## C# || How To Find The Minimum ASCII Delete Sum for Two Strings Using C#

The following is a module with functions which demonstrates how to find the minimum ASCII delete sum for two strings using C#.

1. Minimum Delete Sum – Problem Statement

Given two strings s1 and s2, return 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. ```

2. Minimum Delete Sum – Solution

The following is a solution which demonstrates how to find the minimum ASCII delete sum for two strings.

``` 2. Minimum Delete Sum - Solution C# // ============================================================================ // Author: Kenneth Perkins // Date: Aug 28, 2023 // Taken From: http://programmingnotes.org/ // File: Solution.cs // Description: Demonstrates how to find minimum ASCII delete sum two string // ============================================================================ public class Solution { public int MinimumDeleteSum(string s1, string s2) { // Make sure s2 is smaller string if (s1.Length < s2.Length) { return MinimumDeleteSum(s2, s1); } // Case for empty s1 int m = s1.Length, n = s2.Length; int[] currRow = new int[n + 1]; for (int j = 1; j <= n; j++) { currRow[j] = currRow[j - 1] + s2[j - 1]; } // Compute answer row-by-row for (int i = 1; i <= m; i++) { int diag = currRow[0]; currRow[0] += s1[i - 1]; for (int j = 1; j <= n; j++) { int answer; // If characters are the same, the answer is top-left-diagonal value if (s1[i - 1] == s2[j - 1]) { answer = diag; } // Otherwise, the answer is minimum of top and left values with // deleted character's ASCII value else { answer = Math.Min( s1[i - 1] + currRow[j], s2[j - 1] + currRow[j - 1] ); } // Before overwriting currRow[j] with answer, save it in diag // for the next column diag = currRow[j]; currRow[j] = answer; } } // Return the answer return currRow[n]; } }// http://programmingnotes.org/ 1234567891011121314151617181920212223242526272829303132333435363738394041424344454647484950515253545556 // ============================================================================//    Author: Kenneth Perkins//    Date:   Aug 28, 2023//    Taken From: http://programmingnotes.org///    File:  Solution.cs//    Description: Demonstrates how to find minimum ASCII delete sum two string// ============================================================================public class Solution {    public int MinimumDeleteSum(string s1, string s2) {        // Make sure s2 is smaller string        if (s1.Length < s2.Length) {            return MinimumDeleteSum(s2, s1);        }         // Case for empty s1        int m = s1.Length, n = s2.Length;        int[] currRow = new int[n + 1];        for (int j = 1; j <= n; j++) {            currRow[j] = currRow[j - 1] + s2[j - 1];        }         // Compute answer row-by-row        for (int i = 1; i <= m; i++) {             int diag = currRow[0];            currRow[0] += s1[i - 1];             for (int j = 1; j <= n; j++) {                 int answer;                 // If characters are the same, the answer is top-left-diagonal value                if (s1[i - 1] == s2[j - 1]) {                    answer = diag;                }                 // Otherwise, the answer is minimum of top and left values with                // deleted character's ASCII value                else {                    answer = Math.Min(                        s1[i - 1] + currRow[j],                        s2[j - 1] + currRow[j - 1]                    );                }                 // Before overwriting currRow[j] with answer, save it in diag                // for the next column                diag = currRow[j];                currRow[j] = answer;            }        }         // Return the answer        return currRow[n];    }}// http://programmingnotes.org/ ```

QUICK NOTES:
The highlighted lines are sections of interest to look out for.

The code is heavily commented, so no further insight is necessary. If you have any questions, feel free to leave a comment below.

Once compiled, you should get this as your output for the example cases:

``` 231 403 ```