You are given a string s consisting of '(' and ')', and an integer k.
A string is k-balanced if it is exactly k consecutive '(' followed by k consecutive ')', i.e., '(' * k + ')' * k.
For example, if k = 3, k-balanced is "((()))".
You must repeatedly remove all non-overlapping k-balanced substrings from s, and then join the remaining parts. Continue this process until no k-balanced substring exists.
Return the final string after all possible removals.
Example 1:
Input: s = "(())", k = 1
Output: ""
Explanation:
k-balanced substring is "()"
| Step | Current s |
k-balanced |
Result s |
|---|---|---|---|
| 1 | (()) |
( |
() |
| 2 | () |
() |
Empty |
Thus, the final string is "".
Example 2:
Input: s = "(()(", k = 1
Output: "(("
Explanation:
k-balanced substring is "()"
| Step | Current s |
k-balanced |
Result s |
|---|---|---|---|
| 1 | (()( |
( |
(( |
| 2 | (( |
- | (( |
Thus, the final string is "((".
Example 3:
Input: s = "((()))()()()", k = 3
Output: "()()()"
Explanation:
k-balanced substring is "((()))"
| Step | Current s |
k-balanced |
Result s |
|---|---|---|---|
| 1 | ((()))()()() |
|
()()() |
| 2 | ()()() |
- | ()()() |
Thus, the final string is "()()()".
Constraints:
2 <= s.length <= 105s consists only of '(' and ')'.1 <= k <= s.length / 2