Find two non-intersecting subarrays having equal sum of all elements raised to the power of 2

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 2^arr[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 such 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: 

• Sort the given array in ascending order.
• Traverse the array and check if any pair of adjacent elements is equal or not.
• If any such pair is found, print “YES”. Otherwise, print “NO”.

Below is the implementation of the above approach:

C++

// 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;
}

Java

// 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

Python3

# 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

C#

// 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

Javascript

<script>
// javascript program for the above approach

    // Function to check if two non-intersecting
    // subarrays with equal sum exists or not
    function findSubarrays(arr , N) {

        // Sort the given array
        arr.sort();
        var i = 0;

        // Traverse the array
        for (i = 0; i < N - 1; i++) {

            // Check for duplicate elements
            if (arr[i] == arr[i + 1]) {
                document.write("YES");
                return;
            }
        }

        // If no duplicate element is
        // present in the array
        document.write("NO");
    }

    // Driver code


        // Given array
        var arr = [ 4, 3, 0, 1, 2, 0 ];

        // Size of array
        var N = arr.length;

        findSubarrays(arr, N);

// This code is contributed by gauravrajput1
</script>

Output:

YES

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