## DEV Community

Abhishek Chaudhary

Posted on

# 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.

SOLUTION:

``````class Solution:
def md(self, word1, word2, i1, i2):
if (i1, i2) in self.cache:
return self.cache[(i1, i2)]
if i1 == len(word1):
return len(word2) - i2
if i2 == len(word2):
return len(word1) - i1
if word1[i1] == word2[i2]:
self.cache[(i1, i2)] = self.md(word1, word2, i1 + 1, i2 + 1)
return self.cache[(i1, i2)]
else:
self.cache[(i1, i2)] = 1 + min(self.md(word1, word2, i1 + 1, i2 + 1), self.md(word1, word2, i1, i2 + 1), self.md(word1, word2, i1 + 1, i2))
return self.cache[(i1, i2)]

def minDistance(self, word1: str, word2: str) -> int:
self.cache = {}
return self.md(word1, word2, 0, 0)
``````