# 72 Edit Distance

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.
 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 `````` ``````class Solution: def minDistance(self, word1: str, word2: str) -> int: m, n = len(word1)+1, len(word2)+1 dp = [ * n for _ in range(m)] for i in range(m): for j in range(n): if i == 0: dp[i][j] = j elif j == 0: dp[i][j] = i else: if word1[i-1] == word2[j-1]: dp[i][j] = min( dp[i-1][j] + 1, dp[i][j-1] + 1, dp[i-1][j-1] ) else: dp[i][j] = 1 + min( dp[i-1][j], dp[i][j-1], dp[i-1][j-1] ) return dp[m-1][n-1] # can reduce space to O(min(m, n))``````