Skip to content
Related Articles

Related Articles

Find two non-intersecting subarrays having equal sum of all elements raised to the power of 2
  • Last Updated : 04 Jan, 2021

Given an array arr[] of positive integers of size N, the task is to check if there exists two non-intersecting subarrays in arr[] such that sum of all possible 2(subarr[i]) and the sum of all possible 2(subarr2[j]) are equal.

Examples:

Input: arr[] = {4, 3, 0, 1, 2, 0}
Output: YES
Explanation: Expressing every array element in the form of 2arr[i], the array is modified to { 16, 8, 1, 2, 4, 1 }.
Therefore, two valid subarrays are { 16 } and { 8, 1, 2, 4, 1 } whose sum are equal.

Input: arr[]={ 3, 4 }
Output: NO

Approach: Since binary representation of all powers of 2 is unique, two sch subarrays can only be obtained if any repeating element is present in that array. Otherwise, it is not possible.



Follow the steps below to solve the problem: 

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to check if two non-intersecting
// subarrays with equal sum exists or not
void findSubarrays(int arr[], int N)
{
    // Sort the given array
    sort(arr, arr + N);
    int i = 0;
 
    // Traverse the array
    for (i = 0; i < N - 1; i++) {
 
        // Check for duplicate elements
        if (arr[i] == arr[i + 1]) {
 
            cout << "YES" << endl;
            return;
        }
    }
 
    // If no duplicate element is
    // present in the array
    cout << "NO" << endl;
}
 
// Driver Code
int main()
{
    // Given array
    int arr[] = { 4, 3, 0, 1, 2, 0 };
 
    // Size of array
    int N = sizeof(arr) / sizeof(arr[0]);
 
    findSubarrays(arr, N);
 
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program for the above approach
import java.util.*;
  
class GFG{
      
// Function to check if two non-intersecting
// subarrays with equal sum exists or not
static void findSubarrays(int arr[], int N)
{
     
    // Sort the given array
    Arrays.sort(arr);
    int i = 0;
  
    // Traverse the array
    for(i = 0; i < N - 1; i++)
    {
         
        // Check for duplicate elements
        if (arr[i] == arr[i + 1])
        {
            System.out.println("YES");
            return;
        }
    }
  
    // If no duplicate element is
    // present in the array
    System.out.println("NO");
}
  
// Driver code
public static void main(String[] args)
{
     
    // Given array
    int[] arr = { 4, 3, 0, 1, 2, 0 };
  
    // Size of array
    int N = arr.length;
  
    findSubarrays(arr, N);
}
}
 
// This code is contributed by susmitakundugoaldanga

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python program for the above approach
 
# Function to check if two non-intersecting
# subarrays with equal sum exists or not
def findSubarrays(arr, N):
   
    # Sort the given array
    arr.sort();
    i = 0;
 
    # Traverse the array
    for i in range(N - 1):
 
        # Check for duplicate elements
        if (arr[i] == arr[i + 1]):
            print("YES");
            return;
 
    # If no duplicate element is
    # present in the array
    print("NO");
 
# Driver code
if __name__ == '__main__':
   
    # Given array
    arr = [4, 3, 0, 1, 2, 0];
 
    # Size of array
    N = len(arr);
 
    findSubarrays(arr, N);
 
# This code is contributed by 29AjayKumar

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program for the above approach
using System;
   
class GFG{
       
// Function to check if two non-intersecting
// subarrays with equal sum exists or not
static void findSubarrays(int[] arr, int N)
{
     
    // Sort the given array
    Array.Sort(arr);
    int i = 0;
   
    // Traverse the array
    for(i = 0; i < N - 1; i++)
    {
         
        // Check for duplicate elements
        if (arr[i] == arr[i + 1])
        {
            Console.WriteLine("YES");
            return;
        }
    }
   
    // If no duplicate element is
    // present in the array
    Console.WriteLine("NO");
}
   
// Driver code
public static void Main()
{
     
    // Given array
    int[] arr = { 4, 3, 0, 1, 2, 0 };
   
    // Size of array
    int N = arr.Length;
   
    findSubarrays(arr, N);
}
}
 
// This code is contributed by sanjoy_62

chevron_right


Output: 

YES

 

Time Complexity: O(NLogN)
Auxiliary Space: O(1)

 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up
Recommended Articles
Page :