Minimum ASCII Delete Sum for Two Strings

Tested Python solution for LeetCode 712 with 20 pytest cases. Generate a practice environment with lcpy.

LeetCode 712, Medium. Topics: String, Dynamic Programming, Longest Common Subsequence. View on LeetCode.

Generate this problem as a practice environment: tested reference solution, 20 parametrized pytest cases, and a playground notebook:

lcpy gen -n 712   # by problem number
lcpy gen -s minimum_ascii_delete_sum_for_two_strings   # by problem name

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Problem

Given two strings s1 and s2, return the lowest ASCII sum of deleted characters to make two strings equal.

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Examples

Input: s1 = "sea", s2 = "eat"
Output: 231
Explanation: Deleting "s" from "sea" adds the ASCII value of "s" (115) to the sum.
Deleting "t" from "eat" adds 116 to the sum.
At the end, both strings are equal, and 115 + 116 = 231 is the minimum sum possible to achieve this.

Input: s1 = "delete", s2 = "leet"
Output: 403
Explanation: Deleting "dee" from "delete" to turn the string into "let", adds 100[d] + 101[e] + 101[e] to the sum.
Deleting "e" from "leet" adds 101[e] to the sum.
At the end, both strings are equal to "let", and the answer is 100 + 101 + 101 + 101 = 403.

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Constraints

• 1 <= s1.length, s2.length <= 1000
• s1 and s2 consist of lowercase English letters.

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Solution

Reference implementation from solution.py on GitHub, full suite in test_solution.py:

class Solution:
    # Time: O(m * n)
    # Space: O(min(m, n))
    def minimum_delete_sum(self, s1: str, s2: str) -> int:
        if len(s1) < len(s2):
            s1, s2 = s2, s1
        prev = [0] * (len(s2) + 1)
        for j in range(1, len(s2) + 1):
            prev[j] = prev[j - 1] + ord(s2[j - 1])
        for i in range(1, len(s1) + 1):
            curr = [prev[0] + ord(s1[i - 1])]
            for j in range(1, len(s2) + 1):
                if s1[i - 1] == s2[j - 1]:
                    curr.append(prev[j - 1])
                else:
                    curr.append(min(prev[j] + ord(s1[i - 1]), curr[j - 1] + ord(s2[j - 1])))
            prev = curr
        return prev[-1]

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Complexity

Time	Space
O(m * n)	O(min(m, n))

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