---
description: stack
---

# LeetCode 1249. Minimum Remove to Make Valid Parentheses

Given a string s of `'('` , `')'` and lowercase English characters.&#x20;

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)"
```

**Constraints:**

* `1 <= s.length <= 10^5`
* `s[i]` is one of  `'('` , `')'` and lowercase English letters`.`

## Solution

[English Version in Youtube](https://youtu.be/S4LLF3uT-EY)

[中文版解答Youtube Link](https://youtu.be/vxMmmvdnuNs)

[中文版解答Bilibili Link](https://www.bilibili.com/video/BV1n64y1v7Nm/)

```
class Solution {
public:
    string minRemoveToMakeValid(string s) {
        stack<int> mystack;
        
        for (int i = 0; i < s.length(); i++) {
            char ch = s[i];
            if (ch == '(') {
                mystack.push(i);
            } else if (ch == ')') {
                if (!mystack.empty()) {
                    mystack.pop();
                } else {
                    s[i] = '?';
                }
            }
        }
        
        while (!mystack.empty()) {
            s[mystack.top()] = '?';
            mystack.pop();
        }
        
        // https://www.geeksforgeeks.org/remove-spaces-from-a-given-string/
        s.erase(remove(s.begin(), s.end(), '?'), s.end());
        return s;
    }
};
```
