Given an array of size N. Find the number of pairs (i, j) such that XOR = 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**

XOR = 0 is only satisfied when . 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 then, it will contribute 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))

Output :

2

** 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 :

# 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))

Output :

2

** 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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