Leetcode 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.

The simplest way to solve this problem is via stack. We need 2 stacks to make sure that there is no closed parenthese appears more than open parenthese or before any open parenthese.

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

        for i in range(len(s)):
            if s[i] == '(':
                stack.append(i)
            elif s[i] == ')':
                if stack:
                    stack.pop()
                else:
                    right_parentheses.append(i)
        
        s_list = list(s)
        final_remove = stack + right_parentheses
        final_remove.sort(reverse=True)

        for index in final_remove:
            s_list.pop(index)

        return "".join(s_list)

The entire idea is easy, but tricky part will be how to remove the invalid character from the given string.

My first try was like

result = ""
for i in range(len(s)):
     if i not in final_remove:
          result += s[i]

It works, but performance is really bad.