# Length of the longest valid substring

Last Updated : 11 Apr, 2024

Given a string consisting of opening and closing parenthesis, find the length of the longest valid parenthesis substring.

Examples:

`Input: ((()Output : 2Explanation : ()Input: )()())Output : 4Explanation: ()() Input:  ()(()))))Output: 6Explanation:  ()(())`
Recommended Practice

A Simple Approach is to find all the substrings of given string. For every string, check if it is a valid string or not. If valid and length is more than maximum length so far, then update maximum length. We can check whether a substring is valid or not in linear time using a stack (See this for details).

Time complexity of this solution is O(n3).
Space Complexity: O(1)

Approach 1 : The approach remains the same as Valid Parenthesis. The only hard part is to calculate the length. That can be done if the stack holds the indices of the not valid braces indices. Eg : “)()())” for this string the stack will hold the indices 0 and 5 as on these indices the braces are not valid.

Let’s get a quick run of the proposed algorithm:

1. at i = 0, open bracket so push the index 0. Stack contains: 0
2. at i = 1, open bracket so push the index 1. Stack contains: 0 1
3. at i = 2, closed bracket. Stack is not empty and on the top that is an index of open bracket. So pop. Stack contains only 0
4. at i = 3, again open and at i = 4 close bracket to again a push and then pop. Stack contains : 0
5. at i = 5, a close bracket. Stack is not empty but on top it’s a close bracket index. So push index 5. Stack contains : 0 5

Now if you see closely. These are the indices of the point where the not valid cases happen. Now in between these 2 indices there must be a valid string.
NOTE : it might happen that after the index 5, there is a valid string that won’t be in the stack to calculate. But it can easily be calculated if while popping out. Similarly on the left of index 0 there can be a valid string ( think given string “()() ) ()() )” to understand).

To handle the right side of index 5, “last” is initialized with N and on each iteration last is updated with now which is the current index. And to handle the left of index 0 (as mentioned) will return max(result, last). [ Cause if standing on index i, the length of the left part is i ]

So to find out the longest valid parenthesis these differences will contribute. To understand it better follow the below code :

C++ ```// C++ program to find length of the // longest valid substring #include <bits/stdc++.h> using namespace std; int findMaxLen(string str) { int n = str.length(); // Create a stack stack<int> stk; // Traverse all characters of given string for (int i = 0; i < n; i++) { // If opening bracket, push index of it if (str[i] == '(') stk.push(i); // If closing bracket, i.e.,str[i] = ')' else { // If the stack is not empty and on the top // that is and index of a open bracket then pop if (!stk.empty() and str[stk.top()] == '(') stk.pop(); // If stack is empty. push current index as // base for next valid substring (if any) else stk.push(i); } } // Initialize the result // 'last' is initialized // to calculate the distance int result = 0, last = n; while(!stk.empty()) { int now = stk.top(); stk.pop(); // take the maximum in result result = max(result, last-now-1); // update the last index // with current index last = now; } // return the maximum of last and result return max(result, last); } // Driver code int main() { string str = "((()()"; // Function call cout << findMaxLen(str) << endl; str = "()(()))))"; // Function call cout << findMaxLen(str) << endl; return 0; } ``` Java ```import java.util.Stack; public class Main { public static int findMaxLen(String str) { int n = str.length(); // Create a stack Stack<Integer> stk = new Stack<>(); // Traverse all characters of given string for (int i = 0; i < n; i++) { // If opening bracket, push index of it if (str.charAt(i) == '(') stk.push(i); // If closing bracket else { // If the stack is not empty and on the top // that is and index of an open bracket then pop if (!stk.isEmpty() && str.charAt(stk.peek()) == '(') stk.pop(); // If stack is empty, push current index as // base for next valid substring (if any) else stk.push(i); } } // Initialize the result, 'last' is initialized // to calculate the distance int result = 0, last = n; while (!stk.isEmpty()) { int now = stk.pop(); // take the maximum in result result = Math.max(result, last - now - 1); // update the last index with current index last = now; } // return the maximum of last and result return Math.max(result, last); } // Driver code public static void main(String[] args) { String str = "((()()"; // Function call System.out.println(findMaxLen(str)); str = "()(()))))"; // Function call System.out.println(findMaxLen(str)); } } ``` JavaScript ```function GFG(str) { const n = str.length; // Create an array to act as a stack const stack = []; // Traverse all characters of the given string for (let i = 0; i < n; i++) { if (str[i] === '(') stack.push(i); else { // If the stack is not empty and on top // that is and index of the open bracket then pop if (stack.length && str[stack[stack.length - 1]] === '(') stack.pop(); else stack.push(i); } } let result = 0, last = n; // Iterate through the stack to the calculate the maximum valid substring length while(stack.length) { const now = stack.pop(); // take the maximum in the result result = Math.max(result, last - now - 1); last = now; } // Return the maximum of the last and result return Math.max(result, last); } // Driver code const str1 = "((()()"; console.log(GFG(str1)); const str2 = "()(()))))"; console.log(GFG(str2)); ```

Output
```4
6
```
`import java.util.Stack;public class LongestValidParenthesis {    public static int longestValidParentheses(String s) {        int n = s.length();        Stack<Integer> stack = new Stack<>();        stack.push(-1); // Push -1 as a base index to handle the edge case when there's no valid prefix        int result = 0;        int last = -1; // To track the last not valid index        for (int i = 0; i < n; i++) {            char c = s.charAt(i);            if (c == '(') {                stack.push(i); // Push the index of opening parenthesis            } else {                stack.pop(); // Pop the index of matching opening parenthesis                if (stack.isEmpty()) {                    stack.push(i); // Push the current index if stack is empty                } else {                    // Calculate the length of valid substring                    result = Math.max(result, i - stack.peek());                }            }        }        // If stack is not empty, calculate the length of valid substring from the last invalid index till the end        while (!stack.isEmpty()) {            int now = stack.pop();            result = Math.max(result, last - now);            last = now; // Update the last not valid index        }        return result;    }    public static void main(String[] args) {        String[] testCases = {"((()", ")()())", "()(()))))"};        for (String testCase : testCases) {            System.out.println("Longest valid parenthesis for \"" + testCase + "\": " + longestValidParentheses(testCase));        }    }}`

Time Complexity: O(N), here N is the length of string.
Auxiliary Space: O(N)

In the previous approach without using 2 different loops the whole approach can also be done in a single loop.

Below is the implementation of the above algorithm.

C++ ```// C++ program to find length of the // longest valid substring #include <bits/stdc++.h> using namespace std; int findMaxLen(string str) { int n = str.length(); // Create a stack and push -1 as // initial index to it. stack<int> stk; stk.push(-1); // Initialize result int result = 0; // Traverse all characters of given string for (int i = 0; i < n; i++) { // If opening bracket, push index of it if (str[i] == '(') stk.push(i); // If closing bracket, i.e.,str[i] = ')' else { // Pop the previous opening // bracket's index if (!stk.empty()) { stk.pop(); } // Check if this length formed with base of // current valid substring is more than max // so far if (!stk.empty()) result = max(result, i - stk.top()); // If stack is empty. push current index as // base for next valid substring (if any) else stk.push(i); } } return result; } // Driver code int main() { string str = "((()()"; // Function call cout << findMaxLen(str) << endl; str = "()(()))))"; // Function call cout << findMaxLen(str) << endl; return 0; } ``` Java ```// Java program to find length of the longest valid // substring import java.util.Stack; class Test { // method to get length of the longest valid static int findMaxLen(String str) { int n = str.length(); // Create a stack and push -1 // as initial index to it. Stack<Integer> stk = new Stack<>(); stk.push(-1); // Initialize result int result = 0; // Traverse all characters of given string for (int i = 0; i < n; i++) { // If opening bracket, push index of it if (str.charAt(i) == '(') stk.push(i); // // If closing bracket, i.e.,str[i] = ')' else { // Pop the previous // opening bracket's index if(!stk.empty()) stk.pop(); // Check if this length // formed with base of // current valid substring // is more than max // so far if (!stk.empty()) result = Math.max(result, i - stk.peek()); // If stack is empty. push // current index as base // for next valid substring (if any) else stk.push(i); } } return result; } // Driver code public static void main(String[] args) { String str = "((()()"; // Function call System.out.println(findMaxLen(str)); str = "()(()))))"; // Function call System.out.println(findMaxLen(str)); } } ``` Python3 ```# Python program to find length of the longest valid # substring def findMaxLen(string): n = len(string) # Create a stack and push -1 # as initial index to it. stk = [] stk.append(-1) # Initialize result result = 0 # Traverse all characters of given string for i in range(n): # If opening bracket, push index of it if string[i] == '(': stk.append(i) # If closing bracket, i.e., str[i] = ')' else: # Pop the previous opening bracket's index if len(stk) != 0: stk.pop() # Check if this length formed with base of # current valid substring is more than max # so far if len(stk) != 0: result = max(result, i - stk[len(stk)-1]) # If stack is empty. push current index as # base for next valid substring (if any) else: stk.append(i) return result # Driver code string = "((()()" # Function call print (findMaxLen(string)) string = "()(()))))" # Function call print (findMaxLen(string)) # This code is contributed by Bhavya Jain # This code is modified by Susobhan Akhuli ``` C# ```// C# program to find length of // the longest valid substring using System; using System.Collections.Generic; class GFG { // method to get length of // the longest valid public static int findMaxLen(string str) { int n = str.Length; // Create a stack and push -1 as // initial index to it. Stack<int> stk = new Stack<int>(); stk.Push(-1); // Initialize result int result = 0; // Traverse all characters of // given string for (int i = 0; i < n; i++) { // If opening bracket, push // index of it if (str[i] == '(') { stk.Push(i); } else // If closing bracket, // i.e.,str[i] = ')' { // Pop the previous opening // bracket's index if (stk.Count > 0) stk.Pop(); // Check if this length formed // with base of current valid // substring is more than max // so far if (stk.Count > 0) { result = Math.Max(result, i - stk.Peek()); } // If stack is empty. push current // index as base for next valid // substring (if any) else { stk.Push(i); } } } return result; } // Driver Code public static void Main(string[] args) { string str = "((()()"; // Function call Console.WriteLine(findMaxLen(str)); str = "()(()))))"; // Function call Console.WriteLine(findMaxLen(str)); } } // This code is contributed by Shrikant13 ``` JavaScript ``` <script> // JavaScript Program for the above approach function findMaxLen(str) { let n = str.length; // Create a stack and push -1 as // initial index to it. let stk = []; stk.push(-1); // Initialize result let result = 0; // Traverse all characters of given string for (let i = 0; i < n; i++) { // If opening bracket, push index of it if (str.charAt(i) == '(') { stk.push(i); } // If closing bracket, i.e.,str[i] = ')' else { // Pop the previous opening // bracket's index if (stk.length != 0) { stk.pop(); } // Check if this length formed with base of // current valid substring is more than max // so far if (stk.length != 0) { result = Math.max(result, i - stk[stk.length - 1]); } // If stack is empty. push current index as // base for next valid substring (if any) else { stk.push(i); } } } return result; } // Driver code let str = "((()()"; // Function call document.write(findMaxLen(str) + "<br>"); str = "()(()))))"; // Function call document.write(findMaxLen(str) + "<br>"); // This code is contributed by Potta Lokesh </script> ```

Output
```4
6
```

Time Complexity: O(N), here N is the length of string.
Auxiliary Space: O(N)

Explanation with example:

`Input: str = "(()()"Initialize result as 0 and stack with one item -1.For i = 0, str[0] = '(', we push 0 in stackFor i = 1, str[1] = '(', we push 1 in stackFor i = 2, str[2] = ')', currently stack has [-1, 0, 1], we pop from the stack and the stacknow is [-1, 0] and length of current valid substring becomes 2 (we get this 2 by subtracting stack top from current index).Since the current length is more than the current result, we update the result.For i = 3, str[3] = '(', we push again, stack is [-1, 0, 3].For i = 4, str[4] = ')', we pop from the stack, stack becomes [-1, 0] and length of current valid substring becomes 4 (we get this 4 by subtracting stack top from current index). Since current length is more than current result,we update result. `

Another Efficient Approach can solve the problem in O(N) time. The idea is to maintain an array that stores the length of the longest valid substring ending at that index. We iterate through the array and return the maximum value.

Below is the implementations of the above algorithm.

C++ ```// C++ program to find length of the longest valid // substring #include <bits/stdc++.h> using namespace std; int findMaxLen(string s) { if (s.length() <= 1) return 0; // Initialize curMax to zero int curMax = 0; vector<int> longest(s.size(), 0); // Iterate over the string starting from second index 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(longest[i], curMax); } } return curMax; } // Driver code int main() { string str = "((()()"; // Function call cout << findMaxLen(str) << endl; str = "()(()))))"; // Function call cout << findMaxLen(str) << endl; return 0; } // This code is contributed by Vipul Lohani ``` Java ```// Java program to find length of the longest valid // subString import java.util.*; class GFG { static int findMaxLen(String s) { if (s.length() <= 1) return 0; // Initialize curMax to zero int curMax = 0; int[] longest = new int[s.length()]; // Iterate over the String starting from second index for (int i = 1; i < s.length(); i++) { if (s.charAt(i) == ')' && i - longest[i - 1] - 1 >= 0 && s.charAt(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 = Math.max(longest[i], curMax); } } return curMax; } // Driver code public static void main(String[] args) { String str = "((()()"; // Function call System.out.print(findMaxLen(str) +"\n"); str = "()(()))))"; // Function call System.out.print(findMaxLen(str) +"\n"); } } // This code is contributed by aashish1995 ``` Python3 ```# Python3 program to find length of # the longest valid substring def findMaxLen(s): if (len(s) <= 1): return 0 # Initialize curMax to zero curMax = 0 longest = [0] * (len(s)) # Iterate over the string starting # from second index for i in range(1, len(s)): if ((s[i] == ')' and i - longest[i - 1] - 1 >= 0 and s[i - longest[i - 1] - 1] == '(')): longest[i] = longest[i - 1] + 2 if (i - longest[i - 1] - 2 >= 0): longest[i] += (longest[i - longest[i - 1] - 2]) else: longest[i] += 0 curMax = max(longest[i], curMax) return curMax # Driver Code if __name__ == '__main__': Str = "((()()" # Function call print(findMaxLen(Str)) Str = "()(()))))" # Function call print(findMaxLen(Str)) # This code is contributed by PranchalK ``` C# ```// C# program to find length of the longest valid // subString using System; public class GFG { static int findMaxLen(String s) { if (s.Length <= 1) return 0; // Initialize curMax to zero int curMax = 0; int[] longest = new int[s.Length]; // Iterate over the String starting from second index 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 = Math.Max(longest[i], curMax); } } return curMax; } // Driver code public static void Main(String[] args) { String str = "((()()"; // Function call Console.Write(findMaxLen(str) + "\n"); str = "()(()))))"; // Function call Console.Write(findMaxLen(str) + "\n"); } } // This code is contributed by aashish1995 ``` JavaScript ```<script> // javascript program to find length of the longest valid // subString function findMaxLen( s) { if (s.length <= 1) return 0; // Initialize curMax to zero var curMax = 0; var longest = Array(s.length).fill(0); // Iterate over the String starting from second index for (var 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 = Math.max(longest[i], curMax); } } return curMax; } // Driver code var str = "((()()"; // Function call document.write(findMaxLen(str) + "<br\>"); str = "()(()))))"; // Function call document.write(findMaxLen(str) + "<br\>"); // This code is contributed by umadevi9616 </script> ```

Output
```4
6
```

Time Complexity: O(N), here N is the length of string.
Auxiliary Space: O(N)

Another approach in O(1) auxiliary space and O(N) Time complexity:

1. The idea to solve this problem is to traverse the string on and keep track of the count of open parentheses and close parentheses with the help of two counters left and right respectively.
2. First, the string is traversed from the left towards the right and for every “(” encountered, the left counter is incremented by 1 and for every “)” the right counter is incremented by 1.
3. Whenever the left becomes equal to right, the length of the current valid string is calculated and if it greater than the current longest substring, then value of required longest substring is updated with current string length.
4. If the right counter becomes greater than the left counter, then the set of parentheses has become invalid and hence the left and right counters are set to 0.
5. After the above process, the string is similarly traversed from right to left and similar procedure is applied.

Below is the implementation of the above approach:

C++ ```#include <bits/stdc++.h> using namespace std; // Function to return the length of // the longest valid substring int solve(string s, int n) { // Variables for left and right counter. // maxlength to store the maximum length found so far int left = 0, right = 0, maxlength = 0; // Iterating the string from left to right for (int i = 0; i < n; i++) { // If "(" is encountered, // then left counter is incremented // else right counter is incremented if (s[i] == '(') left++; else right++; // Whenever left is equal to right, it signifies // that the subsequence is valid and if (left == right) maxlength = max(maxlength, 2 * right); // Resetting the counters when the subsequence // becomes invalid else if (right > left) left = right = 0; } left = right = 0; // Iterating the string from right to left for (int i = n - 1; i >= 0; i--) { // If "(" is encountered, // then left counter is incremented // else right counter is incremented if (s[i] == '(') left++; else right++; // Whenever left is equal to right, it signifies // that the subsequence is valid and if (left == right) maxlength = max(maxlength, 2 * left); // Resetting the counters when the subsequence // becomes invalid else if (left > right) left = right = 0; } return maxlength; } // A much shorter and concise version of the above code int solve2(string s, int n) { int left = 0, right = 0, maxlength = 0, t = 2; while (t--) { left = 0; right = 0; for (int i = 0; i < n; i++) { if (s[i] == '(') left++; else right++; if (left == right) { maxlength = max(maxlength, 2 * left); } // when travelling from 0 to n-1 if (t % 2 == 1 && right > left) { left = 0; right = 0; } // when travelling from n-1 to 0 if (t % 2 == 0 && left > right) { left = 0; right = 0; } } // now we need to do the same thing from the other // side; reverse(s.begin(), s.end()); } return maxlength; } // Driver code int main() { // Function call string str = "((()()()()(((())"; cout << solve(str, str.length()); return 0; } ``` Java ```import java.util.Scanner; import java.util.Arrays; class GFG { // Function to return the length // of the longest valid substring public static int solve(String s, int n) { // Variables for left and right // counter maxlength to store // the maximum length found so far int left = 0, right = 0; int maxlength = 0; // Iterating the string from left to right for (int i = 0; i < n; i++) { // If "(" is encountered, then // left counter is incremented // else right counter is incremented if (s.charAt(i) == '(') left++; else right++; // Whenever left is equal to right, // it signifies that the subsequence // is valid and if (left == right) maxlength = Math.max(maxlength, 2 * right); // Resetting the counters when the // subsequence becomes invalid else if (right > left) left = right = 0; } left = right = 0; // Iterating the string from right to left for (int i = n - 1; i >= 0; i--) { // If "(" is encountered, then // left counter is incremented // else right counter is incremented if (s.charAt(i) == '(') left++; else right++; // Whenever left is equal to right, // it signifies that the subsequence // is valid and if (left == right) maxlength = Math.max(maxlength, 2 * left); // Resetting the counters when the // subsequence becomes invalid else if (left > right) left = right = 0; } return maxlength; } // Driver code public static void main(String args[]) { String str = "((()()()()(((())"; // Function call System.out.print(solve(str, str.length())); } } ``` Python3 ```# Python3 program to implement the above approach # Function to return the length of # the longest valid substring def solve(s, n): # Variables for left and right counter. # maxlength to store the maximum length found so far left = 0 right = 0 maxlength = 0 # Iterating the string from left to right for i in range(n): # If "(" is encountered, # then left counter is incremented # else right counter is incremented if (s[i] == '('): left += 1 else: right += 1 # Whenever left is equal to right, it signifies # that the subsequence is valid and if (left == right): maxlength = max(maxlength, 2 * right) # Resetting the counters when the subsequence # becomes invalid elif (right > left): left = right = 0 left = right = 0 # Iterating the string from right to left for i in range(n - 1, -1, -1): # If "(" is encountered, # then left counter is incremented # else right counter is incremented if (s[i] == '('): left += 1 else: right += 1 # Whenever left is equal to right, it signifies # that the subsequence is valid and if (left == right): maxlength = max(maxlength, 2 * left) # Resetting the counters when the subsequence # becomes invalid elif (left > right): left = right = 0 return maxlength # Driver code # Function call s = "((()()()()(((())" print(solve(s, len(s))) ``` C# ```// C# program to implement the above approach using System; public class GFG { // Function to return the length // of the longest valid substring public static int solve(String s, int n) { // Variables for left and right // counter maxlength to store // the maximum length found so far int left = 0, right = 0; int maxlength = 0; // Iterating the string from left to right for (int i = 0; i < n; i++) { // If "(" is encountered, then // left counter is incremented // else right counter is incremented if (s[i] == '(') left++; else right++; // Whenever left is equal to right, // it signifies that the subsequence // is valid and if (left == right) maxlength = Math.Max(maxlength, 2 * right); // Resetting the counters when the // subsequence becomes invalid else if (right > left) left = right = 0; } left = right = 0; // Iterating the string from right to left for (int i = n - 1; i >= 0; i--) { // If "(" is encountered, then // left counter is incremented // else right counter is incremented if (s[i] == '(') left++; else right++; // Whenever left is equal to right, // it signifies that the subsequence // is valid and if (left == right) maxlength = Math.Max(maxlength, 2 * left); // Resetting the counters when the // subsequence becomes invalid else if (left > right) left = right = 0; } return maxlength; } // Driver code public static void Main(String []args) { String str = "((()()()()(((())"; // Function call Console.Write(solve(str, str.Length)); } } ``` JavaScript ```// Function to return the length of // the longest valid substring function solve(s, n) { let left = 0, right = 0, maxlength = 0; // Iterating the string from left to right for (let i = 0; i < n; i++) { if (s[i] === '(') left++; else right++; if (left === right) maxlength = Math.max(maxlength, 2 * right); else if (right > left) left = right = 0; } left = right = 0; // Iterating the string from right to left for (let i = n - 1; i >= 0; i--) { if (s[i] === '(') left++; else right++; if (left === right) maxlength = Math.max(maxlength, 2 * left); else if (left > right) left = right = 0; } return maxlength; } // A much shorter and concise version of the above code function solve2(s, n) { let left = 0, right = 0, maxlength = 0; for (let t = 0; t < 2; t++) { left = 0; right = 0; for (let i = 0; i < n; i++) { if (s[i] === '(') left++; else right++; if (left === right) maxlength = Math.max(maxlength, 2 * left); if ((t % 2 === 1 && right > left) || (t % 2 === 0 && left > right)) { left = 0; right = 0; } } // now we need to do the same thing from the other side; s = s.split('').reverse().join(''); } return maxlength; } // Driver code let str = "((()()()()(((())"; console.log(solve(str, str.length)); ```

Output
`8`

Time Complexity: O(N), here N is the length of string.
Auxiliary Space: O(1)

Another Approach (Using Memoization):

Intuition:
The problem statement is asking for the length of the longest valid parentheses substring. One way to think about this problem is that for every ‘(‘ we encounter, we need a corresponding ‘)’ somewhere else in the string to form a valid parentheses pair. Therefore, a valid substring must start with an ‘(‘ and end with a ‘)’, and any number of valid parentheses pairs can be in between.

Approach:
The approach used here is to use a stack to keep track of the indexes of the characters in the input string. When a ‘(‘ character is encountered, its index is pushed onto the stack. When a ‘)’ character is encountered, the top index on the stack is popped. The difference between the current index and the top index on the stack represents the length of a valid substring ending at the current index. If the stack is empty after a ‘)’ is popped, it means that no matching ‘(‘ has been found, so the current index is pushed onto the stack as the new base for future valid substrings. By doing so, the solution keeps track of the potential valid parentheses starts and ends, and makes use of the property that any valid parentheses substring must be closed by an earlier opened one. Finally, the algorithm returns the max length at the end of the loop.

Idea : try to break in smaller sub problem .
case 1: string ends at “()”
longestParenEnding(0, i) = longestParenEnding(0, i – 2) + 2

case 2: string ends with “))” for example “()(())”
longestParenEnding(0, i) =

case 3: string ends with “(“
longestParenEnding(0, i) = 0

Below is the implementation of the above approach:

C++ ```// C++ program to implement the above approach #include <bits/stdc++.h> using namespace std; // Function to return the length of // the longest valid substring int lonParen(int i, string& s, vector<int>& memo) { // base condition if (i <= 0) { return 0; } // checking if value already present in the dp array if (memo[i] != -1) { return memo[i]; } // check for beginning bracket if (s[i] == '(') { memo[i] = 0; } // check if beginning and ending brackets satisfy else if (s[i] == ')' && s[i - 1] == '(') { memo[i] = lonParen(i - 2, s, memo) + 2; } // check if the bracket at the ith position is same as // bracket at i-1th position else if (s[i] == ')' && s[i - 1] == ')') { int len = lonParen(i - 1, s, memo); if (i - 1 - len >= 0 && s[i - 1 - len] == '(') { memo[i] = len + 2 + lonParen(i - len - 2, s, memo); } else { // if none of the condition satisfy store 0 in // the dp array memo[i] = 0; } } // return the value at the last index return memo[i]; } int longestValidParentheses(string s) { int n = s.size(), maxLen = 0; // dp vector for storing the results vector<int> memo(n, -1); // getting the maximum length for (int i = 0; i < n; i++) { maxLen = max(maxLen, lonParen(i, s, memo)); } // return the maximum length return maxLen; } // Driver code int main() { // Function call cout << longestValidParentheses("((()()()()(((())"); return 0; } ``` Java ```import java.util.Arrays; public class Main { // Function to return the length of the longest valid substring public static int longestValidParentheses(String s) { int n = s.length(); int maxLen = 0; int[] memo = new int[n]; Arrays.fill(memo, -1); // Iterate through each character in the input string for (int i = 0; i < n; i++) { // Update maxLen with the maximum length of valid parentheses encountered so far maxLen = Math.max(maxLen, lonParen(i, s, memo)); } return maxLen; // Return the length of the longest valid substring } // Function to calculate the length of the longest valid parentheses substring ending at index i public static int lonParen(int i, String s, int[] memo) { if (i <= 0) { return 0; // If index i is less than or equal to 0, the substring cannot be valid } if (memo[i] != -1) { return memo[i]; // If the length has already been memoized, return the memoized value } if (s.charAt(i) == '(') { memo[i] = 0; // If the current character is '(', the substring cannot be valid } else if (s.charAt(i) == ')' && s.charAt(i - 1) == '(') { // If the current character is ')' and the previous character is '(', we have a valid pair // The length of the valid substring ending at index i is the length of the valid substring ending at index i-2 plus 2 memo[i] = lonParen(i - 2, s, memo) + 2; } else if (s.charAt(i) == ')' && s.charAt(i - 1) == ')') { // If the current character is ')' and the previous character is ')' // We need to check if there is a valid substring ending at index i-1 int len = lonParen(i - 1, s, memo); if (i - 1 - len >= 0 && s.charAt(i - 1 - len) == '(') { // If there is a valid substring ending at index i-1 and the character before it is '(', we have a valid pair // The length of the valid substring ending at index i is the length of the valid substring ending at index i-1 // plus 2 plus the length of the valid substring ending at index i-len-2 memo[i] = len + 2 + lonParen(i - len - 2, s, memo); } else { memo[i] = 0; // If there is no valid substring ending at index i-1, the substring cannot be valid } } return memo[i]; // Return the length of the longest valid parentheses substring ending at index i } public static void main(String[] args) { System.out.println(longestValidParentheses("((()()()()(((())")); // Test the function with an example input } } ``` Python3 ```def lonParen(i, s, memo): # base condition if i <= 0: return 0 # checking if value already present in the dp array if memo[i] != -1: return memo[i] # check for beginning bracket if s[i] == '(': memo[i] = 0 # check if beginning and ending brackets satisfy elif s[i] == ')' and s[i - 1] == '(': memo[i] = lonParen(i - 2, s, memo) + 2 # check if the bracket at the ith position is same as # bracket at i-1th position elif s[i] == ')' and s[i - 1] == ')': length = lonParen(i - 1, s, memo) if i - 1 - length >= 0 and s[i - 1 - length] == '(': memo[i] = length + 2 + lonParen(i - length - 2, s, memo) else: # if none of the condition satisfy store 0 in # the dp array memo[i] = 0 # return the value at the last index return memo[i] def longestValidParentheses(s): n = len(s) maxLen = 0 # dp list for storing the results memo = [-1] * n # getting the maximum length for i in range(n): maxLen = max(maxLen, lonParen(i, s, memo)) # return the maximum length return maxLen # Driver code print(longestValidParentheses("((()()()()(((())")) ``` C# ```using System; class Program { // Function to return the length of the longest valid substring static int LongestValidParentheses(int i, string s, int[] memo) { // Base condition if (i <= 0) { return 0; } // Checking if value already present in the memo array if (memo[i] != -1) { return memo[i]; } // Check for beginning bracket if (s[i] == '(') { memo[i] = 0; } // Check if beginning and ending brackets satisfy else if (s[i] == ')' && s[i - 1] == '(') { memo[i] = LongestValidParentheses(i - 2, s, memo) + 2; } // Check if the bracket at the ith position is the same as the bracket at i-1th position else if (s[i] == ')' && s[i - 1] == ')') { int len = LongestValidParentheses(i - 1, s, memo); if (i - 1 - len >= 0 && s[i - 1 - len] == '(') { memo[i] = len + 2 + LongestValidParentheses(i - len - 2, s, memo); } else { // If none of the conditions satisfy, store 0 in the memo array memo[i] = 0; } } // Return the value at the last index return memo[i]; } static int LongestValidParentheses(string s) { int n = s.Length, maxLen = 0; // Memo array for storing the results int[] memo = new int[n]; for (int i = 0; i < n; i++) { memo[i] = -1; } // Getting the maximum length for (int i = 0; i < n; i++) { maxLen = Math.Max(maxLen, LongestValidParentheses(i, s, memo)); } // Return the maximum length return maxLen; } // Driver code static void Main() { // Function call Console.WriteLine(LongestValidParentheses("((()()()()(((())")); } } ``` JavaScript ```function lonParen(i, s, memo) { // Base condition if (i <= 0) { return 0; } // Checking if value already present in the memo array if (memo[i] !== -1) { return memo[i]; } // Check for beginning bracket if (s.charAt(i) === '(') { memo[i] = 0; } // Check if beginning and ending brackets satisfy else if (s.charAt(i) === ')' && s.charAt(i - 1) === '(') { memo[i] = lonParen(i - 2, s, memo) + 2; } // Check if the bracket at the ith position is the same as // the bracket at i-1th position else if (s.charAt(i) === ')' && s.charAt(i - 1) === ')') { let length = lonParen(i - 1, s, memo); if (i - 1 - length >= 0 && s.charAt(i - 1 - length) === '(') { memo[i] = length + 2 + lonParen(i - length - 2, s, memo); } else { // If none of the conditions satisfy, store 0 in the memo array memo[i] = 0; } } // Return the value at the last index return memo[i]; } function longestValidParentheses(s) { let n = s.length; let maxLen = 0; // Memo array for storing the results let memo = Array(n).fill(-1); // Getting the maximum length for (let i = 0; i < n; i++) { maxLen = Math.max(maxLen, lonParen(i, s, memo)); } // Return the maximum length return maxLen; } // Driver code console.log(longestValidParentheses("((()()()()(((())")); ```

Output
`8`

Time complexity: O(n),Here, The algorithm has a time complexity of O(n) because it simply iterates the string once.
Auxiliary Space: O(n),Space complexity is O(n) because it uses a stack to keep track of the indexes of the characters in the input string.

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