In the following solution `DP[i][j]`

denotes the longest common subsequence in `s.substring(i)`

and `t.substring(j)`

```
public int minDistance(String s, String t) {
int dp[][] = new int[s.length() + 1][t.length() + 1];
for (int i = s.length() - 1; i >= 0; i--)
for (int j = t.length() - 1; j >= 0; j--)
dp[i][j] = s.charAt(i) == t.charAt(j) ? dp[i + 1][j + 1] + 1 : Math.max(dp[i + 1][j], dp[i][j + 1]);
return s.length() + t.length() - 2 * dp[0][0];
}
```