# LC 72. Edit Distance

### [Problem link](https://leetcode.com/problems/edit-distance/)

###### tags: `leedcode` `medium` `c++` `DP`

Given two strings <code>word1</code> and <code>word2</code>, return the minimum number of operations required to convert <code>word1</code> to <code>word2</code>.

You have the following three operations permitted on a word:

- Insert a character
- Delete a character
- Replace a character

**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:**

- <code>0 <= word1.length, word2.length <= 500</code>
- <code>word1</code> and <code>word2</code> consist of lowercase English letters.

## Solution 1
#### C++

class Solution {
public:
    int minDistance(string word1, string word2) {
        vector<vector<int>> dp(word1.size() + 1, vector<int> (word2.size() + 1));
        for (int i = 0; i <= word1.size(); i++) {
            dp[i][0] = i;
        }
        for (int i = 0; i <= word2.size(); i++) {
            dp[0][i] = i;
        }
        for (int i = 0; i < word1.size(); i++) {
            for (int j = 0; j < word2.size(); j++) {
                if (word1[i] == word2[j]) {
                    dp[i + 1][j + 1] = dp[i][j];
                } else {
                    dp[i + 1][j + 1] = min({dp[i][j] + 1, dp[i + 1][j] + 1, dp[i][j + 1] + 1});
                }
            }
        }
        return dp.back().back();
    }
};

>### Complexity
>m = word1.length
>n = word2.length
>| | Time Complexity | Space Complexity |
>| ----------- | --------------- | ---------------- |
>| Solution 1 | O(mn) | O(mn) |

## Note
sol1: