# [72\. Edit Distance](https://leetcode.com/problems/edit-distance/)

- 2 維 DP 矩陣，注意大小是 $(m + 1) \cdot (n + 1)$。
- 不需要編輯的狀況：`dp[i][j] = dp[i - 1][j - 1]`
- 狀態轉移：`dp[i][j] = 1 + min({dp[i - 1][j - 1], dp[i - 1][j], dp[i][j - 1]})`

:::spoiler Solution

class Solution {
public:
    int minDistance(string word1, string word2) {
        int m = word1.size(), n = word2.size();
        vector<vector<int>> dp(m + 1, vector<int>(n + 1));

        for (int i = 0; i <= m; ++i) dp[i][0] = i;
        for (int j = 0; j <= n; ++j) dp[0][j] = j;
        
        for (int i = 1; i <= m; ++i)
        {
            for (int j = 1; j <= n; ++j)
            {
                if (word1[i - 1] == word2[j - 1])
                {
                    // 不需要編輯的狀況
                    dp[i][j] = dp[i - 1][j - 1];
                }
                else
                {
                    // 新增 / 刪除 / 取代 三種情況
                    // + 1 代表做一次的操作
                    dp[i][j] = 1 + min({dp[i - 1][j - 1], dp[i - 1][j], dp[i][j - 1]});
                }
            }
        }
        return dp[m][n];
    }
};

- 時間複雜度：$O(M \cdot N)$
- 空間複雜度：$O(M \cdot N)$
:::