LeetCode - 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.

class Solution {
    public String minRemoveToMakeValid(String s) {
        int[] arr = new int[s.length()];
        Stack<Integer> stack = new Stack<>();
        
        for (int i = 0; i < s.length(); i ++) {
            if (s.charAt(i) == ‘(‘) {
                stack.push(i);
            } 
            else if (s.charAt(i) == ‘)‘) {
                if (stack.isEmpty()) {
                    arr[i] = 1;
                }
                else {
                    stack.pop();
                }
            }
        }
        for (Integer item: stack) {
            arr[item] = 1;
        }
        
        StringBuilder sb = new StringBuilder();
        for (int k = 0; k < s.length(); k++) {
            if (arr[k] != 1) {
                sb.append(s.charAt(k));
            }
        }
        return sb.toString();
    }
}