Tags: "leetcode", "stack", access_time 2-min read

Minimum Remove to Make Valid Parentheses

Created: January 23, 2020 by [lek-tin]

Last updated: January 23, 2020

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:

1. It is the empty string, contains only lowercase characters, or
2. It can be written as AB (A concatenated with B), where A and B are valid strings, or
3. 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.

Example 4

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

Constraints:

1. 1 <= s.length <= 10^5
2. s[i] is one of '(' , ')' and lowercase English letters.

Solution

class Solution:
    def minRemoveToMakeValid(self, s: str) -> str:
        stack = []
        indices = []
        for i, char in enumerate(s):
            if char != '(' and char != ')':
                continue
            elif char == '(':
                stack.append(i)
            elif stack:
                stack.pop()
            else:
                indices.append(i)

        indices += stack

        newStr = ""

        for i, char in enumerate(s):
            if i not in indices:
                newStr += char

        return newStr