# 1249. Minimum Remove to Make Valid Parentheses

<https://leetcode.com/problems/minimum-remove-to-make-valid-parentheses>

## Description

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)"
```

**Constraints:**

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

## ac

```java
class Solution {
    public String minRemoveToMakeValid(String s) {
        // Edge cases
        if (s == null || s.length() == 0) return s;
        
        Stack<Integer> left = new Stack<>();
        Stack<Integer> right = new Stack<>();
        HashSet<Integer> toRemove = new HashSet<>();
        
        for (int i = 0; i < s.length(); i++) {
            if (s.charAt(i) == '(') {
                left.push(i);
            } else if (s.charAt(i) == ')') {
                right.push(i);
            }
             
            if (right.size() > left.size()) {
                toRemove.add(right.pop());
            }
        }
        
        while (left.size() > right.size()) {
            toRemove.add(left.pop());
        }
        
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < s.length(); i++) {
            if (toRemove.contains(i)) continue;
            sb.append(s.charAt(i));
        }
        
        return sb.toString();
    }
}

// O(N) time, O(N) space. Parentheses->left&right. Similar to # 022-Generate Parentheses.
```


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