.length(), m = t.length();
+
+        // 暴力搜索
+        int res = editDistanceDFS(s, t, n, m);
+        System.out.println("将 " + s + " 更改为 " + t + " 最少需要编辑 " + res + " 步");
+
+        // 记忆化搜索
+        int[][] mem = new int[n + 1][m + 1];
+        for (int[] row : mem)
+            Arrays.fill(row, -1);
+        res = editDistanceDFSMem(s, t, mem, n, m);
+        System.out.println("将 " + s + " 更改为 " + t + " 最少需要编辑 " + res + " 步");
+
+        // 动态规划
+        res = editDistanceDP(s, t);
+        System.out.println("将 " + s + " 更改为 " + t + " 最少需要编辑 " + res + " 步");
+
+        // 状态压缩后的动态规划
+        res = editDistanceDPComp(s, t);
+        System.out.println("将 " + s + " 更改为 " + t + " 最少需要编辑 " + res + " 步");
+    }
+}
diff --git a/codes/python/chapter_dynamic_programming/edit_distance.py b/codes/python/chapter_dynamic_programming/edit_distance.py
new file mode 100644
index 000000000..f13923c91
--- /dev/null
+++ b/codes/python/chapter_dynamic_programming/edit_distance.py
@@ -0,0 +1,123 @@
+"""
+File: edit_distancde.py
+Created Time: 2023-07-04
+Author: Krahets (krahets@163.com)
+"""
+
+
+def edit_distance_dfs(s: str, t: str, i: int, j: int) -> int:
+    """编辑距离：暴力搜索"""
+    # 若 s 和 t 都为空，则返回 0
+    if i == 0 and j == 0:
+        return 0
+    # 若 s 为空，则返回 t 长度
+    if i == 0:
+        return j
+    # 若 t 为空，则返回 s 长度
+    if j == 0:
+        return i
+    # 若两字符相等，则直接跳过此两字符
+    if s[i - 1] == t[j - 1]:
+        return edit_distance_dfs(s, t, i - 1, j - 1)
+    # 最少编辑步数 = 插入、删除、替换这三种操作的最少编辑步数 + 1
+    insert = edit_distance_dfs(s, t, i, j - 1)
+    delete = edit_distance_dfs(s, t, i - 1, j)
+    replace = edit_distance_dfs(s, t, i - 1, j - 1)
+    # 返回最少编辑步数
+    return min(insert, delete, replace) + 1
+
+
+def edit_distance_dfs_mem(s: str, t: str, mem: list[list[int]], i: int, j: int) -> int:
+    """编辑距离：记忆化搜索"""
+    # 若 s 和 t 都为空，则返回 0
+    if i == 0 and j == 0:
+        return 0
+    # 若 s 为空，则返回 t 长度
+    if i == 0:
+        return j
+    # 若 t 为空，则返回 s 长度
+    if j == 0:
+        return i
+    # 若已有记录，则直接返回之
+    if mem[i][j] != -1:
+        return mem[i][j]
+    # 若两字符相等，则直接跳过此两字符
+    if s[i - 1] == t[j - 1]:
+        return edit_distance_dfs_mem(s, t, mem, i - 1, j - 1)
+    # 最少编辑步数 = 插入、删除、替换这三种操作的最少编辑步数 + 1
+    insert = edit_distance_dfs_mem(s, t, mem, i, j - 1)
+    delete = edit_distance_dfs_mem(s, t, mem, i - 1, j)
+    replace = edit_distance_dfs_mem(s, t, mem, i - 1, j - 1)
+    # 记录并返回最少编辑步数
+    mem[i][j] = min(insert, delete, replace) + 1
+    return mem[i][j]
+
+
+def edit_distance_dp(s: str, t: str) -> int:
+    """编辑距离：动态规划"""
+    n, m = len(s), len(t)
+    dp = [[0] * (m + 1) for _ in range(n + 1)]
+    # 状态转移：首行首列
+    for i in range(1, n + 1):
+        dp[i][0] = i
+    for j in range(1, m + 1):
+        dp[0][j] = j
+    # 状态转移：其余行列
+    for i in range(1, n + 1):
+        for j in range(1, m + 1):
+            if s[i - 1] == t[j - 1]:
+                # 若两字符相等，则直接跳过此两字符
+                dp[i][j] = dp[i - 1][j - 1]
+            else:
+                # 最少编辑步数 = 插入、删除、替换这三种操作的最少编辑步数 + 1
+                dp[i][j] = min(dp[i][j - 1], dp[i - 1][j], dp[i - 1][j - 1]) + 1
+    return dp[n][m]
+
+
+def edit_distance_dp_comp(s: str, t: str) -> int:
+    """编辑距离：状态压缩后的动态规划"""
+    n, m = len(s), len(t)
+    dp = [0] * (m + 1)
+    # 状态转移：首行
+    for j in range(1, m + 1):
+        dp[j] = j
+    # 状态转移：其余行
+    for i in range(1, n + 1):
+        # 状态转移：首列
+        leftup = dp[0]  # 暂存 dp[i-1, j-1]
+        dp[0] += 1
+        # 状态转移：其余列
+        for j in range(1, m + 1):
+            temp = dp[j]
+            if s[i - 1] == t[j - 1]:
+                # 若两字符相等，则直接跳过此两字符
+                dp[j] = leftup
+            else:
+                # 最少编辑步数 = 插入、删除、替换这三种操作的最少编辑步数 + 1
+                dp[j] = min(dp[j - 1], dp[j], leftup) + 1
+            leftup = temp  # 更新为下一轮的 dp[i-1, j-1]
+    return dp[m]
+
+
+"""Driver Code"""
+if __name__ == "__main__":
+    s = "bag"
+    t = "pack"
+    n, m = len(s), len(t)
+
+    # 暴力搜索
+    res = edit_distance_dfs(s, t, n, m)
+    print(f"将 {s} 更改为 {t} 最少需要编辑 {res} 步")
+
+    # 记忆化搜索
+    mem = [[-1] * (m + 1) for _ in range(n + 1)]
+    res = edit_distance_dfs_mem(s, t, mem, n, m)
+    print(f"将 {s} 更改为 {t} 最少需要编辑 {res} 步")
+
+    # 动态规划
+    res = edit_distance_dp(s, t)
+    print(f"将 {s} 更改为 {t} 最少需要编辑 {res} 步")
+
+    # 状态压缩后的动态规划
+    res = edit_distance_dp_comp(s, t)
+    print(f"将 {s} 更改为 {t} 最少需要编辑 {res} 步")
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+++ b/docs/chapter_dynamic_programming/edit_distance_problem.md
@@ -0,0 +1,245 @@
+# 编辑距离问题
+
+编辑距离，也被称为 Levenshtein 距离，是两个字符串之间互相转换的最小修改次数，通常用于在信息检索和自然语言处理中度量两个序列的相似度。
+
+!!! question
+
+    输入两个字符串 $s$ 和 $t$ ，返回将 $s$ 转换为 $t$ 所需的最少编辑步数。
+    
+    你可以在一个字符串中进行三种编辑操作：插入一个字符、删除一个字符、替换字符为任意一个字符。
+
+如下图所示，将 `kitten` 转换为 `sitting` 需要编辑 3 步，包括 2 次替换操作与 1 次添加操作；将 `hello` 转换为 `algo` 需要 3 步，包括 2 次替换操作和 1 次删除操作。
+
+![编辑距离的示例数据](edit_distance_problem.assets/edit_distance_example.png)
+
+**编辑距离问题可以很自然地用决策树模型来解释**。字符串对应树节点，一轮决策（一次编辑操作）对应树的一条边。
+
+如下图所示，在不限制操作的情况下，每个节点都可以派生出许多条边，每条边对应一种操作。实际上，从 `hello` 转换到 `algo` 有许多种可能的路径，下图展示的是最短路径。从决策树的角度看，本题目标是求解节点 `hello` 和节点 `algo` 之间的最短路径。
+
+![基于决策树模型表示编辑距离问题](edit_distance_problem.assets/edit_distance_decision_tree.png)
+
+**第一步：思考每轮的决策，定义状态，从而得到 $dp$ 表**
+
+每一轮的决策是对字符串 $s$ 进行一次编辑操作。
+
+我们希望在编辑操作的过程中，问题的规模逐渐缩小，这样才能构建子问题。设字符串 $s$ 和 $t$ 的长度分别为 $n$ 和 $m$ ，我们先考虑两字符串尾部的字符 $s[n-1]$ 和 $t[m-1]$ ：
+
+- 若 $s[n-1]$ 和 $t[m-1]$ 相同，我们可以直接跳过它们，接下来考虑 $s[n-2]$ 和 $t[m-2]$ ;
+- 若 $s[n-1]$ 和 $t[m-1]$ 不同，我们需要对 $s$ 进行一次编辑（插入、删除、替换），使得两字符串尾部的字符相同，从而可以跳过它们，考虑规模更小的问题；
+
+也就是说，我们在字符串 $s$ 中进行的每一轮决策（编辑操作），都会使得 $s$ 和 $t$ 中剩余的待匹配字符发生变化。因此，状态定义为当前在 $s$ , $t$ 中考虑的第 $i$ , $j$ 个字符，记为 $[i, j]$ 。
+
+状态 $[i, j]$ 对应的子问题：**将 $s$ 的前 $i$ 个字符更改为 $t$ 的前 $j$ 个字符所需的最少编辑步数**。
+
+至此得到一个尺寸为 $(i+1) \times (j+1)$ 的二维 $dp$ 表。
+
+**第二步：找出最优子结构，进而推导出状态转移方程**
+
+考虑子问题 $dp[i, j]$ ，其对应的两个字符串的尾部字符为 $s[i-1]$ 和 $t[j-1]$ ，可根据不同编辑操作分为三种情况：
+
+1. 在 $s$ 尾部添加 $t[j-1]$ ，则剩余子问题 $dp[i, j-1]$ ；
+2. 删除 $s[i-1]$ ，则剩余子问题 $dp[i-1, j]$ ；
+3. 将 $s[i-1]$ 替换为 $t[j-1]$ ，则剩余子问题 $dp[i-1, j-1]$ ；
+
+![编辑距离的状态转移](edit_distance_problem.assets/edit_distance_state_transfer.png)
+
+根据以上分析，可得最优子结构：$dp[i, j]$ 的最少编辑步数等于 $dp[i, j-1]$ , $dp[i-1, j]$ , $dp[i-1, j-1]$ 三者中的最少编辑步数，再加上本次编辑的步数 $1$ 。对应的状态转移方程为：
+
+$$
+dp[i, j] = \min(dp[i, j-1], dp[i-1, j], dp[i-1, j-1]) + 1
+$$
+
+请注意，**当 $s[i-1]$ 和 $t[j-1]$ 相同时，无需编辑当前字符**，此时状态转移方程为：
+
+$$
+dp[i, j] = dp[i-1, j-1]
+$$
+
+**第三步：确定边界条件和状态转移顺序**
+
+当两字符串都为空时，编辑步数为 $0$ ，即 $dp[0, 0] = 0$ 。当 $s$ 为空但 $t$ 不为空时，最少编辑步数等于 $t$ 的长度，即 $dp[0, j] = j$ 。当 $s$ 不为空但 $t$ 为空时，等于 $s$ 的长度，即 $dp[i, 0] = i$ 。
+
+观察状态转移方程，解 $dp[i, j]$ 依赖左方、上方、左上方的解，因此通过两层循环正序遍历整个 $dp$ 表即可。
+
+=== "Java"
+
+    ```java title="edit_distance.java"
+    [class]{edit_distance}-[func]{editDistanceDP}
+    ```
+
+=== "C++"
+
+    ```cpp title="edit_distance.cpp"
+    [class]{}-[func]{editDistanceDP}
+    ```
+
+=== "Python"
+
+    ```python title="edit_distance.py"
+    [class]{}-[func]{edit_distance_dp}
+    ```
+
+=== "Go"
+
+    ```go title="edit_distance.go"
+    [class]{}-[func]{editDistanceDP}
+    ```
+
+=== "JavaScript"
+
+    ```javascript title="edit_distance.js"
+    [class]{}-[func]{editDistanceDP}
+    ```
+
+=== "TypeScript"
+
+    ```typescript title="edit_distance.ts"
+    [class]{}-[func]{editDistanceDP}
+    ```
+
+=== "C"
+
+    ```c title="edit_distance.c"
+    [class]{}-[func]{editDistanceDP}
+    ```
+
+=== "C#"
+
+    ```csharp title="edit_distance.cs"
+    [class]{edit_distance}-[func]{editDistanceDP}
+    ```
+
+=== "Swift"
+
+    ```swift title="edit_distance.swift"
+    [class]{}-[func]{editDistanceDP}
+    ```
+
+=== "Zig"
+
+    ```zig title="edit_distance.zig"
+    [class]{}-[func]{editDistanceDP}
+    ```
+
+=== "Dart"
+
+    ```dart title="edit_distance.dart"
+    [class]{}-[func]{editDistanceDP}
+    ```
+
+如下图所示，编辑距离问题的状态转移过程与背包问题非常类似，都可以看作是填写一个二维网格的过程。
+
+=== "<1>"
+    ![编辑距离的动态规划过程](edit_distance_problem.assets/edit_distance_dp_step1.png)
+
+=== "<2>"
+    ![edit_distance_dp_step2](edit_distance_problem.assets/edit_distance_dp_step2.png)
+
+=== "<3>"
+    ![edit_distance_dp_step3](edit_distance_problem.assets/edit_distance_dp_step3.png)
+
+=== "<4>"
+    ![edit_distance_dp_step4](edit_distance_problem.assets/edit_distance_dp_step4.png)
+
+=== "<5>"
+    ![edit_distance_dp_step5](edit_distance_problem.assets/edit_distance_dp_step5.png)
+
+=== "<6>"
+    ![edit_distance_dp_step6](edit_distance_problem.assets/edit_distance_dp_step6.png)
+
+=== "<7>"
+    ![edit_distance_dp_step7](edit_distance_problem.assets/edit_distance_dp_step7.png)
+
+=== "<8>"
+    ![edit_distance_dp_step8](edit_distance_problem.assets/edit_distance_dp_step8.png)
+
+=== "<9>"
+    ![edit_distance_dp_step9](edit_distance_problem.assets/edit_distance_dp_step9.png)
+
+=== "<10>"
+    ![edit_distance_dp_step10](edit_distance_problem.assets/edit_distance_dp_step10.png)
+
+=== "<11>"
+    ![edit_distance_dp_step11](edit_distance_problem.assets/edit_distance_dp_step11.png)
+
+=== "<12>"
+    ![edit_distance_dp_step12](edit_distance_problem.assets/edit_distance_dp_step12.png)
+
+=== "<13>"
+    ![edit_distance_dp_step13](edit_distance_problem.assets/edit_distance_dp_step13.png)
+
+=== "<14>"
+    ![edit_distance_dp_step14](edit_distance_problem.assets/edit_distance_dp_step14.png)
+
+=== "<15>"
+    ![edit_distance_dp_step15](edit_distance_problem.assets/edit_distance_dp_step15.png)
+
+下面考虑状态压缩，将 $dp$ 表的第一维删除。由于 $dp[i,j]$ 是由上方 $dp[i-1, j]$ 、左方 $dp[i, j-1]$ 、左上方状态 $dp[i-1, j-1]$ 转移而来，而正序遍历会丢失左上方 $dp[i-1, j-1]$ ，倒序遍历无法提前构建 $dp[i, j-1]$ ，因此两种遍历顺序都不可取。
+
+为解决此问题，我们可以使用一个变量 `leftup` 来暂存左上方的解 $dp[i-1, j-1]$ ，这样便只用考虑左方和上方的解，与完全背包问题的情况相同，可使用正序遍历。
+
+=== "Java"
+
+    ```java title="edit_distance.java"
+    [class]{edit_distance}-[func]{editDistanceDPComp}
+    ```
+
+=== "C++"
+
+    ```cpp title="edit_distance.cpp"
+    [class]{}-[func]{editDistanceDPComp}
+    ```
+
+=== "Python"
+
+    ```python title="edit_distance.py"
+    [class]{}-[func]{edit_distance_dp_comp}
+    ```
+
+=== "Go"
+
+    ```go title="edit_distance.go"
+    [class]{}-[func]{editDistanceDPComp}
+    ```
+
+=== "JavaScript"
+
+    ```javascript title="edit_distance.js"
+    [class]{}-[func]{editDistanceDPComp}
+    ```
+
+=== "TypeScript"
+
+    ```typescript title="edit_distance.ts"
+    [class]{}-[func]{editDistanceDPComp}
+    ```
+
+=== "C"
+
+    ```c title="edit_distance.c"
+    [class]{}-[func]{editDistanceDPComp}
+    ```
+
+=== "C#"
+
+    ```csharp title="edit_distance.cs"
+    [class]{edit_distance}-[func]{editDistanceDPComp}
+    ```
+
+=== "Swift"
+
+    ```swift title="edit_distance.swift"
+    [class]{}-[func]{editDistanceDPComp}
+    ```
+
+=== "Zig"
+
+    ```zig title="edit_distance.zig"
+    [class]{}-[func]{editDistanceDPComp}
+    ```
+
+=== "Dart"
+
+    ```dart title="edit_distance.dart"
+    [class]{}-[func]{editDistanceDPComp}
+    ```
diff --git a/mkdocs.yml b/mkdocs.yml
index 2ddc0a45c..0974342d3 100644
--- a/mkdocs.yml
+++ b/mkdocs.yml
@@ -219,6 +219,7 @@ nav:
     - 13.3. &nbsp; DP 解题思路（New）: chapter_dynamic_programming/dp_solution_pipeline.md
     - 13.4. &nbsp; 0-1 背包问题（New）: chapter_dynamic_programming/knapsack_problem.md
     - 13.5. &nbsp; 完全背包问题（New）: chapter_dynamic_programming/unbounded_knapsack_problem.md
+    - 13.6. &nbsp; 编辑距离问题（New）: chapter_dynamic_programming/edit_distance_problem.md
   - 14. &nbsp; &nbsp; 附录:
     - 14.1. &nbsp; 编程环境安装: chapter_appendix/installation.md
     - 14.2. &nbsp; 一起参与创作: chapter_appendix/contribution.md