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Find number of pairs in an array such that their XOR is 0

Difficulty Level : Easy
Last Updated : 21 Jan, 2021

Given an array $A[ ]$ of size N. Find the number of pairs (i, j) such that $A_i$ XOR $A_j$ = 0, and 1 <= i < j <= N.
Examples :

Input : A[] = {1, 3, 4, 1, 4}
Output : 2
Explanation : Index (0, 3) and (2, 4)

Input : A[] = {2, 2, 2}
Output : 3

First Approach : Sorting
$A_i$ XOR $A_j$ = 0 is only satisfied when $A_i = A_j$ . Therefore, we will first sort the array and then count the frequency of each element. By combinatorics, we can observe that if frequency of some element is $count$ then, it will contribute $count*(count-1)/2$ to the answer.
Below is the implementation of above approach:

C++

// C++ program to find number 
// of pairs in an array such
// that their XOR is 0
#include <bits/stdc++.h>
using namespace std;
 
// Function to calculate the
// count
int calculate(int a[], int n)
{
    // Sorting the list using
    // built in function
    sort(a, a + n);
 
    int count = 1;
    int answer = 0;
 
    // Traversing through the
    // elements
    for (int i = 1; i < n; i++) {
     
        if (a[i] == a[i - 1]){
 
            // Counting frequency of each 
            // elements
            count += 1;
             
        } 
        else
        {
            // Adding the contribution of
            // the frequency to the answer
            answer = answer + (count * (count - 1)) / 2;
            count = 1;
        }
    }
 
    answer = answer + (count * (count - 1)) / 2;
 
    return answer;
}
 
// Driver Code
int main()
{
 
    int a[] = { 1, 2, 1, 2, 4 };
    int n = sizeof(a) / sizeof(a[0]);
 
    // Print the count
    cout << calculate(a, n);
    return 0;
}
 
// This article is contributed by Sahil_Bansall.

Java

// Java program to find number 
// of pairs in an array such
// that their XOR is 0
import java.util.*;
 
class GFG 
{
    // Function to calculate 
    // the count
    static int calculate(int a[], int n)
    {
        // Sorting the list using
        // built in function
        Arrays.sort(a);
     
        int count = 1;
        int answer = 0;
     
        // Traversing through the
        // elements
        for (int i = 1; i < n; i++) 
        {
         
            if (a[i] == a[i - 1])
            {
                // Counting frequency of each 
                // elements
                count += 1;
                 
            } 
            else
            {
                // Adding the contribution of
                // the frequency to the answer
                answer = answer + (count * (count - 1)) / 2;
                count = 1;
            }
        }
     
        answer = answer + (count * (count - 1)) / 2;
     
        return answer;
    }
     
    // Driver Code
    public static void main (String[] args) 
    {
        int a[] = { 1, 2, 1, 2, 4 };
        int n = a.length;
     
        // Print the count
        System.out.println(calculate(a, n));
    }
}
 
// This code is contributed by Ansu Kumari.

Python3

# Python3 program to find number of pairs
# in an array such that their XOR is 0
 
# Function to calculate the count
def calculate(a) :
 
    # Sorting the list using
    # built in function
    a.sort()
     
    count = 1
    answer = 0
     
    # Traversing through the elements
    for i in range(1, len(a)) :
 
        if a[i] == a[i - 1] :
             
            # Counting frequncy of each elements
            count += 1
 
        else :
 
            # Adding the contribution of
            # the frequency to the answer
            answer = answer + count * (count - 1) // 2
            count = 1
 
    answer = answer + count * (count - 1) // 2
     
    return answer
 
 
# Driver Code
if __name__ == '__main__':
     
    a = [1, 2, 1, 2, 4]
 
    # Print the count
    print(calculate(a))

C#

// C# program to find number 
// of pairs in an array such
// that their XOR is 0
using System;
 
class GFG 
{
    // Function to calculate 
    // the count
    static int calculate(int []a, int n)
    {
        // Sorting the list using
        // built in function
        Array.Sort(a);
     
        int count = 1;
        int answer = 0;
     
        // Traversing through the
        // elements
        for (int i = 1; i < n; i++) 
        {
         
            if (a[i] == a[i - 1])
            {
                // Counting frequency of each 
                // elements
                count += 1;
                 
            } 
            else
            {
                // Adding the contribution of
                // the frequency to the answer
                answer = answer + (count * (count - 1)) / 2;
                count = 1;
            }
        }
     
        answer = answer + (count * (count - 1)) / 2;
     
        return answer;
    }
     
    // Driver Code
    public static void Main () 
    {
        int []a = { 1, 2, 1, 2, 4 };
        int n = a.Length;
     
        // Print the count
        Console.WriteLine(calculate(a, n));
    }
}
 
// This code is contributed by vt_m.

PHP

<?php
// PHP program to find number 
// of pairs in an array such
// that their XOR is 0
 
// Function to calculate 
// the count
function calculate($a, $n)
{
     
    // Sorting the list using
    // built in function
    sort($a);
 
    $count = 1;
    $answer = 0;
 
    // Traversing through the
    // elements
    for ($i = 1; $i < $n; $i++)
    {
     
        if ($a[$i] == $a[$i - 1])
        {
 
            // Counting frequency of 
            // each elements
            $count += 1;
             
        } 
         
        else
        {
             
            // Adding the contribution of
            // the frequency to the answer
            $answer = $answer + ($count * 
                       ($count - 1)) / 2;
            $count = 1;
        }
    }
 
    $answer = $answer + ($count * 
               ($count - 1)) / 2;
 
    return $answer;
}
 
    // Driver Code
    $a = array(1, 2, 1, 2, 4);
    $n = count($a);
 
    // Print the count
    echo calculate($a, $n);
 
// This code is contributed by anuj_67.
?>

Output :

Time Complexity : O(N Log N)
Second Approach : Hashing (Index Mapping)
Solution is handy, if we can count the frequency of each element in the array. Index mapping technique can be used to count the frequency of each element.
Below is the implementation of above approach :

C++

// C++ program to find number of pairs
// in an array such that their XOR is 0
#include <bits/stdc++.h>
using namespace std;
 
// Function to calculate the answer
int calculate(int a[], int n){
     
    // Finding the maximum of the array
    int *maximum = max_element(a, a + n);
 
    // Creating frequency array
    // With initial value 0
    int frequency[*maximum + 1] = {0};
     
    // Traversing through the array 
    for(int i = 0; i < n; i++)
    { 
        // Counting frequency
        frequency[a[i]] += 1;
    }
    int answer = 0;
     
    // Traversing through the frequency array
    for(int i = 0; i < (*maximum)+1; i++)
    { 
        // Calculating answer
        answer = answer + frequency[i] * (frequency[i] - 1) ;
    }
    return answer/2;
}
 
// Driver Code
int main()
{
   int a[] = {1, 2, 1, 2, 4};
   int n = sizeof(a) / sizeof(a[0]); 
    
   // Function calling
   cout << (calculate(a,n));
}
 
// This code is contributed by Smitha

Java

// Java program to find number of pairs 
// in an array such that their XOR is 0 
import java.util.*;
 
class GFG 
{
 
    // Function to calculate the answer 
    static int calculate(int a[], int n) 
    {
 
        // Finding the maximum of the array 
        int maximum = Arrays.stream(a).max().getAsInt();
 
        // Creating frequency array 
        // With initial value 0 
        int frequency[] = new int[maximum + 1];
 
        // Traversing through the array 
        for (int i = 0; i < n; i++) 
        {
             
            // Counting frequency 
            frequency[a[i]] += 1;
        }
        int answer = 0;
 
        // Traversing through the frequency array 
        for (int i = 0; i < (maximum) + 1; i++) 
        {
             
            // Calculating answer 
            answer = answer + frequency[i] * (frequency[i] - 1);
        }
        return answer / 2;
    }
 
    // Driver Code 
    public static void main(String[] args) 
    {
        int a[] = {1, 2, 1, 2, 4};
        int n = a.length;
 
        // Function calling 
        System.out.println(calculate(a, n));
    }
}
 
// This code is contributed by 29AjayKumar

Python 3

# Python3 program to find number of pairs
# in an array such that their XOR is 0
 
# Function to calculate the answer
def calculate(a) :
     
    # Finding the maximum of the array
    maximum = max(a)
     
    # Creating frequency array
    # With initial value 0
    frequency = [0 for x in range(maximum + 1)]
     
    # Traversing through the array 
    for i in a :
          
        # Counting frequency
        frequency[i] += 1
     
    answer = 0
     
    # Traversing through the frequency array
    for i in frequency :
         
        # Calculating answer
        answer = answer + i * (i - 1) // 2
     
    return answer
 
# Driver Code
a = [1, 2, 1, 2, 4]
print(calculate(a))

C#

// C# program to find number of pairs 
// in an array such that their XOR is 0 
using System;
using System.Linq;
class GFG 
{
 
    // Function to calculate the answer 
    static int calculate(int []a, int n) 
    {
 
        // Finding the maximum of the array 
        int maximum = a.Max();
 
        // Creating frequency array 
        // With initial value 0 
        int []frequency = new int[maximum + 1];
 
        // Traversing through the array 
        for (int i = 0; i < n; i++) 
        {
             
            // Counting frequency 
            frequency[a[i]] += 1;
        }
        int answer = 0;
 
        // Traversing through the frequency array 
        for (int i = 0; i < (maximum) + 1; i++) 
        {
             
            // Calculating answer 
            answer = answer + frequency[i] * 
                             (frequency[i] - 1);
        }
        return answer / 2;
    }
 
    // Driver Code 
    public static void Main(String[] args) 
    {
        int []a = {1, 2, 1, 2, 4};
        int n = a.Length;
 
        // Function calling 
        Console.WriteLine(calculate(a, n));
    }
}
 
// This code is contributed by PrinciRaj1992

PHP

<?php
// PHP program to find number 
// of pairs in an array such 
// that their XOR is 0
 
// Function to calculate the answer
function calculate($a, $n)
{
     
    // Finding the maximum of the array
    $maximum = max($a);
     
    // Creating frequency array
    // With initial value 0
    $frequency = array_fill(0, $maximum + 1, 0);
     
    // Traversing through the array 
    for($i = 0; $i < $n; $i++)
    { 
        // Counting frequency
        $frequency[$a[$i]] += 1;
    }
    $answer = 0;
     
    // Traversing through 
    // the frequency array
    for($i = 0; $i < ($maximum) + 1; $i++)
    { 
        // Calculating answer
        $answer = $answer + $frequency[$i] *
                        ($frequency[$i] - 1);
    }
    return $answer / 2;
}
 
// Driver Code
$a = array(1, 2, 1, 2, 4);
$n = count($a);
// Function calling
echo (calculate($a,$n));
 
// This code is contributed by Smitha
?>

Output :

Time Complexity : O(N)
Note : Index Mapping method can only be used when the numbers in the array are not large. In such cases, sorting method can be used.

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