72.Edit Distance
===

###### tags: `Hard`,`String`,`DP`

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

### 題目描述

Given two strings `word1` and `word2`, return *the minimum number of operations required to convert* `word1` *to* `word2`.

You have the following three operations permitted on a word:

* Insert a character
* Delete a character
* Replace a character

### 範例

**Example 1:**

Input: word1 = "horse", word2 = "ros"
Output: 3
Explanation: 
horse -> rorse (replace 'h' with 'r')
rorse -> rose (remove 'r')
rose -> ros (remove 'e')

**Example 2:**

Input: word1 = "intention", word2 = "execution"
Output: 5
Explanation: 
intention -> inention (remove 't')
inention -> enention (replace 'i' with 'e')
enention -> exention (replace 'n' with 'x')
exention -> exection (replace 'n' with 'c')
exection -> execution (insert 'u')

**Constraints**:

* 0 <= `word1.length`, `word2.length` <= 500
* `word1` and `word2` consist of lowercase English letters.

### 解答
#### C++
##### 思路:
1. 把word1變成word2, 有增刪改三種動作可以做, 求最小的操作次數
2. 思考基本case
2.1. 當word1 or word2其中之一為空, 則次數為非空者的長度
2.2. 當前雙方第一個字符相同, 則考慮word1[1:]及word2[1:] substring的minDistance
2.3. 當前雙方第一個字符不同, 三種動作可做:
* 增: 接下來只需考慮word1[:] 及word2[1:]的maxDistance
* 刪: 接下來只需考慮word1[1:]及word2[:]的maxDistance
* 改: 接下來只需考慮word1[1:]及word2[1:]的maxDistance
3. 避免解重複substring的maxDistance用DP
4. 用類似Longest common subsequence作法, 每次多考慮word1 or word2 一個字符, 根據2.的條件計算DP

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, 0));
        // If one of word is empty
        for(int i = 0; i <= m; i++)
            dp[i][0] = i;
        for(int i = 0; i <= n; i++)
            dp[0][i] = i;
        
        for(int i = 1; i <= m; i++)
        {
            for(int j = 1; j <= n; j++)
            {
                if(word1[i-1]==word2[j-1])
                    dp[i][j] = dp[i-1][j-1];
                else
                {
                    dp[i][j] = 1 + min({
                        dp[i][j-1], // insert
                        dp[i-1][j], // delete
                        dp[i-1][j-1] // replace
                    });
                }
            }
        }
        return dp[m][n];
    }
};

Time: $O(mn)$
Extra Space: $O(mn)$

> [name=XD] [time= Feb 27, 2023]

### Reference

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