1249. Minimum Remove to Make Valid Parentheses

Given a string s of '(' , ')' and lowercase English characters.

Your task is to remove the minimum number of parentheses ('(' or ')', in any positions ) so that the resulting parentheses string is valid and return any valid string.

Formally, a parentheses string is valid if and only if:

It is the empty string, contains only lowercase characters, or It can be written as AB (A concatenated with B), where A and B are valid strings, or It can be written as (A), where A is a valid string.

Example 1:

Input: s = "lee(t(c)o)de)"
Output: "lee(t(c)o)de"
Explanation: "lee(t(co)de)" , "lee(t(c)ode)" would also be accepted.

Example 2:

Input: s = "a)b(c)d"
Output: "ab(c)d"

Example 3:

Input: s = "))(("
Output: ""
Explanation: An empty string is also valid.

Constraints:

1 <= s.length <= 105
s[i] is either'(' , ')', or lowercase English letter.

Complexity

Time complexity:

• O(n)

Space complexity:

• O(n)

Code

class Solution:
    def minRemoveToMakeValid(self, s: str) -> str:
        left_count = 0
        right_count = 0
        stack = []

        for ch in s:
            if ch == '(':
                left_count += 1
            elif ch == ')':
                right_count += 1
            if right_count > left_count:
                right_count -= 1
            else:
                stack.append(ch)

        result = ""

        while stack:
            current_char = stack.pop()
            if left_count > right_count and current_char == '(':
                left_count -= 1
            else:
                result += current_char

        return result[::-1]