Minimum ASCII Delete Sum for Two Strings

Description

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].

Solution

Let dp[i + 1][j + 1] be the minimum delete sum of s1[0..i] and s2[0..j].

• If s1[i] == s2[j], then

  • dp[i + 1][j + 1] = dp[i][j]

• Otherwise, delete either s1[i] or s2[j] or both. So,

  • dp[i + 1][j + 1] = min(dp[i][j + 1] + s1[i], dp[i + 1][j] + s2[j])

  • Both dp[i][j + 1] and dp[i][j + 1] contain the case where both s1[i] and s2[j] are deleted.

Let $n_1,n_2$ be the length of s1 and s2, respectively.

The time complexity is $O(n_1 n_2)$. The space complexity is also $O(n_1 n_2)$, but it can be optimized to $O(\min(n_1, n_2))$.

class Solution {
public:
    int minimumDeleteSum(string s1, string s2) {
        int n1 = s1.size();
        int n2 = s2.size();
        vector<vector<int>> dp(n1 + 1, vector<int>(n2 + 1));
        dp[0][0] = 0;
        for(int i = 0; i < n1; ++i){
            dp[i + 1][0] = dp[i][0] + s1[i];
        }
        for(int j = 0; j < n2; ++j){
            dp[0][j + 1] = dp[0][j] + s2[j];
        }
        for(int i = 0; i < n1; ++i){
            for(int j = 0; j < n2; ++j){
                if(s1[i] == s2[j]){
                    dp[i + 1][j + 1] = dp[i][j];
                }else{
                    dp[i + 1][j + 1] = min(dp[i][j + 1] + s1[i], dp[i + 1][j] + s2[j]);
                }
            }
        }
        return dp[n1][n2];
    }
};