# Minimum Remove to Make Valid Parentheses

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

Solution:

```class Solution {
public String minRemoveToMakeValid(String s) {
Deque<Integer> stack = new ArrayDeque<>();
Set<Integer> remove = new HashSet<>();
for (int i = 0; i < s.length(); i ++) {
char c = s.charAt(i);
if (c == '(') {
stack.push(i);
}
if (c == ')') {
if (stack.isEmpty()) {
} else {
stack.pop();
}
}
}
while (!stack.isEmpty()) {