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 <= 1000s1ands2consist 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];
};
Input
s1 = "sea", s2 = "eat"
Output
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 = [[0] * 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]
Input
s1 = "sea", s2 = "eat"
Output
231