Minimum Remove to Make Valid Parentheses

Published: Sep 7, 2022

Medium Stack String


Parentheses related string problems can be solved by a stack in general. The string might include other characters. However, focusing parentheses only would be enough in many cases.

Problem Description

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.


  • 1 <= s.length <= 10**5
  • s[i] is either’(‘ , ‘)’, or lowercase English letter.


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: ""


Using the stack, check the validity of parentheses. At the same time, save invalid indices in a set. The final step is to construct a valid string while omitting invalid indices.


class MinimumRemoveToMakeValidParentheses:
    def minRemoveToMakeValid(self, s: str) -> str:
        n, stack, ss = len(s), [], set()
        for i in range(n):
            if s[i] == '(':
            elif s[i] == ')':
                if stack:
        ss = ss.union(set(stack))
        result = []
        for i in range(n):
            if i not in ss:
        return ''.join(result)


  • Time: O(n)
  • Space: O(n)