712. Minimum ASCII Delete Sum for Two Strings

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].

 

Approach #1: DFS + Memeory. [C++]

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
class Solution {
public:
    int minimumDeleteSum(string s1, string s2) {
        int l1 = s1.length();
        int l2 = s2.length();
        memo = vector<vector<int>>(l1+1, vector<int>(l2+1, INT_MAX));
        return solve(s1, l1, s2, l2);
    }
     
private:
    vector<vector<int>> memo;
    int solve(const string& s1, int i, const string& s2, int j) {
        if (i == 0 && j == 0) return 0;
        if (memo[i][j] != INT_MAX) return memo[i][j];
        if (i == 0) return memo[i][j] = solve(s1, i, s2, j-1) + s2[j-1];
        if (j == 0) return memo[i][j] = solve(s1, i-1, s2, j) + s1[i-1];
        if (s1[i-1] == s2[j-1]) return memo[i][j] = solve(s1, i-1, s2, j-1);
        return memo[i][j] = min(solve(s1, i-1, s2, j) + s1[i-1],
                                solve(s1, i, s2, j-1) + s2[j-1]);
    }
};

  

Approach #2: DP. [Java]

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
class Solution {
    public int minimumDeleteSum(String s1, String s2) {
        int l1 = s1.length();
        int l2 = s2.length();
        int[][] dp = new int[l1+1][l2+1];
        for (int i = 1; i <= l1; ++i)
            dp[i][0] = dp[i-1][0] + s1.charAt(i-1);
        for (int j = 1; j <= l2; ++j)
            dp[0][j] = dp[0][j-1] + s2.charAt(j-1);
        for (int i = 1; i <= l1; ++i)
            for (int j = 1; j <= l2; ++j)
                if (s1.charAt(i-1) == s2.charAt(j-1))
                    dp[i][j] = dp[i-1][j-1];
                else
                    dp[i][j] = Math.min(dp[i-1][j] + s1.charAt(i-1),
                                   dp[i][j-1] + s2.charAt(j-1));
        return dp[l1][l2];
    }
}

  

Reference:

http://zxi.mytechroad.com/blog/dynamic-programming/leetcode-712-minimum-ascii-delete-sum-for-two-strings/

 

posted @   Veritas_des_Liberty  阅读(228)  评论(0编辑  收藏  举报
编辑推荐:
· 为什么说在企业级应用开发中,后端往往是效率杀手?
· 用 C# 插值字符串处理器写一个 sscanf
· Java 中堆内存和栈内存上的数据分布和特点
· 开发中对象命名的一点思考
· .NET Core内存结构体系(Windows环境)底层原理浅谈
阅读排行:
· 为什么说在企业级应用开发中,后端往往是效率杀手?
· DeepSeek 解答了困扰我五年的技术问题。时代确实变了!
· 本地部署DeepSeek后,没有好看的交互界面怎么行!
· 趁着过年的时候手搓了一个低代码框架
· 推荐一个DeepSeek 大模型的免费 API 项目!兼容OpenAI接口!
历史上的今天:
2018-03-14 C - Trailing Zeroes (III)(二分)
2018-03-14 B - Pie (二分)
点击右上角即可分享
微信分享提示