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)

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