Short Python based on LCS

  • 0

    This problem is exactly the same as longest common subsequence problem. Using 2D DP to find the length of LCS L, and the answer is just n1 + n2 - L*2.

    class Solution(object):
        def minDistance(self, word1, word2):
            :type word1: str
            :type word2: str
            :rtype: int
            n1, n2 = len(word1), len(word2)
            dp = [[0] * (n2+1) for _ in range(n1+1)]
            for i in range(n1):
                for j in range(n2):
                    dp[i+1][j+1] = dp[i][j] + 1 if word1[i] == word2[j] else max(dp[i+1][j], dp[i][j+1])
            return n1 + n2 - dp[n1][n2]*2

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