leetcode: 72. Edit Distance

Q

Given two words word1 and word2, find the minimum number of steps required to convert word1 to word2. (each operation is counted as 1 step.)

You have the following 3 operations permitted on a word:

a) Insert a character
b) Delete a character
c) Replace a character

AC

class Solution(object):    def minDistance(self, word1, word2):        """        :type word1: str        :type word2: str        :rtype: int        """        distance = [[i] for i in xrange(len(word1) + 1)]        distance[0] = [j for j in xrange(len(word2) + 1)]        for i in xrange(1, len(word1) + 1):            for j in xrange(1, len(word2) + 1):                insert = distance[i][j - 1] + 1                delete = distance[i - 1][j] + 1                replace = distance[i - 1][j - 1]                if word1[i - 1] != word2[j - 1]:                    replace += 1                distance[i].append(min(insert, delete, replace))        return distance[-1][-1]# Time:  O(n * m)# Space: O(n + m)class Solution2(object):    # @return an integer    def minDistance(self, word1, word2):        if len(word1) < len(word2):            return self.minDistance(word2, word1)        distance = [i for i in xrange(len(word2) + 1)]        for i in xrange(1, len(word1) + 1):            pre_distance_i_j = distance[0]            distance[0] = i            for j in xrange(1, len(word2) + 1):                insert = distance[j - 1] + 1                delete = distance[j] + 1                replace = pre_distance_i_j                if word1[i - 1] != word2[j - 1]:                    replace += 1                pre_distance_i_j = distance[j]                distance[j] = min(insert, delete, replace)        return distance[-1]# Time:  O(n * m)# Space: O(n * m)class Solution3(object):    def minDistance(self, word1, word2):        distance = [[i] for i in xrange(len(word1) + 1)]        distance[0] = [j for j in xrange(len(word2) + 1)]        for i in xrange(1, len(word1) + 1):            for j in xrange(1, len(word2) + 1):                insert = distance[i][j - 1] + 1                delete = distance[i - 1][j] + 1                replace = distance[i - 1][j - 1]                if word1[i - 1] != word2[j - 1]:                    replace += 1                distance[i].append(min(insert, delete, replace))        return distance[-1][-1]if __name__ == "__main__":    assert Solution().minDistance("Rabbit", "Racket") == 3    assert Solution().minDistance("Rabbit", "Rabket") == 2    assert Solution().minDistance("Rabbit", "Rabbitt") == 1

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