32. Longest Valid Parentheses

Explanation:

To solve this problem, we can use a stack to keep track of the indices of the opening parentheses. We iterate through the input string and whenever we encounter an opening parenthesis, we push its index onto the stack. When we encounter a closing parenthesis, we check if the stack is empty. If it is empty, we update the starting index for the next valid substring. If the stack is not empty, we pop the top index from the stack and calculate the length of the current valid substring.

We also need to handle cases where there are unmatched parentheses. To do this, we initialize the starting index as -1 and update it whenever we encounter a closing parenthesis with an empty stack.

Finally, the length of the longest valid substring will be the maximum difference between the current index and the starting index.

Time Complexity: O(n) - where n is the length of the input string. Space Complexity: O(n) - for the stack to store indices.

:

View Leetcode Problem

class Solution {
    public int longestValidParentheses(String s) {
        int maxLen = 0;
        Stack<Integer> stack = new Stack<>();
        stack.push(-1);
        
        for (int i = 0; i < s.length(); i++) {
            if (s.charAt(i) == '(') {
                stack.push(i);
            } else {
                stack.pop();
                if (stack.isEmpty()) {
                    stack.push(i);
                } else {
                    maxLen = Math.max(maxLen, i - stack.peek());
                }
            }
        }
        
        return maxLen;
    }
}

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