LC-0712 Medium LeetCode

712. Minimum ASCII Delete Sum for Two Strings

Read the full problem statement on LeetCode.

Difficulty: medium Acceptance: 66% Topics: String, Dynamic Programming

Reading material

Strings 12 min 1-D Dynamic Programming 16 min

Reference solution (spoiler · python)

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

class Solution(object):
    def minimumDeleteSum(self, s1, s2):
        """
        :type s1: str
        :type s2: str
        :rtype: int
        """
        dp = [[0] * (len(s2)+1) for _ in xrange(2)]
        for j in xrange(len(s2)):
            dp[0][j+1] = dp[0][j] + ord(s2[j])

        for i in xrange(len(s1)):
            dp[(i+1)%2][0] = dp[i%2][0] + ord(s1[i])
            for j in xrange(len(s2)):
                if s1[i] == s2[j]:
                    dp[(i+1)%2][j+1] = dp[i%2][j]
                else:
                    dp[(i+1)%2][j+1] = min(dp[i%2][j+1] + ord(s1[i]), \
                                           dp[(i+1)%2][j] + ord(s2[j]))

        return dp[len(s1)%2][-1]


# Time:  O(m * n)
# Space: O(m * n)
class Solution2(object):
    def minimumDeleteSum(self, s1, s2):
        """
        :type s1: str
        :type s2: str
        :rtype: int
        """
        dp = [[0] * (len(s2)+1) for _ in xrange(len(s1)+1)]
        for i in xrange(len(s1)):
            dp[i+1][0] = dp[i][0] + ord(s1[i])
        for j in xrange(len(s2)):
            dp[0][j+1] = dp[0][j] + ord(s2[j])

        for i in xrange(len(s1)):
            for j in xrange(len(s2)):
                if s1[i] == s2[j]:
                    dp[i+1][j+1] = dp[i][j]
                else:
                    dp[i+1][j+1] = min(dp[i][j+1] + ord(s1[i]), \
                                       dp[i+1][j] + ord(s2[j]))

        return dp[-1][-1]

Solution from kamyu104/LeetCode-Solutions · MIT