Given a string consisting of opening and closing parenthesis, find the length of the longest valid parenthesis substring.
Examples:
Input: ((()
Output : 2
Explanation : ()
Input: )()())
Output : 4
Explanation: ()()
Input: ()(()))))
Output: 6
Explanation: ()(())
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;
}
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));
}
}
def find_max_len(s):
n = len(s) # Get the length of the input string
stk = [] # Create an empty stack to store indices of '('
result = 0 # Variable to store the maximum length of valid substring
last = n # Variable to keep track of the last index
# Traverse the input string
for i in range(n):
if s[i] == '(': # If current character is '('
stk.append(i) # Push its index onto the stack
else:
# If stack is not empty and top of stack contains '('
if stk and s[stk[-1]] == '(':
stk.pop() # Pop the index of '(' from the stack
else:
# Push current index onto stack as base for next valid substring
stk.append(i)
# After traversing the string, calculate the length of longest valid substring
while stk:
now = stk.pop() # Pop the index of '(' from stack
# Calculate the length of valid substring
result = max(result, last - now - 1)
last = now # Update last index
# Return the maximum length of valid substring
return max(result, last)
# Driver code
if __name__ == "__main__":
s = "((()()" # Test string 1
# Output: 4 (Length of the longest valid substring is 4)
print(find_max_len(s))
s = "()(()))))" # Test string 2
# Output: 6 (Length of the longest valid substring is 6)
print(find_max_len(s))
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));
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 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));
}
}
# 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# 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
<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>
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 stack
For i = 1, str[1] = '(', we push 1 in stack
For i = 2, str[2] = ')', currently stack has
[-1, 0, 1], we pop from the stack and the stack
now 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 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 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# 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
<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>
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:
- 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.
- 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.
- 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.
- 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.
- 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;
}
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 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# 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));
}
}
// 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));
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;
}
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
}
}
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("((()()()()(((())"))
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("((()()()()(((())"));
}
}
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("((()()()()(((())"));
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.
Please Login to comment...