#!/bin/env python
# -*- coding:utf-8 -*-

"""
给定两个单词 word1 和 word2，计算出将 word1 转换成 word2 所使用的最少操作数 。

你可以对一个单词进行如下三种操作：

插入一个字符
删除一个字符
替换一个字符
示例 1:

输入: word1 = "horse", word2 = "ros"
输出: 3
解释:
horse -> rorse (将 'h' 替换为 'r')
rorse -> rose (删除 'r')
rose -> ros (删除 'e')
示例 2:

输入: word1 = "intention", word2 = "execution"
输出: 5
解释:
intention -> inention (删除 't')
inention -> enention (将 'i' 替换为 'e')
enention -> exention (将 'n' 替换为 'x')
exention -> exection (将 'n' 替换为 'c')
exection -> execution (插入 'u')
"""


class Solution(object):
    def minDistance(self, word1, word2):
        """
        :type word1: str
        :type word2: str
        :rtype: int
        """
        """
        dp[i][j] word1 的前 i 个字符替换成 word2 的 前 j 字符需要的最少操作次数.
        dp[len(word1)-1][len(word2)-1] 是最终的解
        """
        m = len(word1)+1
        n = len(word2)+1
        dp = [[0 for _ in range(n)] for _ in range(m)]

        for i in range(n):
            dp[0][i] = i
        for i in range(m):
            dp[i][0] = i

        for i in range(1, m):
            for j in range(1, n):
                if word1[i-1] == word2[j-1]:
                    dp[i][j] = dp[i-1][j-1]
                else:
                    dp[i][j] = min(dp[i][j-1], dp[i-1][j], dp[i-1][j-1]) + 1

        return dp[m-1][n-1]
