Find the XOR of first N Prime Numbers

Namachila 27 seconds ago

0 0 5 minutes read

Given a positive integer N, the task is to find the XOR of the first N prime numbers.
Examples:

Input: N = 3
Output: 4
First 3 prime numbers are 2, 3 and 5.
And 2 ^ 3 ^ 5 = 4
Input: N = 5
Output: 8

Approach:

Create Sieve of Eratosthenes to identify if a number is prime or not in O(1) time.
Run a loop starting from 1 until and unless we find N prime numbers.
XOR all the prime numbers and neglect those which are not prime.
Finally, print the XOR of the 1^st N prime numbers.

Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define MAX 10000
 
// Create a boolean array "prime[0..n]" and initialize
// all entries it as true. A value in prime[i] will
// finally be false if i is Not a prime, else true.
bool prime[MAX + 1];
void SieveOfEratosthenes()
{
    memset(prime, true, sizeof(prime));
 
    prime[1] = false;
 
    for (int p = 2; p * p <= MAX; p++) {
 
        // If prime[p] is not changed, then it is a prime
        if (prime[p] == true) {
 
            // Set all multiples of p to non-prime
            for (int i = p * 2; i <= MAX; i += p)
                prime[i] = false;
        }
    }
}
 
// Function to return the xor of 1st N prime numbers
int xorFirstNPrime(int n)
{
    // Count of prime numbers
    int count = 0, num = 1;
 
    // XOR of prime numbers
    int xorVal = 0;
 
    while (count < n) {
 
        // If the number is prime xor it
        if (prime[num]) {
            xorVal ^= num;
 
            // Increment the count
            count++;
        }
 
        // Get to the next number
        num++;
    }
    return xorVal;
}
 
// Driver code
int main()
{
    // Create the sieve
    SieveOfEratosthenes();
 
    int n = 4;
 
    // Find the xor of 1st n prime numbers
    cout << xorFirstNPrime(n);
 
    return 0;
}

Java

// Java implementation of the approach 
class GFG 
{
static final int MAX = 10000;
 
// Create a boolean array "prime[0..n]" 
// and initialize all entries it as true. 
// A value in prime[i] will finally be false
// if i is Not a prime, else true. 
static boolean prime[] = new boolean [MAX + 1]; 
 
static void SieveOfEratosthenes() 
{ 
    int i ;
    for (i = 0; i < MAX + 1; i++)
    {
        prime[i] = true;
    }
 
    prime[1] = false; 
 
    for (int p = 2; p * p <= MAX; p++) 
    { 
 
        // If prime[p] is not changed, 
        // then it is a prime 
        if (prime[p] == true) 
        { 
 
            // Set all multiples of p to non-prime 
            for (i = p * 2; i <= MAX; i += p) 
                prime[i] = false; 
        } 
    } 
} 
 
// Function to return the xor of 
// 1st N prime numbers 
static int xorFirstNPrime(int n) 
{ 
    // Count of prime numbers 
    int count = 0, num = 1; 
 
    // XOR of prime numbers 
    int xorVal = 0; 
 
    while (count < n)
    { 
 
        // If the number is prime xor it 
        if (prime[num]) 
        { 
            xorVal ^= num; 
 
            // Increment the count 
            count++; 
        } 
 
        // Get to the next number 
        num++; 
    } 
    return xorVal; 
} 
 
// Driver code 
public static void main (String[] args) 
{ 
    // Create the sieve 
    SieveOfEratosthenes(); 
 
    int n = 4; 
 
    // Find the xor of 1st n prime numbers 
    System.out.println(xorFirstNPrime(n)); 
 
} 
}
 
// This code is contributed by AnkitRai01

Python3

# Python3 implementation of the approach
MAX = 10000
 
# Create a boolean array "prime[0..n]" and 
# initialize all entries it as true. 
# A value in prime[i] will finally be false +
# if i is Not a prime, else true.
prime = [True for i in range(MAX + 1)]
 
def SieveOfEratosthenes():
 
    prime[1] = False
 
    for p in range(2, MAX + 1):
 
        # If prime[p] is not changed, 
        # then it is a prime
        if (prime[p] == True):
 
            # Set all multiples of p to non-prime
            for i in range(2 * p, MAX + 1, p):
                prime[i] = False
 
# Function to return the xor of 
# 1st N prime numbers
def xorFirstNPrime(n):
     
    # Count of prime numbers
    count = 0
    num = 1
 
    # XOR of prime numbers
    xorVal = 0
 
    while (count < n):
 
        # If the number is prime xor it
        if (prime[num]):
            xorVal ^= num
 
            # Increment the count
            count += 1
 
        # Get to the next number
        num += 1
 
    return xorVal
 
# Driver code
 
# Create the sieve
SieveOfEratosthenes()
 
n = 4
 
# Find the xor of 1st n prime numbers
print(xorFirstNPrime(n))
 
# This code is contributed by Mohit Kumar

C#

// C# implementation of the approach 
using System;
 
class GFG
{
     
static int MAX = 10000;
 
// Create a boolean array "prime[0..n]" 
// and initialize all entries it as true. 
// A value in prime[i] will finally be false
// if i is Not a prime, else true. 
static bool []prime = new bool [MAX + 1]; 
 
static void SieveOfEratosthenes() 
{ 
    int i ;
    for (i = 0; i < MAX + 1; i++)
    {
        prime[i] = true;
    }
 
    prime[1] = false; 
 
    for (int p = 2; p * p <= MAX; p++) 
    { 
 
        // If prime[p] is not changed, 
        // then it is a prime 
        if (prime[p] == true) 
        { 
 
            // Set all multiples of p to non-prime 
            for (i = p * 2; i <= MAX; i += p) 
                prime[i] = false; 
        } 
    } 
} 
 
// Function to return the xor of 
// 1st N prime numbers 
static int xorFirstNPrime(int n) 
{ 
    // Count of prime numbers 
    int count = 0, num = 1; 
 
    // XOR of prime numbers 
    int xorVal = 0; 
 
    while (count < n)
    { 
 
        // If the number is prime xor it 
        if (prime[num]) 
        { 
            xorVal ^= num; 
 
            // Increment the count 
            count++; 
        } 
 
        // Get to the next number 
        num++; 
    } 
    return xorVal; 
} 
 
// Driver code 
static public void Main ()
{
     
    // Create the sieve 
    SieveOfEratosthenes(); 
    int n = 4; 
 
    // Find the xor of 1st n prime numbers 
    Console.Write(xorFirstNPrime(n)); 
} 
}
 
// This code is contributed by Sachin

Javascript

<script>
 
    // Javascript implementation of the approach 
     
    let MAX = 10000; 
   
    // Create a boolean array "prime[0..n]"  
    // and initialize all entries it as true.  
    // A value in prime[i] will finally be false 
    // if i is Not a prime, else true.  
    let prime = new Array(MAX + 1);  
       
    function SieveOfEratosthenes()  
    {  
        let i; 
        for (i = 0; i < MAX + 1; i++) 
        { 
            prime[i] = true; 
        } 
       
        prime[1] = false;  
       
        for (let p = 2; p * p <= MAX; p++)  
        {  
       
            // If prime[p] is not changed,  
            // then it is a prime  
            if (prime[p] == true)  
            {  
       
                // Set all multiples of p to non-prime  
                for (i = p * 2; i <= MAX; i += p)  
                    prime[i] = false;  
            }  
        }  
    }  
       
    // Function to return the xor of  
    // 1st N prime numbers  
    function xorFirstNPrime(n)  
    {  
        // Count of prime numbers  
        let count = 0, num = 1;  
       
        // XOR of prime numbers  
        let xorVal = 0;  
       
        while (count < n) 
        {  
       
            // If the number is prime xor it  
            if (prime[num])  
            {  
                xorVal ^= num;  
       
                // Increment the count  
                count++;  
            }  
       
            // Get to the next number  
            num++;  
        }  
        return xorVal;  
    }
     
    // Create the sieve  
    SieveOfEratosthenes();  
    let n = 4;  
   
    // Find the xor of 1st n prime numbers  
    document.write(xorFirstNPrime(n));  
     
</script>

Output:

Time Complexity: O(n + MAX*log(log(MAX)))

Auxiliary Space: O(MAX)

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!