LeetCode in Kotlin
1750. Minimum Length of String After Deleting Similar Ends
Medium
Given a string s consisting only of characters 'a', 'b', and 'c'. You are asked to apply the following algorithm on the string any number of times:
- Pick a non-empty prefix from the string
swhere all the characters in the prefix are equal. - Pick a non-empty suffix from the string
swhere all the characters in this suffix are equal. - The prefix and the suffix should not intersect at any index.
- The characters from the prefix and suffix must be the same.
- Delete both the prefix and the suffix.
Return the minimum length of s after performing the above operation any number of times (possibly zero times).
Example 1:
Input: s = “ca”
Output: 2
Explanation: You can’t remove any characters, so the string stays as is.
Example 2:
Input: s = “cabaabac”
Output: 0
Explanation: An optimal sequence of operations is:
-
Take prefix = “c” and suffix = “c” and remove them, s = “abaaba”.
-
Take prefix = “a” and suffix = “a” and remove them, s = “baab”.
-
Take prefix = “b” and suffix = “b” and remove them, s = “aa”.
-
Take prefix = “a” and suffix = “a” and remove them, s = “”.
Example 3:
Input: s = “aabccabba”
Output: 3
Explanation: An optimal sequence of operations is:
-
Take prefix = “aa” and suffix = “a” and remove them, s = “bccabb”.
-
Take prefix = “b” and suffix = “bb” and remove them, s = “cca”.
Constraints:
1 <= s.length <= 105sonly consists of characters'a','b', and'c'.
Solution
class Solution {
fun minimumLength(s: String): Int {
var i = 0
var j: Int = s.length - 1
if (s[i] == s[j]) {
while (i < j && s[i] == s[j]) {
val c: Char = s[i]
i++
while (c == s[i] && i < j) {
i++
}
j--
while (c == s[j] && i < j) {
j--
}
}
}
return if (i <= j) s.substring(i, j).length + 1 else 0
}
}