Minimum Remove to Make Valid Parentheses

Published: Sep 7, 2022

Introduction

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.

Constraints:

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

https://leetcode.com/problems/minimum-remove-to-make-valid-parentheses/

Examples

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: ""

Analysis

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.

Solution

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

Complexities

• Time: O(n)
• Space: O(n)