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Given two words word1 and word2, find the minimum number of operations required to convert word1 to word2.

You have the following 3 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')

My code:

import numpy as np
class Solution:
    def minDistance(self, word1, word2):
        """
        :type word1: str
        :type word2: str
        :rtype: int
        """
        if len(word1)==0:
            return len(word2)
        if len(word2)==0:
            return len(word1)
        self.matrix=np.zeros([len(word1),len(word2)])-1
        count=self.common(word1,word2,len(word1)-1,len(word2)-1)
        #print(self.matrix)
        #print(len(word2))
        #all_length=max([len(word1),len(word2)])
        return int(count)
    def common(self,word1,word2,i,j):
        if i<0:
            return j+1
        if j<0:
            return i+1
        if self.matrix[i,j]!=-1:
            return self.matrix[i,j]
        if word1[i]==word2[j]:
            count0=self.common(word1,word2,i-1,j-1)
            self.matrix[i,j]=count0
            return count0
        else:
            count0=self.common(word1,word2,i-1,j-1)+1
        count1=1+self.common(word1,word2,i-1,j)
        count2=1+self.common(word1,word2,i,j-1)
        self.matrix[i,j]=min([count0,count1,count2])
        return self.matrix[i,j]

Solution: https://leetcode.com/problems/edit-distance/