LeetCode in Kotlin
1021. Remove Outermost Parentheses
Easy
A valid parentheses string is either empty "", "(" + A + ")", or A + B, where A and B are valid parentheses strings, and + represents string concatenation.
- For example,
"","()","(())()", and"(()(()))"are all valid parentheses strings.
A valid parentheses string s is primitive if it is nonempty, and there does not exist a way to split it into s = A + B, with A and B nonempty valid parentheses strings.
Given a valid parentheses string s, consider its primitive decomposition: s = P1 + P2 + ... + Pk, where Pi are primitive valid parentheses strings.
Return s after removing the outermost parentheses of every primitive string in the primitive decomposition of s.
Example 1:
Input: s = “(()())(())”
Output: “()()()”
Explanation: The input string is “(()())(())”, with primitive decomposition “(()())” + “(())”. After removing outer parentheses of each part, this is “()()” + “()” = “()()()”.
Example 2:
Input: s = “(()())(())(()(()))”
Output: “()()()()(())”
Explanation: The input string is “(()())(())(()(()))”, with primitive decomposition “(()())” + “(())” + “(()(()))”. After removing outer parentheses of each part, this is “()()” + “()” + “()(())” = “()()()()(())”.
Example 3:
Input: s = “()()”
Output: “”
Explanation: The input string is “()()”, with primitive decomposition “()” + “()”. After removing outer parentheses of each part, this is “” + “” = “”.
Constraints:
1 <= s.length <= 105s[i]is either'('or')'.sis a valid parentheses string.
Solution
class Solution {
fun removeOuterParentheses(s: String): String {
val primitives: MutableList<String> = ArrayList()
var i = 1
while (i < s.length) {
val initialI = i - 1
var left = 1
while (i < s.length && left > 0) {
if (s[i] == '(') {
left++
} else {
left--
}
i++
}
primitives.add(s.substring(initialI, i))
i++
}
val sb = StringBuilder()
for (primitive in primitives) {
sb.append(primitive, 1, primitive.length - 1)
}
return sb.toString()
}
}