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.

Example 4:

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


  • 1 <= s.length <= 105
  • s[i] is either'(' , ')', or lowercase English letter.


class Solution(object):
    def minRemoveToMakeValid(self, s):
        stack, ret = [], ['']*len(s)        
        for i,c in enumerate(s):
            if c=='(':
            elif c==')':
                if stack:
                    ret[left] = s[left]
                    ret[i] = c
                ret[i] = c
        return ''.join(ret)


class Solution:
    def minRemoveToMakeValid(self, s):
        stack, res = [], ""
        for i, ch in enumerate(s):
            if ch in {'(', ')'}: 
                if ch == ')':
                    if stack and stack[-1][1] == '(':
                stack.append((i, ch))
        # Whatever left in the stack are the ones to remove.
        # Initialize a set for constant-time lookup. 
        idx_set = set([i for i, _ in stack])
        return ''.join([ch for i, ch in enumerate(s) if i not in idx_set])


class Solution:
    def minRemoveToMakeValid(self, s):
        # stack = string segments for outStr, cur = current string segment
        stack, cur = [], ''
        for c in s:
            if c == '(':
                # add current string segment to stack, cos we're now at new stack
                stack += [cur]
                cur = ''
            elif c == ')':
                # since segments are only added via '(', this means that the previous segment was preceded by a '('
                if stack:
                    cur = stack.pop() + '(' + cur + ')' 
                cur += c
        # add all the string segments together
        while stack:
            cur = stack.pop() + cur
        return cur

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