Add longest-valid-parentheses #3038

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	/**
	* @file
	* @details
	* Given a string containing just the characters '(' and ')',
	* find the length of the longest valid (well-formed) parentheses substring.
	*
	* ### Approach
	* This solution uses Dynamic Programming.
	* We maintain an array `longest[]` where `longest[i]` represents
	* the length of the longest valid parentheses substring ending at index `i`.
	*
	* - If `s[i] == '('`, then `longest[i] = 0` (a valid substring cannot end with '(').
	* - If `s[i] == ')'`:
	* - If `s[i - 1] == '('`, then `longest[i] = longest[i - 2] + 2`
	* - Else if `s[i - 1] == ')'` and `s[i - longest[i - 1] - 1] == '('`,
	* then `longest[i] = longest[i - 1] + 2 + longest[i - longest[i - 1] - 2]`
	*
	* Example:
	* Input: "()(())"
	* At i = 5, longest = [0, 2, 0, 0, 2, 0]
	* longest[5] = longest[4] + 2 + longest[1] = 6
	*
	* @note
	* Time Complexity: O(n)
	* Space Complexity: O(n)
	*/

	#include <bits/stdc++.h>
	using namespace std;

	/**
	* @brief Dynamic Programming based solution to find the longest valid parentheses substring.
	* @param s Input string containing '(' and ')'
	* @return Length of the longest valid parentheses substring
	*/
	int longestValidParentheses(const string &s) {
	if (s.length() <= 1) return 0;

	int curMax = 0;
	vector<int> longest(s.size(), 0);

	for (int i = 1; i < s.length(); i++) {
	if (s[i] == ')') {
	if (s[i - 1] == '(') {
	longest[i] = (i - 2 >= 0 ? longest[i - 2] + 2 : 2);
	} else if (i - longest[i - 1] - 1 >= 0 && s[i - longest[i - 1] - 1] == '(') {
	longest[i] = longest[i - 1] + 2 +
	((i - longest[i - 1] - 2 >= 0) ? longest[i - longest[i - 1] - 2] : 0);
	}
	curMax = max(curMax, longest[i]);
	}
	}
	return curMax;
	}

	/**
	* @brief Concise version
	* Same logic as above but written in a more compact form.
	*/
	int longestValidParenthesesConcise(const string &s) {
	if (s.length() <= 1) return 0;

	int curMax = 0;
	vector<int> longest(s.size(), 0);

	for (int i = 1; i < s.length(); i++) {
	if (s[i] == ')' && i - longest[i - 1] - 1 >= 0 && s[i - longest[i - 1] - 1] == '(') {
	longest[i] = longest[i - 1] + 2 +
	((i - longest[i - 1] - 2 >= 0) ? longest[i - longest[i - 1] - 2] : 0);
	curMax = max(curMax, longest[i]);
	}
	}
	return curMax;
	}

	/**
	* @brief Driver code for demonstration and basic testing.
	*/
	int main() {
	vector<string> test_cases = {
	"(()", // expected 2
	")()())", // expected 4
	"()(())", // expected 6
	"((((((", // expected 0
	"()(()))))" // expected 6
	};

	cout << "Testing Longest Valid Parentheses using DP:\n";
	for (const auto &s : test_cases) {
	cout << "Input: " << s
	<< " \| Output: " << longestValidParentheses(s) << endl;
	}

	cout << "\nTesting Concise Version:\n";
	for (const auto &s : test_cases) {
	cout << "Input: " << s
	<< " \| Output: " << longestValidParenthesesConcise(s) << endl;
	}

	return 0;
	}

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