Problem description:

A parentheses string is valid if and only if:

  • It is the empty string,
  • 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.

You are given a parentheses string s. In one move, you can insert a parenthesis at any position of the string.

  • For example, if s = "()))", you can insert an opening parenthesis to be "(**(**)))" or a closing parenthesis to be "())**)**)".

Return the minimum number of moves required to make s valid.

Example 1:

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Input: s = "())"
Output: 1

Example 2:

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Input: s = "((("
Output: 3

Example 3:

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Input: s = "()"
Output: 0

Example 4:

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Input: s = "()))(("
Output: 4

Constraints:

  • 1 <= s.length <= 1000
  • s[i] is either '(' or ')'.

Solution:

Count invalid left and right 

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class Solution:
    def minAddToMakeValid(self, s: str) -> int:
        l, r = 0, 0
        for c in s:
            if c == '(':
                l += 1
            elif l > 0:
                l -= 1
            else:
                r += 1
        return l + r

Stack

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class Solution:
    def minAddToMakeValid(self, s: str) -> int:
        stack = list()
        result = 0
        for i in range(len(s)):
            if s[i] == "(":
                stack.append(s[i])
            else:
                if stack:
                    stack.pop()
                else:
                    result += 1
                    
        return result+len(stack)

time complexity: $O()$
space complexity: $O()$
reference:
related problem: