## Algorithm

Problem Name: 712. Minimum ASCII Delete Sum for Two Strings

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.
```

Constraints:

• `1 <= s1.length, s2.length <= 1000`
• `s1` and `s2` consist of lowercase English letters.

## Code Examples

### #1 Code Example with Javascript Programming

```Code - Javascript Programming```

``````
const minimumDeleteSum = function(s1, s2) {
const l1 = s1.length;
const l2 = s2.length;
const dp = [];
for (let i = 0; i <= l1; i++) {
dp[i] = [];
}
let sum = 0;
for (let i = 0; i <= l1; i++) {
for (let j = 0; j <= l2; j++) {
if (i === 0 || j === 0) {
sum = 0;
for (let k = 0; k < Math.max(i, j); k++) {
sum += i > j ? s1.charCodeAt(k) : s2.charCodeAt(k);
}
dp[i][j] = sum;
} else {
if (s1[i - 1] === s2[j - 1]) {
dp[i][j] = dp[i - 1][j - 1];
} else {
dp[i][j] = Math.min(
s1.charCodeAt(i - 1) + dp[i - 1][j],
s2.charCodeAt(j - 1) + dp[i][j - 1],
s1.charCodeAt(i - 1) + s2.charCodeAt(j - 1) + dp[i - 1][j - 1]
);
}
}
}
}
return dp[l1][l2];
};
``````
Copy The Code &

Input

cmd
s1 = "sea", s2 = "eat"

Output

cmd
231

### #2 Code Example with Python Programming

```Code - Python Programming```

``````
class Solution:
def minimumDeleteSum(self, s1, s2):
l1, l2 = len(s1) + 1, len(s2) + 1
d = [ * l2 for i in range(l1)]
for i in range(l1):
for j in range(l2):
c1, c2 = ord(s1[i - 1]), ord(s2[j - 1])
if not i * j:
d[i][j] = d[i - 1][j] + c1 if i else d[i][j - 1] + c2 if j else 0
elif s1[i - 1] == s2[j - 1]:
d[i][j] = d[i - 1][j - 1]
else:
d[i][j] = min(d[i - 1][j] + c1, d[i][j - 1] + c2, d[i - 1][j - 1] + c1 + c2)
return d[-1][-1]
``````
Copy The Code &

Input

cmd
s1 = "sea", s2 = "eat"

Output

cmd
231