Minimum Add to Make Parentheses Valid

Given a string Sof'('and')'parentheses, we add the minimum number of parentheses ('('or')', and in any positions ) so that the resulting parentheses string is valid.

Formally, a parentheses string is valid if and only if:

• It is the empty string, 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.

Given a parentheses string, return the minimum number of parentheses we must add to make the resulting string valid.

Example 1:

Input: 
"())"
Output: 
1

Example 2:

Input: 
"((("
Output: 
3

Example 3:

Input: 
"()"
Output: 
0

Example 4:

Input: 
"()))(("
Output: 
4

分析

注意这里不能单纯比较()个数，也要比较出场顺序。所以counter是算顺序，）stk是算残余多少的 （

class Solution:
    def minAddToMakeValid(self, S):
        """
        :type S: str
        :rtype: int
        """
        left = right = 0
        for i in S:
            if i == '(':
                left +=1
            elif left<=0:#都平衡了，再多个)就不行了 所以要<=
                right+=1
            else:
                left -= 1
        return left + right