LeetCode 1249: Minimum Remove to Make Valid Parentheses Solution

Master LeetCode problem 1249 (Minimum Remove to Make Valid Parentheses), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1249. Minimum Remove to Make Valid Parentheses

Medium

String Stack

Problem Explanation

Explanation:

To solve this problem, we can use a stack to keep track of the indices of opening parentheses ( that are not matched with a closing parenthesis ). We will also use a set to keep track of the indices of closing parentheses ) that are not matched with an opening parenthesis (.

Iterate through the input string s to find and store the indices of mismatched parentheses in the stack and set.
Construct a new string by ignoring the characters at the indices stored in the stack and set.
Return the constructed string as the minimum removed valid parentheses string.

Time Complexity: O(n) where n is the length of the input string s. Space Complexity: O(n) for the stack and set.

Solution Code

class Solution {
    public String minRemoveToMakeValid(String s) {
        Stack<Integer> stack = new Stack<>();
        Set<Integer> set = new HashSet<>();
        
        for (int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            if (c == '(') {
                stack.push(i);
            } else if (c == ')') {
                if (!stack.isEmpty()) {
                    stack.pop();
                } else {
                    set.add(i);
                }
            }
        }
        
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < s.length(); i++) {
            if (!stack.isEmpty() && stack.peek() == i) {
                stack.pop();
            } else if (set.contains(i)) {
                continue;
            } else {
                sb.append(s.charAt(i));
            }
        }
        
        return sb.toString();
    }
}

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