XOR of all Prime numbers in an Array at positions divisible by K

Last Updated : 16 Nov, 2021

Given an array arr of integers of size N and an integer K, the task is to find the XOR of all the numbers which are prime and at a position divisible by K.

Examples:

Input: arr[] = {2, 3, 5, 7, 11, 8}, K = 2
Output: 4
Explanation:
Positions which are divisible by K are 2, 4, 6 and the elements are 3, 7, 8.
3 and 7 are primes among these.
Therefore XOR = 3 ^ 7 = 4

Input: arr[] = {1, 2, 3, 4, 5}, K = 3
Output: 3

Naive Approach: Traverse the array and for every position i which is divisible by k, check if it is prime or not. If it is a prime then compute the XOR of this number with the previous answer.

Efficient Approach: An efficient approach is to use Sieve Of Eratosthenes. Sieve array can be used to check a number is prime or not in O(1) time.

Below is the implementation of the above approach:

C++

// C++ program to find XOR of
// all Prime numbers in an Array
// at positions divisible by K

#include <bits/stdc++.h>
using namespace std;
#define MAX 1000005

void SieveOfEratosthenes(vector<bool>& prime)
{
    // 0 and 1 are not prime numbers
    prime[1] = false;
    prime[0] = false;

    for (int p = 2; p * p < MAX; p++) {

        // If prime[p] is not changed,
        // then it is a prime
        if (prime[p] == true) {

            // Update all multiples of p
            for (int i = p * 2;
                 i < MAX; i += p)
                prime[i] = false;
        }
    }
}

// Function to find the required XOR
void prime_xor(int arr[], int n, int k)
{

    vector<bool> prime(MAX, true);

    SieveOfEratosthenes(prime);

    // To store XOR of the primes
    long long int ans = 0;

    // Traverse the array
    for (int i = 0; i < n; i++) {

        // If the number is a prime
        if (prime[arr[i]]) {

            // If index is divisible by k
            if ((i + 1) % k == 0) {
                ans ^= arr[i];
            }
        }
    }

    // Print the xor
    cout << ans << endl;
}

// Driver code
int main()
{
    int arr[] = { 2, 3, 5, 7, 11, 8 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int K = 2;

    // Function call
    prime_xor(arr, n, K);

    return 0;
}

Java

// Java program to find XOR of
// all Prime numbers in an Array
// at positions divisible by K
class GFG {

    static int MAX = 1000005;
    static boolean prime[] = new boolean[MAX];
    
    static void SieveOfEratosthenes(boolean []prime)
    {
        // 0 and 1 are not prime numbers
        prime[1] = false;
        prime[0] = false;
    
        for (int p = 2; p * p < MAX; p++) {
    
            // If prime[p] is not changed,
            // then it is a prime
            if (prime[p] == true) {
    
                // Update all multiples of p
                for (int i = p * 2;i < MAX; i += p)
                    prime[i] = false;
            }
        }
    }
    
    // Function to find the required XOR
    static void prime_xor(int arr[], int n, int k)
    {
        
        for(int i = 0; i < MAX; i++)
            prime[i] = true;
    
        SieveOfEratosthenes(prime);
    
        // To store XOR of the primes
        int ans = 0;
    
        // Traverse the array
        for (int i = 0; i < n; i++) {
    
            // If the number is a prime
            if (prime[arr[i]]) {
    
                // If index is divisible by k
                if ((i + 1) % k == 0) {
                    ans ^= arr[i];
                }
            }
        }
    
        // Print the xor
        System.out.println(ans);
    }
    
    // Driver code
    public static void main (String[] args) 
    {
        int arr[] = { 2, 3, 5, 7, 11, 8 };
        int n = arr.length;
        int K = 2;
    
        // Function call
        prime_xor(arr, n, K);
    
    }
}

// This code is contributed by Yash_R

Python3

# Python3 program to find XOR of 
# all Prime numbers in an Array 
# at positions divisible by K 
MAX = 1000005 

def SieveOfEratosthenes(prime) : 

    # 0 and 1 are not prime numbers 
    prime[1] = False; 
    prime[0] = False; 

    for p in range(2, int(MAX ** (1/2))) :

        # If prime[p] is not changed, 
        # then it is a prime 
        if (prime[p] == True) :

            # Update all multiples of p 
            for i in range(p * 2, MAX, p) :
                prime[i] = False; 

# Function to find the required XOR 
def prime_xor(arr, n, k) : 

    prime = [True]*MAX ; 

    SieveOfEratosthenes(prime); 

    # To store XOR of the primes 
    ans = 0; 

    # Traverse the array 
    for i in range(n) :

        # If the number is a prime 
        if (prime[arr[i]]) :

            # If index is divisible by k 
            if ((i + 1) % k == 0) :
                ans ^= arr[i]; 

    # Print the xor 
    print(ans); 

# Driver code 
if __name__ == "__main__" : 

    arr = [ 2, 3, 5, 7, 11, 8 ]; 
    n = len(arr); 
    K = 2; 

    # Function call 
    prime_xor(arr, n, K); 

# This code is contributed by Yash_R

// C# program to find XOR of
// all Prime numbers in an Array
// at positions divisible by K
using System;

class GFG {

    static int MAX = 1000005;
    static bool []prime = new bool[MAX];
    
    static void SieveOfEratosthenes(bool []prime)
    {
        // 0 and 1 are not prime numbers
        prime[1] = false;
        prime[0] = false;
    
        for (int p = 2; p * p < MAX; p++) {
    
            // If prime[p] is not changed,
            // then it is a prime
            if (prime[p] == true) {
    
                // Update all multiples of p
                for (int i = p * 2;i < MAX; i += p)
                    prime[i] = false;
            }
        }
    }
    
    // Function to find the required XOR
    static void prime_xor(int []arr, int n, int k)
    {
        
        for(int i = 0; i < MAX; i++)
            prime[i] = true;
    
        SieveOfEratosthenes(prime);
    
        // To store XOR of the primes
        int ans = 0;
    
        // Traverse the array
        for (int i = 0; i < n; i++) {
    
            // If the number is a prime
            if (prime[arr[i]]) {
    
                // If index is divisible by k
                if ((i + 1) % k == 0) {
                    ans ^= arr[i];
                }
            }
        }
    
        // Print the xor
        Console.WriteLine(ans);
    }
    
    // Driver code
    public static void Main (string[] args) 
    {
        int []arr = { 2, 3, 5, 7, 11, 8 };
        int n = arr.Length;
        int K = 2;
    
        // Function call
        prime_xor(arr, n, K);
    
    }
}

// This code is contributed by Yash_R

JavaScript

<script>

// Javascript program to find XOR of
// all Prime numbers in an Array
// at positions divisible by K
const MAX = 1000005;

function SieveOfEratosthenes(prime)
{
    
    // 0 and 1 are not prime numbers
    prime[1] = false;
    prime[0] = false;

    for(let p = 2; p * p < MAX; p++)
    {
        
        // If prime[p] is not changed,
        // then it is a prime
        if (prime[p] == true)
        {
            
            // Update all multiples of p
            for(let i = p * 2; i < MAX; i += p)
                prime[i] = false;
        }
    }
}

// Function to find the required XOR
function prime_xor(arr, n, k)
{
    let prime = new Array(MAX).fill(true);

    SieveOfEratosthenes(prime);

    // To store XOR of the primes
    let ans = 0;

    // Traverse the array
    for(let i = 0; i < n; i++) 
    {
        
        // If the number is a prime
        if (prime[arr[i]])
        {
            
            // If index is divisible by k
            if ((i + 1) % k == 0)
            {
                ans ^= arr[i];
            }
        }
    }
    
    // Print the xor
    document.write(ans);
}

// Driver code
let arr = [ 2, 3, 5, 7, 11, 8 ];
let n = arr.length;
let K = 2;

// Function call
prime_xor(arr, n, K);

// This code is contributed by subham348

</script>

Time Complexity: O(N log (log N))

Auxiliary Space: O(?n)

Number of 0s and 1s at prime positions in the given array

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XOR of all Prime numbers in an Array at positions divisible by K

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