# Minimum Add to Make Parentheses Valid

Given a string `S`of`'('`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
```