LeetCode 712. Minimum ASCII Delete Sum for Two Strings

Posted 2022-09-26 dylan-java-nyc

原题链接在这里：https://leetcode.com/problems/minimum-ascii-delete-sum-for-two-strings/

题目：

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].

题解：

For two strings, dp[i][j] denotes s1 up to i and s2 up to j, minimum ASCII delete sum for two substrings.

There could be 2 cases:

case 1:

s1.charAt(i) == s2.charAt(j), then it doesn‘t need to delete last characters, dp[i][j] = dp[i-1][j-1].

case 2:

s1.charAt(i) != s2.charAt(j). then minimum delete could comes from 2 possibilities, take the minimum:

• delete s1.charAt(i). dp[i][j] = dp[i-1][j] + s1.charAt(i).
• delete s2.charAt(j). dp[i][j] = dp[i][j-1] + s2.charAt(j).

Time Complexity: O(m*n). m = s1.length(). n = s2.length().

Space: O(m*n).

AC Java:

 1 class Solution 
 2     public int minimumDeleteSum(String s1, String s2) 
 3         int m = s1.length();
 4         int n = s2.length();
 5         int [][] dp = new int[m+1][n+1];
 6         
 7         for(int i = 1; i<=m; i++)
 8             dp[i][0] = dp[i-1][0]+s1.charAt(i-1);
 9         
10         
11         for(int j = 1; j<=n; j++)
12             dp[0][j] = dp[0][j-1] + s2.charAt(j-1);
13         
14         
15         for(int i = 1; i<=m; i++)
16             for(int j = 1; j<=n; j++)
17                 if(s1.charAt(i-1) == s2.charAt(j-1))
18                     dp[i][j] = dp[i-1][j-1];
19                 else
20                     dp[i][j] = Math.min(dp[i-1][j]+s1.charAt(i-1), dp[i][j-1]+s2.charAt(j-1));
21                 
22             
23         
24         
25         return dp[m][n];
26     
27 

类似Longest Common Subsequence, Delete Operation for Two Strings, Edit Distance.