Leetcode-Minimum ASCII Delete Sum for Two Strings

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本文介绍了一种利用动态规划解决两个字符串通过删除字符使二者相等，并求这些被删除字符ASCII值之和最小值的问题。文章详细阐述了算法的设计思路、实现过程及时间与空间复杂度。

Problem
Analysis
code

Problem

Given two strings s1, s2, find the lowest ASCII sum of deleted characters to make two strings equal.
Example 1:

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.

Example 2:

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.
If instead we turned both strings into “lee” or “eet”, we would get answers of 433 or 417, which are higher.

Note:

0 < s1.length, s2.length <= 1000.
All elements of each string will have an ASCII value in [97, 122].

具体详见Leetcode

Analysis

类似的题目

这个问题和两道题有点像，第一道是最长公共子串问题，第二道是字符串的Edit distance。此题又有点像最长公共子序列，因为是可以不连续的序列，不像子串问题一样一定需要连续的。Edit Distance 有三种情况，第一种是替换，第二种是增加，第三种是删减。而此题只有删减一种情况。所以此题可以使用动态规划的方法。

算法

设 $matrix[i][j]$ 为s1.substr(0,i)和s2.substr(0,j)有相同子序列的字符的ASCII码值的总和。

则 $matrix[i][j] = max(matrix[i-1][j-1]+same(i,j)*ASCII(s1[i-1]),matrix[i-1][j],matrix[i][j-1]);$

最后的答案：ASCII(s1+s2)-2*matrix[s1.size()][s2.size()]

Complexity

时间复杂度： $O(n^2)$
空间复杂度： $O(n^2)$

code

class Solution {
public:
    int minimumDeleteSum(string s1, string s2) {
        int ** matrix = new int*[s1.size()+1];
        for (int i = 0; i <= s1.size(); i++)
            matrix[i] = new int[s2.size()+1];

        for (int i = 0; i <= s1.size(); i++) {
            matrix[i][0] = 0;
        }

        for (int i = 1; i <= s2.size(); i++) {
            matrix[0][i] = 0;
        }
        int sum = 0;
        for (int i = 0; i < s1.size(); i++) {
            sum += (s1[i] - 'a'+97);
        }
        for (int i = 0; i < s2.size(); i++) {
            sum += (s2[i] - 'a'+97);
        }
        for (int i = 1; i <= s1.size(); i++) {
            for (int j = 1; j <= s2.size(); j++) {
                int sim = (s1[i-1] == s2[j-1]) ? (s1[i-1]-'a'+97) : 0;
                matrix[i][j] = max(matrix[i-1][j-1]+sim,matrix[i-1][j]);
                matrix[i][j] = max(matrix[i][j-1],matrix[i][j]);
            }
        }

        int result = sum - 2*matrix[s1.size()][s2.size()];

        for (int i = 0; i <= s1.size(); i++)
            delete []matrix[i];
        delete matrix;
        return result;
    }
};