This page explains Java solution to problem `Edit Distance`

using `Dynamic programming`

.

Given two words `word1`

and `word2`

, find the minimum number of operations required to convert `word1`

to `word2`

.

You have the following 3 operations permitted on a word:

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

Input: word1 = "horse", word2 = "ros"Output: 3Explanation:

horse -> rorse (replace 'h' with 'r')

rorse -> rose (remove 'r')

rose -> ros (remove 'e')

Input: word1 = "intention", word2 = "execution"Output: 5Explanation:

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')

```
package com.vc.hard;
class EditDistance {
public int minDistance(String word1, String word2) {
if(word1 == null && word2 == null) return 0;
if(word1 == null) return word2.length();
if(word2 == null) return word1.length();
int m = word1.length();
int n = word2.length();
int[][] dp = new int[m + 1][n + 1];
for(int i = 0; i <= m; i++) {
for(int j = 0; j <= n; j++) {
if(i == 0 && j == 0) dp[i][j] = 0;
else if(i == 0) dp[i][j] = j;
else if(j == 0) dp[i][j] = i;
else {
if(word1.charAt(i - 1) != word2.charAt(j - 1))
dp[i][j] = Math.min(dp[i - 1][j - 1], Math.min(dp[i][j - 1], dp[i - 1][j])) + 1;
else
dp[i][j] = dp[i - 1][j - 1];
}
}
}
return dp[m][n];
}
}
```

O(M * N)

Where M is length of word1

And N is length of word2

O(M * N)