leetcode - 1249. Minimum Remove to Make Valid Parentheses

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.

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 <= 10^5
s[i] is either'(' , ')', or lowercase English letter.

Solution

List

Go through the string from left to right, use a left_cnt to keep track of how many ( we have visited, and decrease when there’s a ), discard redundant ) during this process.

Go through the string again, this time from right to left, use a right_cnt to keep track of how may ) we have, and decrease when there’s a (, discard ( this time.

Time complexity: $o(n)$
Space complexity: $o(n)$

Stack

Use a stack to store all the (, and pop when there’s a ), at the end, all the remaining (s are those we need to discard.

Time complexity: $o(n)$
Space complexity: $o(n)$

Code

List

class Solution:
    def minRemoveToMakeValid(self, s: str) -> str:
        left_cnt = 0
        res = []
        for i in range(len(s)):
            if s[i] not in ('(', ')'):
                res += s[i]
            elif s[i] == '(':
                res += s[i]
                left_cnt += 1
            elif s[i] == ')' and left_cnt > 0:
                res += s[i]
                left_cnt -= 1
        right_cnt = 0
        new_res = ''
        for i in range(len(res) - 1, -1, -1):
            if res[i] not in ('(', ')'):
                new_res += res[i]
            elif res[i] == ')':
                new_res += res[i]
                right_cnt += 1
            elif res[i] == '(' and right_cnt > 0:
                new_res += res[i]
                right_cnt -= 1
        return new_res[::-1]

Stack

class Solution:
    def minRemoveToMakeValid(self, s: str) -> str:
        stack = []
        discard = []
        for i in range(len(s)):
            if s[i] == '(':
                stack.append(i)
            elif s[i] == ')':
                if stack:
                    stack.pop()
                else:
                    discard.append(i)
        discard += stack
        res = []
        for i in range(len(s)):
            if i not in discard:
                res.append(s[i])
        return ''.join(res)