# Edit Distance

Updated: Mar 25, 2021

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

**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.

**Solution:**

```
class Solution {
public int minDistance(String word1, String word2) {
int m = word1.length();
int n = word2.length();
int[][] dp = new int[m + 1][n + 1];
//first row
for(int i = 0; i <= m; i++)
dp[i][0] = i;
//first column
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.charAt(i - 1) == word2.charAt(j - 1))
dp[i][j] = dp[i - 1][j - 1];
else {
dp[i][j] = 1 + Math.min(dp[i - 1][j - 1], Math.min(dp[i][j - 1], dp[i - 1][j]));
}
}
}
return dp[m][n];
}
}
```