# 712 Minimum ASCII Delete Sum for Two Strings

## 712. [Minimum ASCII Delete Sum for Two Strings](https://leetcode.com/problems/minimum-ascii-delete-sum-for-two-strings/description/)

## 1. Question

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]`.

## 2. Implementation

**(1) DP**

```java
class Solution {
    public int minimumDeleteSum(String s1, String s2) {
        int m = s1.length(), n = s2.length();
        int[][] dp = new int[m + 1][n + 1];

        for (int i = 0; i <= m; i++) {
            for (int j = 0; j <= n; j++) {
                if (i == 0 && j == 0) {
                    dp[i][j] = 0;
                }
                else if (i == 0) {
                    dp[i][j] = dp[i][j - 1] + (int)s2.charAt(j - 1);
                }
                else if (j == 0) {
                    dp[i][j] = dp[i - 1][j] + (int)s1.charAt(i - 1);
                }
                else if (s1.charAt(i - 1) == s2.charAt(j - 1)) {
                    dp[i][j] = dp[i - 1][j - 1];
                }
                else {
                    int cost1 = (int)s1.charAt(i - 1);
                    int cost2 = (int)s2.charAt(j - 1);
                    dp[i][j] = Math.min(dp[i - 1][j] + cost1, dp[i][j - 1] + cost2);
                }
            }
        }
        return dp[m][n];
    }
}
```

## 3. Time & Space Complexity

**DP:** 时间复杂度O(mn), 空间复杂度O(mn)
