1092. Shortest Common Supersequence

# Time:  O(m * n)
# Space: O(m * n)

class Solution(object):
    def shortestCommonSupersequence(self, str1, str2):
        """
        :type str1: str
        :type str2: str
        :rtype: str
        """
        dp = [[0 for _ in xrange(len(str2)+1)] for _ in xrange(2)]
        bt = [[None for _ in xrange(len(str2)+1)] for _ in xrange(len(str1)+1)]
        for i, c in enumerate(str1):
            bt[i+1][0] = (i, 0, c)
        for j, c in enumerate(str2):
            bt[0][j+1] = (0, j, c)
        for i in xrange(len(str1)):
            for j in xrange(len(str2)):
                if dp[i % 2][j+1] > dp[(i+1) % 2][j]:
                    dp[(i+1) % 2][j+1] = dp[i % 2][j+1]
                    bt[i+1][j+1] = (i, j+1, str1[i])
                else:
                    dp[(i+1) % 2][j+1] = dp[(i+1) % 2][j]
                    bt[i+1][j+1] = (i+1, j, str2[j])
                if str1[i] != str2[j]:
                    continue
                if dp[i % 2][j]+1 > dp[(i+1) % 2][j+1]:
                    dp[(i+1) % 2][j+1] = dp[i % 2][j]+1
                    bt[i+1][j+1] = (i, j, str1[i])
        
        i, j = len(str1), len(str2)
        result = []
        while i != 0 or j != 0:
            i, j, c = bt[i][j]
            result.append(c)
        result.reverse()
        return "".join(result)

Solution from kamyu104/LeetCode-Solutions · MIT