first find the longest common sequence

then the solution is the total length minus length of LCS

```
class Solution {
public:
int minDistance(string word1, string word2) {
int len1 = word1.length(), len2 = word2.length();
vector<vector<int>>dp(len1+1, vector<int>(len2+1, 0));
for(int i = 1; i <= len1; i++)
{
for(int j = 1; j <= len2; j++)
{
dp[i][j] = max(dp[i-1][j-1], max(dp[i-1][j], dp[i][j-1]));
if(word1[i-1] == word2[j-1]) dp[i][j] = dp[i-1][j-1] + 1;
}
}
return len1 + len2 - 2*dp[len1][len2];
}
};
```