XOR of all Prime numbers in an Array

Given an array of integers arr[]. The task is to find the bitwise XOR of all the prime numbers present in the array.

Examples:

Input: arr[] = {2, 5, 8, 4, 3}
Output: 4

Input: arr[] = {7, 12, 2, 6, 11}
Output: 14

Approach:

Below is the implementation of the above approach:

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// C++ program to find Xor of all
// Prime numbers in array
  
#include <bits/stdc++.h>
using namespace std;
  
bool prime[100005];
  
void SieveOfEratosthenes(int n)
{
  
    memset(prime, true, sizeof(prime));
  
    // false here indicates
    // that it is not prime
    prime[1] = false;
  
    for (int p = 2; p * p <= n; p++) {
  
        // If prime[p] is not changed,
        // then it is a prime
        if (prime[p]) {
  
            // Update all multiples of p,
            // set them to non-prime
            for (int i = p * 2; i <= n; i += p)
                prime[i] = false;
        }
    }
}
  
// Function to compute xor of all
// prime elements
int xorPrimes(int arr[], int n)
{
    SieveOfEratosthenes(100005);
  
    int xorVal = 0;
  
    for (int i = 0; i < n; i++) {
  
        // if the element is prime
        if (prime[arr[i]])
            xorVal = xorVal ^ arr[i];
    }
  
    return xorVal;
}
  
// Driver code
int main()
{
  
    int arr[] = { 4, 3, 2, 6, 100, 17 };
    int n = sizeof(arr) / sizeof(arr[0]);
  
    cout << xorPrimes(arr, n);
  
    return 0;
}
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// Java program to find Xor of all
// Prime numbers in array 
import java.util.Arrays;
  
  
class GFG 
{
    static boolean prime[] = new boolean[100005];
  
    static void SieveOfEratosthenes(int n) 
    {
        Arrays.fill(prime, true);
  
        // false here indicates
        // that it is not prime
        prime[1] = false;
  
        for (int p = 2; p * p < n; p++)
        {
  
            // If prime[p] is not changed,
            // then it is a prime
            if (prime[p])
            {
                // Update all multiples of p,
                // set them to non-prime
                for (int i = p * 2; i < n; i += p)
                {
                    prime[i] = false;
                }
            }
        }
    }
  
    // Function to compute xor of all
    // prime elements
    static int xorPrimes(int arr[], int n) 
    {
        SieveOfEratosthenes(100005);
        int xorVal = 0;
        for (int i = 0; i < n; i++) 
        {
            // if the element is prime
            if (prime[arr[i]]) 
            {
                xorVal = xorVal ^ arr[i];
            }
        }
        return xorVal;
    }
  
    // Driver code
    public static void main(String[] args) 
    {
        int arr[] = {4, 3, 2, 6, 100, 17};
        int n = arr.length;
        System.out.println(xorPrimes(arr, n));
    }
}
  
// This code is contributed by 
// Rajput-Ji
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# Python3 program to find Xor of 
# all Prime numbers in array 
  
prime = [True] * (100005)
  
def SieveOfEratosthenes(n): 
   
    # False here indicates 
    # that it is not prime 
    prime[1] = False
    p = 2
      
    while p*p <= n:
  
        # If prime[p] is not changed, 
        # then it is a prime 
        if prime[p]:  
  
            # Update all multiples of p, 
            # set them to non-prime 
            for i in range(p * 2, n+1, p): 
                prime[i] = False
                  
        p += 1
           
# Function to compute xor 
# of all prime elements 
def xorPrimes(arr, n): 
   
    SieveOfEratosthenes(100004
  
    xorVal = 0 
    for i in range(0, n):  
  
        # if the element is prime 
        if prime[arr[i]]: 
            xorVal = xorVal ^ arr[i] 
       
    return xorVal 
   
# Driver code 
if __name__ == "__main__"
   
    arr = [4, 3, 2, 6, 100, 17]  
    n = len(arr) 
  
    print(xorPrimes(arr, n))
  
# This code is contributed by Rituraj Jain
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// C# program to find Xor of all
// Prime numbers in array 
using System;
  
class GFG 
{
    static bool []prime = new bool[100005];
  
    static void SieveOfEratosthenes(int n) 
    {
        for(int i = 0; i < 100005; i++)
            prime[i] = true;
  
        // false here indicates
        // that it is not prime
        prime[1] = false;
  
        for (int p = 2; p * p < n; p++)
        {
  
            // If prime[p] is not changed,
            // then it is a prime
            if (prime[p])
            {
                // Update all multiples of p,
                // set them to non-prime
                for (int i = p * 2; i < n; i += p)
                {
                    prime[i] = false;
                }
            }
        }
    }
  
    // Function to compute xor of all
    // prime elements
    static int xorPrimes(int []arr, int n) 
    {
        SieveOfEratosthenes(100005);
        int xorVal = 0;
        for (int i = 0; i < n; i++) 
        {
            // if the element is prime
            if (prime[arr[i]]) 
            {
                xorVal = xorVal ^ arr[i];
            }
        }
        return xorVal;
    }
  
    // Driver code
    public static void Main() 
    {
        int []arr = {4, 3, 2, 6, 100, 17};
        int n = arr.Length;
        Console.WriteLine(xorPrimes(arr, n));
    }
}
  
/* This code contributed by PrinciRaj1992 */
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Output:
16

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