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LeetCode 72 Edit Distance - Hard

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

·2 min read·#LeetCode#Hard#Python

72. Edit Distance — Hard

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Problem

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')

Solution

class Solution:
    def minDistance(self, word1: str, word2: str) -> int:
        ### DP problem:
        m = len(word1)
        n = len(word2)
        
        f = [[0] * (n + 1) for _ in range(m + 1)]
        
        for i in range(m + 1):
            for j in range(n + 1):
                ### when word1 is None, insert, insert, insert ... to word1 till word1 is word2
                ### after this step we need to jump to next, use continue to avoid the overwrite
                if i == 0:
                    f[0][j] = j
                    continue
                ### when word2 is None, delete, delete, delete ... from word1 till word1 is word2
                ### after this step we need to jump to next, use continue to avoid the overwrite
                if j == 0:
                    f[i][0] = i
                    continue
                if word1[i - 1] != word2[j - 1]:
                    ### delete, insert, replace
                    f[i][j] = min(f[i - 1][j], f[i][j - 1], f[i - 1][j - 1]) + 1
                else: ### no need to operate when equal
                    f[i][j] = f[i - 1][j - 1]
                
        return f[m][n]
 
 
### 滚动数组 空间优化：
class Solution:
    def minDistance(self, word1: str, word2: str) -> int:
        ### DP problem:
        m = len(word1)
        n = len(word2)
        
        f = [[0] * (n + 1) for _ in range(2)]
        old, now = 0, 0
        for i in range(m + 1):
            old = now 
            now = 1 - now
            for j in range(n + 1):
                ### when word1 is None, insert, insert, insert ... to word1 till word1 is word2
                ### after this step we need to jump to next, use continue to avoid the overwrite
                if i == 0:
                    f[now][j] = j
                    continue
                ### when word2 is None, delete, delete, delete ... from word1 till word1 is word2
                ### after this step we need to jump to next, use continue to avoid the overwrite
                if j == 0:
                    f[now][j] = i
                    continue
                if word1[i - 1] != word2[j - 1]:
                    ### delete, insert, replace
                    f[now][j] = min(f[old][j], f[now][j - 1], f[old][j - 1]) + 1
                else: ### no need to operate when equal
                    f[now][j] = f[old][j - 1]
                
        return f[now][n]

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