This problem is exactly the same as longest common subsequence problem. Using 2D DP to find the length of LCS `L`

, and the answer is just `n1 + n2 - L*2`

.

```
class Solution(object):
def minDistance(self, word1, word2):
"""
:type word1: str
:type word2: str
:rtype: int
"""
n1, n2 = len(word1), len(word2)
dp = [[0] * (n2+1) for _ in range(n1+1)]
for i in range(n1):
for j in range(n2):
dp[i+1][j+1] = dp[i][j] + 1 if word1[i] == word2[j] else max(dp[i+1][j], dp[i][j+1])
return n1 + n2 - dp[n1][n2]*2
```