short C++ DP solution, a variant of LCS problem


  • 0
    G
    int minDistance(string word1, string word2) {
        const auto M = word1.size();
        const auto N = word2.size();
        vector<vector<int>> dp(M+1, vector<int>(N+1, 0));
        
        for (auto i = 1; i <= M; i++) {
            for (auto j = 1; j <= N; j++) {
                dp[i][j] = max(dp[i-1][j], dp[i][j-1]);
                if (word1[i-1] == word2[j-1]) {
                    dp[i][j] = max(dp[i][j], dp[i-1][j-1]+1);
                }
            }
        }
        return M + N - dp[M][N] * 2;
    }

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