Longest Valid Parentheses Substring
Last Updated :
11 Feb, 2025
Given a string str consisting of opening and closing parenthesis '(' and ')', the task is to find the length of the longest valid parenthesis substring.
Examples:
Input: str = "((()"
Output: 2
Explanation: Longest Valid Parentheses Substring is "()".
Input: str = ")()())"
Output: 4
Explanation: Longest Valid Parentheses Substring is "()()".
[Expected Approach - 1] Using Stack - O(n) Time and O(n) Space
The idea is to use a stack-based approach to track the indices of unmatched open parentheses. It determines the length of valid (well-formed) parentheses substrings by keeping track of the position of the last unmatched closing parenthesis or the starting position of a valid substring.
Follow the steps below to solve the problem:
- For every opening parenthesis, we push its index onto the stack.
- For every closing parenthesis, we pop the stack.
- If the stack becomes empty after popping, it means we've encountered an unmatched closing parenthesis, so we push the current index to serve as a base for the next potential valid substring.
- If the stack is not empty, we calculate the length of the valid substring by subtracting the index at the top of the stack from the current index.
- A variable maxLength keeps track of the maximum length of valid parentheses encountered during the traversal.
C++
// C++ program to find length of the
// longest valid substring
#include <iostream>
#include <stack>
using namespace std;
int maxLength(string s) {
stack<int> st;
// Push -1 as the initial index to
// handle the edge case
st.push(-1);
int maxLen = 0;
// Traverse the string
for (int i = 0; i < s.length(); i++) {
// If we encounter an opening parenthesis,
// push its index
if (s[i] == '(') {
st.push(i);
} else {
// If we encounter a closing parenthesis,
// pop the stack
st.pop();
// If stack is empty, push the current index
// as a base for the next valid substring
if (st.empty()) {
st.push(i);
} else {
// Update maxLength with the current length
// of the valid parentheses substring
maxLen = max(maxLen, i - st.top());
}
}
}
return maxLen;
}
int main() {
string s = ")()())";
cout << maxLength(s) << endl;
return 0;
}
// Java program to find length of the
// longest valid substring
import java.util.Stack;
class GfG {
// Function to find the length of the
// longest valid parentheses substring
static int maxLength(String s) {
Stack<Integer> stack = new Stack<>();
// Push -1 as the initial index to
// handle the edge case
stack.push(-1);
int maxLen = 0;
// Traverse the string
for (int i = 0; i < s.length(); i++) {
// If we encounter an opening parenthesis,
// push its index
if (s.charAt(i) == '(') {
stack.push(i);
} else {
// If we encounter a closing parenthesis,
// pop the stack
stack.pop();
// If stack is empty, push the current index
// as a base for the next valid substring
if (stack.isEmpty()) {
stack.push(i);
} else {
// Update maxLength with the current length
// of the valid parentheses substring
maxLen = Math.max(maxLen, i - stack.peek());
}
}
}
return maxLen;
}
public static void main(String[] args) {
String s = ")()())";
System.out.println(maxLength(s));
}
}
# Python program to find length of the
# longest valid substring
def maxLength(s):
stack = []
# Push -1 as the initial index to
# handle the edge case
stack.append(-1)
maxLen = 0
# Traverse the string
for i in range(len(s)):
# If we encounter an opening parenthesis,
# push its index
if s[i] == '(':
stack.append(i)
else:
# If we encounter a closing parenthesis,
# pop the stack
stack.pop()
# If stack is empty, push the current index
# as a base for the next valid substring
if not stack:
stack.append(i)
else:
# Update maxLength with the current length
# of the valid parentheses substring
maxLen = max(maxLen, i - stack[-1])
return maxLen
if __name__ == "__main__":
s = ")()())"
print(maxLength(s))
// C# program to find length of the
// longest valid substring
using System;
using System.Collections.Generic;
class GfG {
// Function to find the length of the
// longest valid parentheses substring
static int maxLength(string s) {
Stack<int> stack = new Stack<int>();
// Push -1 as the initial index to handle
// the edge case
stack.Push(-1);
int maxLen = 0;
// Traverse the string
for (int i = 0; i < s.Length; i++) {
// If we encounter an opening parenthesis,
// push its index
if (s[i] == '(') {
stack.Push(i);
} else {
// If we encounter a closing parenthesis,
// pop the stack
stack.Pop();
// If stack is empty, push the current index
// as a base for the next valid substring
if (stack.Count == 0) {
stack.Push(i);
} else {
// Update maxLength with the current length
// of the valid parentheses substring
maxLen = Math.Max(maxLen, i - stack.Peek());
}
}
}
return maxLen;
}
static void Main() {
string s = ")()())";
Console.WriteLine(maxLength(s));
}
}
// JavaScript program to find length of the
// longest valid substring
function maxLength(s) {
let stack = [];
// Push -1 as the initial index to handle
// the edge case
stack.push(-1);
let maxLen = 0;
// Traverse the string
for (let i = 0; i < s.length; i++) {
// If we encounter an opening parenthesis,
// push its index
if (s[i] === '(') {
stack.push(i);
} else {
// If we encounter a closing parenthesis,
// pop the stack
stack.pop();
// If stack is empty, push the current index
// as a base for the next valid substring
if (stack.length === 0) {
stack.push(i);
} else {
// Update maxLength with the current length
// of the valid parentheses substring
maxLen = Math.max(maxLen, i - stack[stack.length - 1]);
}
}
}
return maxLen;
}
// Driver Code
let s = ")()())";
console.log(maxLength(s));
[Expected Approach - 2] Using DP - O(n) Time and O(n) Space
The idea is to solve this problem using dynamic programming (DP) where dp[i] represents the length of the longest valid parentheses substring ending at index i. If a valid substring ends at i, we calculate and store the length of that substring in dp[i].
Follow the steps below to solve the problem:
- If we encounter an opening parenthesis, we can't form a valid substring yet, so we move to the next character.
- If we encounter a closing parenthesis, we check the previous character to determine if it forms a valid substring.
- If the previous character is '(', we have a valid pair, so dp[i] = dp[i-2] + 2 (if i-2 is valid).
- If the previous character is ')', check if the substring before it forms a valid substring. We use dp[i-1] to determine where the valid substring might start.
- Also, store the maximum length of valid parentheses during the traversal.
C++
// C++ program to find length of the
// longest valid substring using DP
#include <iostream>
#include <vector>
using namespace std;
// Function to find the length of the
// longest valid parentheses substring
int maxLength(string s) {
int n = s.length();
vector<int> dp(n, 0);
int maxLen = 0;
// Traverse the string
for (int i = 1; i < n; i++) {
// If we encounter a closing parenthesis
if (s[i] == ')') {
// Check if the previous character is an
// opening parenthesis '('
if (s[i - 1] == '(') {
if (i >= 2) {
dp[i] = dp[i - 2] + 2;
}
else {
dp[i] = 2;
}
}
// Check if the previous character is a
// closing parenthesis ')' and the matching opening
// parenthesis exists before the valid substring
else if (i - dp[i - 1] > 0 && s[i - dp[i - 1] - 1] == '(') {
if (i - dp[i - 1] >= 2) {
dp[i] = dp[i - 1] + dp[i - dp[i - 1] - 2] + 2;
}
else {
dp[i] = dp[i - 1] + 2;
}
}
// Update the maximum length
maxLen = max(maxLen, dp[i]);
}
}
return maxLen;
}
int main() {
string s = ")()())";
cout << maxLength(s) << endl;
return 0;
}
// Java program to find length of the
// longest valid substring using DP
import java.util.*;
class GfG {
// Function to find the length of the
// longest valid parentheses substring
static int maxLength(String s) {
int n = s.length();
int[] dp = new int[n];
int maxLen = 0;
// Traverse the string
for (int i = 1; i < n; i++) {
// If we encounter a closing parenthesis
if (s.charAt(i) == ')') {
// Check if the previous character is an
// opening parenthesis '('
if (s.charAt(i - 1) == '(') {
if (i >= 2) {
dp[i] = dp[i - 2] + 2;
}
else {
dp[i] = 2;
}
}
// Check if the previous character is a
// closing parenthesis ')' and the matching
// opening parenthesis exists before the
// valid substring
else if (i - dp[i - 1] > 0
&& s.charAt(i - dp[i - 1] - 1) == '(') {
if (i - dp[i - 1] >= 2) {
dp[i] = dp[i - 1] + dp[i - dp[i - 1] - 2] + 2;
}
else {
dp[i] = dp[i - 1] + 2;
}
}
// Update the maximum length
maxLen = Math.max(maxLen, dp[i]);
}
}
return maxLen;
}
public static void main(String[] args) {
String s = ")()())";
System.out.println(maxLength(s));
}
}
#Python program to find length of the
# longest valid substring using DP
def maxLength(s):
n = len(s)
dp = [0] * n
maxLen = 0
# Traverse the string
for i in range(1, n):
# If we encounter a closing parenthesis
if s[i] == ')':
# Check if the previous character is an opening
# parenthesis '('
if s[i - 1] == '(':
if i >= 2:
dp[i] = dp[i - 2] + 2
else:
dp[i] = 2
# Check if the previous character is a
# closing parenthesis ')' and the matching opening
# parenthesis exists before the valid substring
elif i - dp[i - 1] > 0 and s[i - dp[i - 1] - 1] == '(':
if i - dp[i - 1] >= 2:
dp[i] = dp[i - 1] + dp[i - dp[i - 1] - 2] + 2
else:
dp[i] = dp[i - 1] + 2
# Update the maximum length
maxLen = max(maxLen, dp[i])
return maxLen
if __name__ == "__main__":
s = ")()())"
print(maxLength(s))
// C# program to find length of the
// longest valid substring using DP
using System;
class GfG {
// Function to find the length of the
// longest valid parentheses substring
static int maxLength(string s) {
int n = s.Length;
int[] dp = new int[n];
int maxLen = 0;
// Traverse the string
for (int i = 1; i < n; i++) {
// If we encounter a closing parenthesis
if (s[i] == ')') {
// Check if the previous character is an
// opening parenthesis '('
if (s[i - 1] == '(') {
if (i >= 2) {
dp[i] = dp[i - 2] + 2;
}
else {
dp[i] = 2;
}
}
// Check if the previous character is a
// closing parenthesis ')' and the matching
// opening parenthesis exists before the
// valid substring
else if (i - dp[i - 1] > 0
&& s[i - dp[i - 1] - 1] == '(') {
if (i - dp[i - 1] >= 2) {
dp[i] = dp[i - 1] + dp[i - dp[i - 1] - 2] + 2;
}
else {
dp[i] = dp[i - 1] + 2;
}
}
// Update the maximum length
maxLen = Math.Max(maxLen, dp[i]);
}
}
return maxLen;
}
static void Main(string[] args) {
string s = ")()())";
Console.WriteLine(maxLength(s));
}
}
// Javascript program to find length of the
// longest valid substring using DP
function maxLength(s) {
const n = s.length;
const dp = new Array(n).fill(0);
let maxLen = 0;
// Traverse the string
for (let i = 1; i < n; i++) {
// If we encounter a closing parenthesis
if (s[i] === ")") {
// Check if the previous character is an opening
// parenthesis '('
if (s[i - 1] === "(") {
if (i >= 2) {
dp[i] = dp[i - 2] + 2;
}
else {
dp[i] = 2;
}
}
// Check if the previous character is a
// closing parenthesis ')' and the matching
// opening parenthesis exists before the valid
// substring
else if (i - dp[i - 1] > 0
&& s[i - dp[i - 1] - 1] === "(") {
if (i - dp[i - 1] >= 2) {
dp[i] = dp[i - 1] + dp[i - dp[i - 1] - 2] + 2;
}
else {
dp[i] = dp[i - 1] + 2;
}
}
// Update the maximum length
maxLen = Math.max(maxLen, dp[i]);
}
}
return maxLen;
}
// Driver Code
const s = ")()())";
console.log(maxLength(s));
[Expected Approach - 3] Using Two Traversals - O(n) Time and O(1) Space
The idea is to solve the problem using two traversals of the string, one from left to right and one from right to left, while keeping track of the number of open and close parentheses using two counters: open and close.
Why there is a Two Traversals?
- Left to right traversal ensures that every valid substring that ends at the rightmost closing parenthesis is counted.
- Right to left traversal ensures that every valid substring that starts from the leftmost opening parenthesis is counted.
Follow the steps below to solve the problem:
- Left to Right Traversal:
- Use two counters, say open to count the number of opening parentheses '(', and close to count the number of closing parentheses ')'.
- For each character, if it's '(', increment open else increment close.
- Whenever open == close, we have found a valid substring, and we calculate the length: 2 * close. Keep track of the maximum valid length.
- If at any point close exceeds open, it means we have too many closing parentheses, and we reset both open and close to 0.
- Right to Left Traversal:
- Similarly, use open and close counters again.
- For each character, if it's '(', increment open else increment close.
- Again, whenever open == close, update the maximum valid length.
- If open exceeds close, reset both counters to 0.
C++
// C++ program to find length of the
// longest valid substring
#include <iostream>
using namespace std;
int maxLength(string s) {
int maxLen = 0;
// Left to Right Traversal
int open = 0, close = 0;
for (char ch : s) {
if (ch == '(') {
open++;
}
else if (ch == ')') {
close++;
}
if (open == close) {
maxLen = max(maxLen, 2 * close);
}
else if (close > open) {
open = close = 0;
}
}
// Right to Left Traversal
open = close = 0;
for (int i = s.size() - 1; i >= 0; i--) {
if (s[i] == '(') {
open++;
}
else if (s[i] == ')') {
close++;
}
if (open == close) {
maxLen = max(maxLen, 2 * open);
}
else if (open > close) {
open = close = 0;
}
}
return maxLen;
}
int main() {
string s = ")()())";
cout << maxLength(s) << endl;
return 0;
}
// C program to find length of the
// longest valid substring
#include <stdio.h>
#include <string.h>
#include <stdbool.h>
int maxLength(char s[]) {
int maxLen = 0;
// Left to Right Traversal
int open = 0, close = 0;
int len = strlen(s);
for (int i = 0; i < len; i++) {
if (s[i] == '(') {
open++;
} else if (s[i] == ')') {
close++;
}
if (open == close) {
maxLen = (maxLen > 2 * close) ? maxLen : 2 * close;
} else if (close > open) {
open = close = 0;
}
}
// Right to Left Traversal
open = close = 0;
for (int i = len - 1; i >= 0; i--) {
if (s[i] == '(') {
open++;
} else if (s[i] == ')') {
close++;
}
if (open == close) {
maxLen = (maxLen > 2 * open) ? maxLen : 2 * open;
} else if (open > close) {
open = close = 0;
}
}
return maxLen;
}
int main() {
char s[] = ")()())";
printf("%d\n", maxLength(s));
return 0;
}
// Java program to find length of the
// longest valid substring
class GfG {
static int maxLength(String s) {
int maxLen = 0;
// Left to Right Traversal
int open = 0, close = 0;
for (char ch : s.toCharArray()) {
if (ch == '(') {
open++;
} else if (ch == ')') {
close++;
}
if (open == close) {
maxLen = Math.max(maxLen, 2 * close);
} else if (close > open) {
open = close = 0;
}
}
// Right to Left Traversal
open = close = 0;
for (int i = s.length() - 1; i >= 0; i--) {
if (s.charAt(i) == '(') {
open++;
} else if (s.charAt(i) == ')') {
close++;
}
if (open == close) {
maxLen = Math.max(maxLen, 2 * open);
} else if (open > close) {
open = close = 0;
}
}
return maxLen;
}
public static void main(String[] args) {
String s = ")()())";
System.out.println(maxLength(s));
}
}
# Python program to find length of the
# longest valid substring
def maxLength(s):
maxLen = 0
# Left to Right Traversal
open = close = 0
for ch in s:
if ch == '(':
open += 1
elif ch == ')':
close += 1
if open == close:
maxLen = max(maxLen, 2 * close)
elif close > open:
open = close = 0
# Right to Left Traversal
open = close = 0
for ch in reversed(s):
if ch == '(':
open += 1
elif ch == ')':
close += 1
if open == close:
maxLen = max(maxLen, 2 * open)
elif open > close:
open = close = 0
return maxLen
if __name__ == "__main__":
s = ")()())"
print(maxLength(s))
// C# program to find length of the
// longest valid substring
using System;
class GfG {
static int maxLength(string s) {
int maxLen = 0;
// Left to Right Traversal
int open = 0, close = 0;
foreach (char ch in s) {
if (ch == '(') {
open++;
} else if (ch == ')') {
close++;
}
if (open == close) {
maxLen = Math.Max(maxLen, 2 * close);
} else if (close > open) {
open = close = 0;
}
}
// Right to Left Traversal
open = close = 0;
for (int i = s.Length - 1; i >= 0; i--) {
if (s[i] == '(') {
open++;
} else if (s[i] == ')') {
close++;
}
if (open == close) {
maxLen = Math.Max(maxLen, 2 * open);
} else if (open > close) {
open = close = 0;
}
}
return maxLen;
}
static void Main(string[] args) {
string s = ")()())";
Console.WriteLine(maxLength(s));
}
}
// JavaScript program to find length of the
// longest valid substring
function maxLength(s) {
let maxLen = 0;
// Left to Right Traversal
let open = 0, close = 0;
for (let i = 0; i < s.length; i++) {
if (s[i] === '(') {
open++;
} else if (s[i] === ')') {
close++;
}
if (open === close) {
maxLen = Math.max(maxLen, 2 * close);
} else if (close > open) {
open = close = 0;
}
}
// Right to Left Traversal
open = close = 0;
for (let i = s.length - 1; i >= 0; i--) {
if (s[i] === '(') {
open++;
} else if (s[i] === ')') {
close++;
}
if (open === close) {
maxLen = Math.max(maxLen, 2 * open);
} else if (open > close) {
open = close = 0;
}
}
return maxLen;
}
// Driver Code
const s = ")()())";
console.log(maxLength(s));
Length of the longest valid substring
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