All pairs whose xor gives unique prime

Given an array arr[], the task is to count all the pairs whose xor gives the unique prime, i.e. no two pairs should give the same prime.
Examples:

Input: arr[] = {2, 3, 4, 5, 6, 7, 8, 9} 
Output: 6 
(2, 5), (2, 7), (2, 9), (4, 6), (4, 7) and (4, 9) are the only pairs whose XORs are primes i.e. 7, 5, 11, 2, 3 and 13 respectively.
Input: arr[] = {10, 12, 23, 45, 5, 6} 
Output: 4 

Approach: Iterating every possible pair and check whether the xor of the current pair is a prime. If its a prime then update count = count + 1 and also save the prime in an unordered_map in order to keep track of the repeating primes. Print the count in the end.
Below is the implementation of the above approach: 

C++

// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;

// Function that returns true if n is prime
bool isPrime(int n)
{
    // Corner cases
    if (n <= 1)
        return false;
    if (n <= 3)
        return true;

    // This is checked so that we can skip
    // middle five numbers in below loop
    if (n % 2 == 0 || n % 3 == 0)
        return false;

    for (int i = 5; i * i <= n; i = i + 6)
        if (n % i == 0 || n % (i + 2) == 0)
            return false;

    return true;
}

// Function to return the count of valid pairs
int countPairs(int a[], int n)
{
    int count = 0;

    unordered_map<int, int> m;

    for (int i = 0; i < n - 1; i++) {
        for (int j = i + 1; j < n; j++) {

            // If xor(a[i], a[j]) is prime and unique
            if (isPrime(a[i] ^ a[j]) && m[a[i] ^ a[j]] == 0) {
                m[(a[i] ^ a[j])]++;
                count++;
            }
        }
    }

    return count;
}

// Driver code
int main()
{
    int a[] = { 10, 12, 23, 45, 5, 6 };

    int n = sizeof(a) / sizeof(a[0]);

    cout << countPairs(a, n);
}

Java

// Java implementation of above approach
import java.util.*;
class solution
{

// Function that returns true if n is prime
static boolean isPrime(int n)
{
    // Corner cases
    if (n <= 1)
        return false;
    if (n <= 3)
        return true;

    // This is checked so that we can skip
    // middle five numbers in below loop
    if (n % 2 == 0 || n % 3 == 0)
        return false;

    for (int i = 5; i * i <= n; i = i + 6)
        if (n % i == 0 || n % (i + 2) == 0)
            return false;

    return true;
}

// Function to return the count of valid pairs
static int countPairs(int a[], int n)
{
    int count = 0;

    Map<Integer, Integer> m=new HashMap< Integer,Integer>();

    for (int i = 0; i < n - 1; i++) {
        for (int j = i + 1; j < n; j++) {

            // If xor(a[i], a[j]) is prime and unique
            if (isPrime(a[i] ^ a[j]) && m.get(a[i] ^ a[j]) == null) {
                m.put((a[i] ^ a[j]),1);
                count++;
            }
        }
    }

    return count;
}

// Driver code
public static void main(String args[])
{
    int a[] = { 10, 12, 23, 45, 5, 6 };

    int n = a.length;

    System.out.println(countPairs(a, n));
}
}
// This code is contributed by Arnab Kundu

Python3

# Python3 implementation of above approach

# Function that returns true if n is prime
def isPrime(n):

    # Corner cases
    if n <= 1:
        return False
    if n <= 3:
        return True

    # This is checked so that we can skip
    # middle five numbers in below loop
    if n % 2 == 0 or n % 3 == 0:
        return False

    i = 5
    while i * i <= n:
        if n % i == 0 or n % (i + 2) == 0:
            return False
        i += 6

    return True

# Function to return the count of valid pairs
def countPairs(a, n):
    count = 0

    m = dict()

    for i in range(n - 1):
        for j in range(i + 1, n):

            # If xor(a[i], a[j]) is prime and unique
            if isPrime(a[i] ^ a[j]) and m.get(a[i] ^ a[j], 0) == 0:
                m[(a[i] ^ a[j])] = 1
                count += 1

    return count

# Driver code
a = [10, 12, 23, 45, 5, 6]

n = len(a)

print(countPairs(a, n))

# This code is contributed by
# Rajnis09

C#

// C# implementation of the approach
using System ;
using System.Collections ;

class solution
{

    // Function that returns true if n is prime
    static bool isPrime(int n)
    {
        // Corner cases
        if (n <= 1)
            return false;
        if (n <= 3)
            return true;

        // This is checked so that we can skip
        // middle five numbers in below loop
        if (n % 2 == 0 || n % 3 == 0)
            return false;

        for (int i = 5; i * i <= n; i = i + 6)
            if (n % i == 0 || n % (i + 2) == 0)
                return false;

        return true;
    }

    // Function to return the count of valid pairs
    static int countPairs(int []a, int n)
    {
        int count = 0;

        Hashtable m=new Hashtable();

        for (int i = 0; i < n - 1; i++) {
            for (int j = i + 1; j < n; j++) {

                // If xor(a[i], a[j]) is prime and unique
                if (isPrime(a[i] ^ a[j]) && m[a[i] ^ a[j]] == null)
                {
                    m.Add((a[i] ^ a[j]),1);
                    count++;
                }
            }
        }

        return count;
    }

    // Driver code
    public static void Main()
    {
        int []a = { 10, 12, 23, 45, 5, 6 };

        int n = a.Length;

        Console.WriteLine(countPairs(a, n));
    }
    // This code is contributed by Ryuga
}

Javascript

<script>
// Javascript implementation of above approach

    // Function that returns true if n is prime
    function isPrime(n)
    {

        // Corner cases
        if (n <= 1)
            return false;
        if (n <= 3)
            return true;

        // This is checked so that we can skip
        // middle five numbers in below loop
        if (n % 2 == 0 || n % 3 == 0)
            return false;

        for (let i = 5; i * i <= n; i = i + 6)
            if (n % i == 0 || n % (i + 2) == 0)
                return false;

        return true;
    }

    // Function to return the count of valid pairs
    function countPairs(a,n)
    {
        let count = 0;
        let m = new Map();
        for (let i = 0; i < n - 1; i++)
        {
            for (let j = i + 1; j < n; j++)
            {

                // If xor(a[i], a[j]) is prime and unique
                if (isPrime(a[i] ^ a[j]) && !m.has(a[i] ^ a[j]))
                {
                    m.set((a[i] ^ a[j]), 1);
                    count++;
                }
            }
        }
        return count;
    }

    // Driver code
    let a = [10, 12, 23, 45, 5, 6];
    let n = a.length;
    document.write(countPairs(a, n));

    // This code is contributed by unknown2108
</script>

Output:

4

Time Complexity: O(n^5/2)

Auxiliary Space: O(n²)