921. Minimum Add to Make Parentheses Valid

Explanation:

To solve this problem, we can use a stack to keep track of the unbalanced parentheses in the string. We iterate through the string, and for each opening parenthesis, we push its index into the stack. If we encounter a closing parenthesis, we check if the stack is empty or the top element in the stack corresponds to an opening parenthesis. If it does, we pop from the stack. At the end, the size of the stack will tell us how many additional parentheses we need to balance the string.

Time complexity: O(n) where n is the length of the input string.
Space complexity: O(n) where n is the length of the input string.

:

View Leetcode Problem

class Solution {
    public int minAddToMakeValid(String s) {
        Stack<Integer> stack = new Stack<>();
        int count = 0;
        
        for (int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            if (c == '(') {
                stack.push(i);
            } else {
                if (!stack.isEmpty() && s.charAt(stack.peek()) == '(') {
                    stack.pop();
                } else {
                    count++;
                }
            }
        }
        
        return count + stack.size();
    }
}

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