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Print all possible pair with prime XOR in the Array

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  • Last Updated : 28 Jan, 2022
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Given an array arr[] of N positive integers. The task is to print all possible pairs such that their XOR is a Prime Number.
Examples: 
 

Input: arr[] = {1, 3, 6, 11} 
Output: (1, 3) (1, 6) (3, 6) (6, 11) 
Explanation: 
The XOR of the above pairs: 
1^3 = 2 
1^6 = 7 
3^6 = 5 
6^11 = 13
Input: arr[] = { 22, 58, 63, 0, 47 } 
Output: (22, 63) (58, 63) (0, 47) 
Explanation: 
The XOR of the above pairs: 
22^33 = 37 
58^63 = 5 
0^47 = 47 
 

 

Approach: 
 

  1. Generate all Prime Numbers using Sieve of Eratosthenes.
  2. For all possible pairs from the given array, check if the XOR of that pair is prime or not.
  3. If the XOR of a pair is prime then print that pair else check for the next pair.

Below is the implementation of the above approach: 
 

CPP




// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
const int sz = 1e5;
bool isPrime[sz + 1];
 
// Function for Sieve of Eratosthenes
void generatePrime()
{
    int i, j;
    memset(isPrime, true, sizeof(isPrime));
 
    isPrime[0] = isPrime[1] = false;
 
    for (i = 2; i * i <= sz; i++) {
 
        // If i is prime, then make all
        // multiples of i false
        if (isPrime[i]) {
            for (j = i * i; j < sz; j += i) {
                isPrime[j] = false;
            }
        }
    }
}
 
// Function to print all Pairs whose
// XOR is prime
void Pair_of_PrimeXor(int A[], int n)
{
 
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
 
            // if A[i]^A[j] is prime,
            // then print this pair
            if (isPrime[(A[i] ^ A[j])]) {
 
                cout << "(" << A[i]
                     << ", " << A[j] << ") ";
            }
        }
    }
}
 
// Driver Code
int main()
{
    int A[] = { 1, 3, 6, 11 };
    int n = sizeof(A) / sizeof(A[0]);
 
    // Generate all the prime number
    generatePrime();
 
    // Function Call
    Pair_of_PrimeXor(A, n);
    return 0;
}

Java




// Java implementation of the above approach
class GFG
{
static int sz = (int) 1e5;
static boolean []isPrime = new boolean[sz + 1];
  
// Function for Sieve of Eratosthenes
static void generatePrime()
{
    int i, j;
    for (i = 2; i  <= sz; i++)
        isPrime[i] = true;
  
    for (i = 2; i * i <= sz; i++) {
  
        // If i is prime, then make all
        // multiples of i false
        if (isPrime[i]) {
            for (j = i * i; j < sz; j += i) {
                isPrime[j] = false;
            }
        }
    }
}
  
// Function to print all Pairs whose
// XOR is prime
static void Pair_of_PrimeXor(int A[], int n)
{
  
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
  
            // if A[i]^A[j] is prime,
            // then print this pair
            if (isPrime[(A[i] ^ A[j])]) {
  
                System.out.print("(" +  A[i]
                    + ", " +  A[j]+ ") ");
            }
        }
    }
}
  
// Driver Code
public static void main(String[] args)
{
    int A[] = { 1, 3, 6, 11 };
    int n = A.length;
  
    // Generate all the prime number
    generatePrime();
  
    // Function Call
    Pair_of_PrimeXor(A, n);
}
}
 
// This code is contributed by sapnasingh4991

Python




# Python implementation of the above approach
sz = 10**5
isPrime = [True]*(sz + 1)
 
# Function for Sieve of Eratosthenes
def generatePrime():
    i, j = 0, 0
    isPrime[0] = isPrime[1] = False
 
    for i in range(2, sz + 1):
        if i * i > sz:
            break
 
        # If i is prime, then make all
        # multiples of i false
        if (isPrime[i]):
            for j in range(i*i, sz, i):
                isPrime[j] = False
 
# Function to print all Pairs whose
# XOR is prime
def Pair_of_PrimeXor(A, n):
 
    for i in range(n):
        for j in range(i + 1, n):
 
            # if A[i]^A[j] is prime,
            # then print this pair
            if (isPrime[(A[i] ^ A[j])]):
 
                print("(",A[i],",",A[j],")",end=" ")
 
# Driver Code
if __name__ == '__main__':
    A = [1, 3, 6, 11]
    n =len(A)
 
    # Generate all the prime number
    generatePrime()
 
    # Function Call
    Pair_of_PrimeXor(A, n)
 
# This code is contributed by mohit kumar 29

C#




// C# implementation of the above approach
using System;
 
class GFG
{
static int sz = (int) 1e5;
static bool []isPrime = new bool[sz + 1];
   
// Function for Sieve of Eratosthenes
static void generatePrime()
{
    int i, j;
    for (i = 2; i  <= sz; i++)
        isPrime[i] = true;
   
    for (i = 2; i * i <= sz; i++) {
   
        // If i is prime, then make all
        // multiples of i false
        if (isPrime[i]) {
            for (j = i * i; j < sz; j += i) {
                isPrime[j] = false;
            }
        }
    }
}
   
// Function to print all Pairs whose
// XOR is prime
static void Pair_of_PrimeXor(int []A, int n)
{
   
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
   
            // if A[i]^A[j] is prime,
            // then print this pair
            if (isPrime[(A[i] ^ A[j])]) {
   
                Console.Write("(" +  A[i]
                    + ", " +  A[j]+ ") ");
            }
        }
    }
}
   
// Driver Code
public static void Main(String[] args)
{
    int []A = { 1, 3, 6, 11 };
    int n = A.Length;
   
    // Generate all the prime number
    generatePrime();
   
    // Function Call
    Pair_of_PrimeXor(A, n);
}
}
 
// This code is contributed by Rajput-Ji

Javascript




<script>
 
// Javascript implementation of
// the above approach
 
const sz = 100000;
let isPrime = new Array(sz + 1).fill(true);
 
// Function for Sieve of Eratosthenes
function generatePrime()
{
    let i, j;
 
    isPrime[0] = isPrime[1] = false;
 
    for (i = 2; i * i <= sz; i++) {
 
        // If i is prime, then make all
        // multiples of i false
        if (isPrime[i]) {
            for (j = i * i; j < sz; j += i) {
                isPrime[j] = false;
            }
        }
    }
}
 
// Function to print all Pairs whose
// XOR is prime
function Pair_of_PrimeXor(A, n)
{
 
    for (let i = 0; i < n; i++) {
        for (let j = i + 1; j < n; j++) {
 
            // if A[i]^A[j] is prime,
            // then print this pair
            if (isPrime[(A[i] ^ A[j])]) {
 
                document.write("(" + A[i]
                     + ", " + A[j] + ") ");
            }
        }
    }
}
 
// Driver Code
    let A = [ 1, 3, 6, 11 ];
    let n = A.length;
 
    // Generate all the prime number
    generatePrime();
 
    // Function Call
    Pair_of_PrimeXor(A, n);
 
</script>

Output: 

(1, 3) (1, 6) (3, 6) (6, 11)

 

Time Complexity: O(N2), where N is the length of the given array.

Auxiliary Space: O(sz)
 


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