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Bitwise XOR of all unordered pairs from a given array
  • Last Updated : 05 May, 2021

Given an array arr[] of size N, the task is to find the bitwise XOR of all possible unordered pairs of the given array. 

Examples:

Input: arr[] = {1, 5, 3, 7}
Output: 0
Explanation:
All possible unordered pairs are (1, 5), (1, 3), (1, 7), (5, 3), (5, 7), (3, 7)
Bitwise XOR of all possible pairs are = 1 ^ 5 ^ 1 ^3 ^ 1 ^ 7 ^ 5 ^ 3 ^ 5^ 7 ^ 3 ^ 7 = 0
Therefore, the required output is 0.

Input: arr[] = {1, 2, 3, 4}
Output: 4

 

Naive approach: The idea is to traverse the array and generate all possible pairs of the given array. Finally, print the Bitwise XOR of each element present in these pairs of the given array. Follow the steps below to solve the problem:



  • Initialize a variable, say totalXOR, to store Bitwise XOR of each element from these pairs.
  • Traverse the given array and generate all possible pairs(arr[i], arr[j]) from the given array.
  • For each pair (arr[i], arr[j]), update the value of totalXOR = (totalXOR ^ arr[i] ^ arr[j]).
  • Finally, print the value of totalXOR.

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 get bitwise XOR
// of all possible pairs of
// the given array
int TotalXorPair(int arr[], int N)
{
    // Stores bitwise XOR
    // of all possible pairs
    int totalXOR = 0;
 
    // Generate all possible pairs
    // and calculate bitwise XOR
    // of all possible pairs
    for (int i = 0; i < N; i++) {
        for (int j = i + 1; j < N;
             j++) {
 
            // Calculate bitwise XOR
            // of each pair
            totalXOR ^= arr[i]
                        ^ arr[j];
        }
    }
    return totalXOR;
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 2, 3, 4 };
    int N = sizeof(arr) / sizeof(arr[0]);
    cout << TotalXorPair(arr, N);
}

Java




// Java program to implement
// the above approach
class GFG{
   
// Function to get bitwise XOR
// of all possible pairs of
// the given array
public static int TotalXorPair(int arr[],
                               int N)
{
  // Stores bitwise XOR
  // of all possible pairs
  int totalXOR = 0;
 
  // Generate all possible pairs
  // and calculate bitwise XOR
  // of all possible pairs
  for (int i = 0; i < N; i++)
  {
    for (int j = i + 1; j < N; j++)
    {
      // Calculate bitwise XOR
      // of each pair
      totalXOR ^= arr[i] ^ arr[j];
    }
  }
   
  return totalXOR;
}
 
// Driver code   
public static void main(String[] args)
{
  int arr[] = {1, 2, 3, 4};
  int N = arr.length;
  System.out.print(TotalXorPair(arr, N));
}
}
 
// This code is contributed by divyeshrabadiya07

Python3




# Python3 program to implement
# the above approach
 
# Function to get bitwise XOR
# of all possible pairs of
# the given array
def TotalXorPair(arr, N):
   
    # Stores bitwise XOR
    # of all possible pairs
    totalXOR = 0;
 
    # Generate all possible pairs
    # and calculate bitwise XOR
    # of all possible pairs
    for i in range(0, N):
        for j in range(i + 1, N):
           
            # Calculate bitwise XOR
            # of each pair
            totalXOR ^= arr[i] ^ arr[j];
 
    return totalXOR;
 
# Driver code
if __name__ == '__main__':
   
    arr = [1, 2, 3, 4];
    N = len(arr);
    print(TotalXorPair(arr, N));
 
# This code is contributed by shikhasingrajput

C#




// C# program to implement
// the above approach
using System;
 
class GFG{
   
// Function to get bitwise XOR
// of all possible pairs of
// the given array
public static int TotalXorPair(int []arr,
                               int N)
{
   
  // Stores bitwise XOR
  // of all possible pairs
  int totalXOR = 0;
 
  // Generate all possible pairs
  // and calculate bitwise XOR
  // of all possible pairs
  for(int i = 0; i < N; i++)
  {
    for(int j = i + 1; j < N; j++)
    {
       
      // Calculate bitwise XOR
      // of each pair
      totalXOR ^= arr[i] ^ arr[j];
    }
  }
  return totalXOR;
}
 
// Driver code   
public static void Main(String[] args)
{
  int []arr = {1, 2, 3, 4};
  int N = arr.Length;
   
  Console.Write(TotalXorPair(arr, N));
}
}
 
// This code is contributed by Princi Singh

Javascript




<script>
 
// Javascript program to implement
// the above approach
 
// Function to get bitwise XOR
// of all possible pairs of
// the given array
function TotalXorPair(arr, N)
{
     
    // Stores bitwise XOR
    // of all possible pairs
    let totalXOR = 0;
     
    // Generate all possible pairs
    // and calculate bitwise XOR
    // of all possible pairs
    for(let i = 0; i < N; i++)
    {
        for(let j = i + 1; j < N; j++)
        {
             
            // Calculate bitwise XOR
            // of each pair
            totalXOR ^= arr[i] ^ arr[j];
        }
    }
    return totalXOR;
}
  
// Driver Code
let arr = [ 1, 2, 3, 4 ];
let N = arr.length;
 
document.write(TotalXorPair(arr, N));
 
// This code is contributed by target_2
 
</script>
Output
4

Time Complexity: O(N2)
Auxiliary Space: O(1)

Efficient Approach: To optimize the above approach, follow the observations below:

Property of Bitwise XOR: 
a ^ a ^ a …….( Even times ) = 0 
a ^ a ^ a …….( Odd times ) = a

Each element of the given array occurs exactly (N – 1) times in all possible pairs.
Therefore, if N is even, then Bitwise XOR of all possible pairs are equal to bitwise XOR of all the array elements.
Otherwise, bitwise XOR of all possible pairs are equal to 0.

Follow the steps below to solve the problem:

  • If N is odd then print 0.
  • If N is even then print the value of bitwise XOR of all the elements of the given array.

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 get bitwise XOR
// of all possible pairs of
// the given array
int TotalXorPair(int arr[], int N)
{
    // Stores bitwise XOR
    // of all possible pairs
    int totalXOR = 0;
 
    // Check if N is odd
    if (N % 2 != 0) {
        return 0;
    }
 
    // If N is even then calculate
    // bitwise XOR of all elements
    // of the given array.
    for (int i = 0; i < N; i++) {
        totalXOR ^= arr[i];
    }
    return totalXOR;
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 2, 3, 4 };
    int N = sizeof(arr) / sizeof(arr[0]);
    cout << TotalXorPair(arr, N);
}

Java




// Java program to implement
// the above approach
class GFG{
   
// Function to get bitwise XOR
// of all possible pairs of
// the given array
public static int TotalXorPair(int arr[],
                               int N)
{
  // Stores bitwise XOR
  // of all possible pairs
  int totalXOR = 0;
 
  // Check if N is odd
  if( N % 2 != 0 )
  {
    return 0;
  }
 
  // If N is even then calculate
  // bitwise XOR of all elements
  // of the array
  for(int i = 0; i < N; i++)
  {
    totalXOR ^= arr[i];
  }
 
  return totalXOR;
}
 
// Driver code   
public static void main(String[] args)
{
  int arr[] = {1, 2, 3, 4};
  int N = arr.length;
  System.out.print(TotalXorPair(arr, N));
}
}
 
// This code is contributed by math_lover

Python3




# Python3 program to implement
# the above appraoch
 
# Function to get bitwise XOR
# of all possible pairs of
# the given array
def TotalXorPair(arr, N):
 
    # Stores bitwise XOR
    # of all possible pairs
    totalXOR = 0
 
    # Check if N is odd
    if (N % 2 != 0):
        return 0
 
    # If N is even then calculate
    # bitwise XOR of all elements
    # of the given array.
    for i in range(N):
        totalXOR ^= arr[i]
 
    return totalXOR
 
# Driver code
if __name__ == '__main__':
 
    arr = [ 1, 2, 3, 4 ]
    N = len(arr)
     
    print(TotalXorPair(arr, N))
 
# This code is contributed by Shivam Singh

C#




// C# program to implement
// the above approach
using System;
 
class GFG{
   
// Function to get bitwise XOR
// of all possible pairs of
// the given array
static int TotalXorPair(int []arr,
                        int N)
{
     
    // Stores bitwise XOR
    // of all possible pairs
    int totalXOR = 0;
     
    // Check if N is odd
    if (N % 2 != 0)
    {
        return 0;
    }
     
    // If N is even then calculate
    // bitwise XOR of all elements
    // of the array
    for(int i = 0; i < N; i++)
    {
        totalXOR ^= arr[i];
    }
    return totalXOR;
}
 
// Driver code   
public static void Main(String[] args)
{
    int []arr = { 1, 2, 3, 4 };
    int N = arr.Length;
     
    Console.Write(TotalXorPair(arr, N));
}
}
 
// This code is contributed by doreamon_

Javascript




<script>
 
// Javascript program to implement
// the above approach
 
// Function to get bitwise XOR
// of all possible pairs of
// the given array
function TotalXorPair(arr, N)
{
    // Stores bitwise XOR
    // of all possible pairs
    let totalXOR = 0;
 
    // Check if N is odd
    if (N % 2 != 0) {
        return 0;
    }
 
    // If N is even then calculate
    // bitwise XOR of all elements
    // of the given array.
    for (let i = 0; i < N; i++) {
        totalXOR ^= arr[i];
    }
    return totalXOR;
}
 
// Driver Code
    let arr = [ 1, 2, 3, 4 ];
    let N = arr.length;
    document.write(TotalXorPair(arr, N));
 
</script>
Output
4

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

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