LeetCode in Kotlin
712. Minimum ASCII Delete Sum for Two Strings
Medium
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.
Solution
class Solution {
fun minimumDeleteSum(s1: String, s2: String): Int {
val len1 = s1.length
val len2 = s2.length
val dp = Array(len1 + 1) { IntArray(len2 + 1) }
var c1: Char
var c2: Char
for (i in 1 until len1 + 1) {
dp[i][0] = dp[i - 1][0] + s1[i - 1].code
}
for (j in 1 until len2 + 1) {
dp[0][j] = dp[0][j - 1] + s2[j - 1].code
}
for (i in 1 until len1 + 1) {
c1 = s1[i - 1]
for (j in 1 until len2 + 1) {
c2 = s2[j - 1]
if (c1 == c2) {
dp[i][j] = dp[i - 1][j - 1]
} else {
dp[i][j] = (dp[i - 1][j] + c1.code).coerceAtMost(dp[i][j - 1] + c2.code)
}
}
}
return dp[len1][len2]
}
}