Minimum Remove to Make Valid Parentheses

1249. 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"

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

/**
 * @param {string} s
 * @return {string}
 */
 var minRemoveToMakeValid = function(s) {
    const out = [...s], open = [];

    for (let i = 0; i < s.length; i++) {
        if (s[i] === '(') {
           open.push(i);
        } else if (s[i] === ')') {
           open.length > 0 ? open.pop() : out[i] = '';
        }
    } 
    open.forEach(idx => out[idx] = '');
    return out.join('');
};