Skip to content
Related Articles
Open in App
Not now

Related Articles

Length of the longest valid substring

Improve Article
Save Article
  • Difficulty Level : Medium
  • Last Updated : 22 Feb, 2023
Improve Article
Save Article

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:  ()(())
Recommended Practice
IPL 2021 – Final
Try It!

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(n2).

Space Complexity: O(1)

An Efficient Solution can solve this problem in O(n) time. The idea is to store indexes of previous starting brackets in a stack. The first element of the stack is a special element that provides index before the beginning of valid substring (base for next valid string). 

1) Create an empty stack and push -1 to it. 
   The first element of the stack is used 
   to provide a base for the next valid string. 

2) Initialize result as 0.

3) If the character is '(' i.e. str[i] == '('), 
   push index'i' to the stack. 
   
2) Else (if the character is ')')
   a) Pop an item from the stack (Most of the 
      time an opening bracket)
   b) If the stack is not empty, then find the
      length of current valid substring by taking 
      the difference between the current index and
      top of the stack. If current length is more 
      than the result, then update the result.
   c) If the stack is empty, push the current index
      as a base for the next valid substring.

3) Return result.

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 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.
1) Create an array longest of length n (size of the input
   string) initialized to zero.
   The array will store the length of the longest valid 
   substring ending at that index.

2) Initialize result as 0.

3) Iterate through the string from second character
   a) If the character is '(' set longest[i]=0 as no 
      valid sub-string will end with '('.
   b) Else
      i) if s[i-1] = '('
            set longest[i] = longest[i-2] + 2
      ii) else
            set longest[i] = longest[i-1] + 2 + 
            longest[i-longest[i-1]-2]

4) In each iteration update result as the maximum of 
   result and longest[i]

5) Return result.
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)
Thanks to Gaurav Ahirwar and Ekta Goel for suggesting above approach.
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++



// 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 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
    cout << solve("((()()()()(((())", 16);
    return 0;
}
Java



// Java program to implement the above approach
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[])
    {
        // Function call
        System.out.print(solve("((()()()()(((())", 16));
    }
}
 
// This code is contributed by SoumikMondal
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
print(solve("((()()()()(((())", 16))
 
# This code is contributed by shubhamsingh10
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)
  {
    // Function call
    Console.Write(solve("((()()()()(((())", 16));
  }
}
 
 
// This code is contributed by Rajput-Ji
Javascript



<script>
 
// JavaScript program to implement the above approach
 
 
// Function to return the length of
// the longest valid substring
function solve(s, n)
{
 
    // Variables for left and right counter.
    // maxlength to store the maximum length found so far
    let left = 0, right = 0, maxlength = 0;
 
    // Iterating the string from left to right
    for (let 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 (let 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
 
 
    // Function call
    document.write( solve("((()()()()(((())", 16));
     
//This code is contributed by Manoj
</script>
Output
8
Time Complexity: O(N), here N is the length of string.
Auxiliary Space: O(1)
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 

My Personal Notes
arrow_drop_up
Related Articles
1.
Find a valid parenthesis sequence of length K from a given valid parenthesis sequence
2.
Generate an N-length string having longest palindromic substring of length K
3.
Longest substring whose any non-empty substring not prefix or suffix of given String
4.
Longest Substring of A that can be changed to Substring of B in at most T cost
5.
Length of the largest substring which have character with frequency greater than or equal to half of the substring
6.
Minimum length of substring whose rotation generates a palindromic substring
7.
Longest Even Length Substring such that Sum of First and Second Half is same
8.
Length of the longest substring that do not contain any palindrome
9.
Length of longest substring having all characters as K
10.
Length of longest substring to be deleted to make a string equal to another string
Previous
Print ancestors of a given binary tree node without recursion
Next

Expression contains redundant bracket or not
Article Contributed By :
https://media.geeksforgeeks.org/auth/avatar.png
GeeksforGeeks
Vote for difficulty
Current difficulty :
Medium





Improved By :
Article Tags :
Practice Tags :
Report Issue
We use cookies to ensure you have the best browsing experience on our website. By using our site, you
acknowledge that you have read and understood our
Cookie Policy &
        Privacy Policy

Lightbox

Start Your Coding Journey Now!