Algorithm question - edit distance (python implementation)

The same problem can be solved using the idea of ​​dynamic programming. The three methods are as follows:

class Solution:
    def minDistance(self, word1: str, word2: str) -> int:
        m = len(word1)
        n = len(word2)
    #纯递归不通过，时间复杂度过高
        # def dp(i,j):
        #     if i == -1:
        #         return j+1
        #     if j==-1:
        #         return i+1
        #     if word1[i]==word2[j]:
        #         return dp(i-1,j-1)
        #     else:
        #         return min(dp(i,j-1)+1,dp(i-1,j)+1,dp(i-1,j-1)+1)
        # return dp(m-1,n-1)

        #备忘录解决，通过
        # memo = dict()
        # def dp(i,j):
        #     if (i,j) in memo:
        #         return memo[(i,j)]
        #     if i == -1:
        #         return j+1
        #     if j==-1:
        #         return i+1
        #     if word1[i]==word2[j]:
        #         memo[(i,j)]=dp(i-1,j-1)
        #     else:
        #         memo[(i,j)] = min(dp(i,j-1)+1,dp(i-1,j)+1,dp(i-1,j-1)+1)
        #     return memo[(i,j)]
        # return dp(m-1,n-1)

        #DP table 解决自底向上
        dp = [[0]*(m+1) for _ in range(n+1)]
        for i in range(m+1):
            dp[0][i]=i
        for j in range(n+1):
            dp[j][0]=j
        for i in range(1,n+1):
            for j in range(1,m+1):
                if word1[j-1]==word2[i-1]:
                    dp[i][j]=dp[i-1][j-1]
                else:
                    dp[i][j]=min(dp[i][j-1]+1,dp[i-1][j-1]+1,dp[i-1][j]+1)

        return dp[n][m]

What should be paid special attention to in the above code is the n and m of the row and column.