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Bitwise XOR of first N natural numbers that are product of two distinct Prime Numbers
  • Last Updated : 16 Jun, 2021

Given a positive integer N, the task is to calculate the Bitwise XOR of first N numbers which are a product of exactly two distinct prime numbers.

Examples:

Input: N = 20
Output: 7
Explanation: The numbers from the range [1, 20] which are a product of exactly two distinct prime numbers are {6, 10, 12, 14, 15, 18, 20}.
Bitwise XOR of these numbers = 6 ^ 10 ^ 12 ^ 14 ^ 15 ^ 18 ^ 20 = 7

Input: N = 50
Output: 26

Naive Approach: The simplest approach is to iterate over each number up to N and find the prime factors of each number using the prime factorization method. The numbers for which the count of distinct prime factors are found to be two, then calculate their XOR with the answer. After checking all the numbers, print the answer obtained. 

Time Complexity: O(N*√N)
Auxiliary Space: O(1)

Efficient Approach: To optimize the above approach, the idea is to use the Sieve of Eratosthenes with a little modification. Follow the steps below to solve the problem:

  • Initialize a variable ans as 0 to store the required result.
  • Create an integer array, arr[] of size N+1, and initialize with all zeros, where arr[i] denotes the number of distinct prime numbers of i.
  • Iterate in the range [2, N] using the variable i and if the value of arr[i] is 0 then, go through all the multiples of i using the variable j and increment arr[j] value by 1 since i is a prime factor of j.
  • Iterate in the range [2, N] using the variable i and if arr[i] is equal to 2, then take XOR of i with ans.
  • Print the value of ans as the result.

Below is the implementation of the above approach:



C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to count prime factors
// using Sieve of Eratosthenes
void sieve(int arr[], int n)
{
    // Iterate in the [2, N]
    for (int i = 2; i <= n; i++) {
 
        // If the current number is prime
        if (arr[i] == 0)
 
            // Iterate over all multiples of i
            for (int j = 2 * i; j <= n; j += i) {
 
                // Increment arr[j] by 1 since
                // i is a prime factor of j
                arr[j]++;
            }
    }
}
 
// Function to find Bitwise XOR
// of first N natural numbers
// satisfying the given condition
void findXOR(int n)
{
    // arr[i]: Stores the number of
    // distinct prime factors of i
    int arr[n + 1] = { 0 };
 
    // Initialize the base cases
    arr[0] = arr[1] = 1;
 
    // Function Call to fill
    // the array, arr[]
    sieve(arr, n);
 
    // Store the required result
    int ans = 0;
 
    // Iterate over the range [2, N]
    for (int i = 2; i <= n; i++) {
 
        // Check if the i-th number has
        // exactly two distinct prime factor
        if (arr[i] == 2) {
 
            // If true, update the answer
            ans = (ans ^ i);
        }
    }
 
    // Print the result
    cout << ans;
}
 
// Driver Code
int main()
{
    // Given Input
    int n = 20;
 
    // Function Call
    findXOR(n);
 
    return 0;
}

Java




// Java program for the above approach
public class GFG {
 
    // Function to count prime factors
    // using Sieve of Eratosthenes
    static void sieve(int arr[], int n)
    {
        // Iterate in the [2, N]
        for (int i = 2; i <= n; i++) {
 
            // If the current number is prime
            if (arr[i] == 0)
 
                // Iterate over all multiples of i
                for (int j = 2 * i; j <= n; j += i) {
 
                    // Increment arr[j] by 1 since
                    // i is a prime factor of j
                    arr[j]++;
                }
        }
    }
 
    // Function to find Bitwise XOR
    // of first N natural numbers
    // satisfying the given condition
    static void findXOR(int n)
    {
       
        // arr[i]: Stores the number of
        // distinct prime factors of i
        int arr[] = new int[n + 1];
 
        // Initialize the base cases
        arr[0] = arr[1] = 1;
 
        // Function Call to fill
        // the array, arr[]
        sieve(arr, n);
 
        // Store the required result
        int ans = 0;
 
        // Iterate over the range [2, N]
        for (int i = 2; i <= n; i++) {
 
            // Check if the i-th number has
            // exactly two distinct prime factor
            if (arr[i] == 2) {
 
                // If true, update the answer
                ans = (ans ^ i);
            }
        }
 
        // Print the result
        System.out.println(ans);
    }
 
    // Driver code
    public static void main(String[] args)
    {
      // Given Input
        int n = 20;
 
        // Function Call
        findXOR(n);
    }
}
 
// This code is contributed by abhinavjain194

Python3




# Python3 program for the above approach
 
# Function to count prime factors
# using Sieve of Eratosthenes
def sieve(arr, n):
     
    # Iterate in the [2, N]
    for i in range(2, n + 1, 1):
         
        # If the current number is prime
        if (arr[i] == 0):
             
            # Iterate over all multiples of i
            for j in range(2 * i, n + 1, i):
                 
                # Increment arr[j] by 1 since
                # i is a prime factor of j
                arr[j] += 1
 
# Function to find Bitwise XOR
# of first N natural numbers
# satisfying the given condition
def findXOR(n):
     
    # arr[i]: Stores the number of
    # distinct prime factors of i
    arr = [0 for i in range(n + 1)]
 
    # Initialize the base cases
    arr[0] = arr[1] = 1
 
    # Function Call to fill
    # the array, arr[]
    sieve(arr, n)
 
    # Store the required result
    ans = 0
 
    # Iterate over the range [2, N]
    for i in range(2, n + 1, 1):
         
        # Check if the i-th number has
        # exactly two distinct prime factor
        if (arr[i] == 2):
             
            # If true, update the answer
            ans = (ans ^ i)
 
    # Print the result
    print(ans)
 
# Driver Code
if __name__ == '__main__':
     
    # Given Input
    n = 20
     
    # Function Call
    findXOR(n)
         
# This code is contributed by SURENDRA_GANGWAR

C#




// C# program for the above approach
using System;
 
class GFG{
 
// Function to count prime factors
// using Sieve of Eratosthenes
static void sieve(int []arr, int n)
{
     
    // Iterate in the [2, N]
    for(int i = 2; i <= n; i++)
    {
         
        // If the current number is prime
        if (arr[i] == 0)
 
            // Iterate over all multiples of i
            for(int j = 2 * i; j <= n; j += i)
            {
                 
                // Increment arr[j] by 1 since
                // i is a prime factor of j
                arr[j]++;
            }
    }
}
 
// Function to find Bitwise XOR
// of first N natural numbers
// satisfying the given condition
static void findXOR(int n)
{
     
    // arr[i]: Stores the number of
    // distinct prime factors of i
    int []arr = new int[n + 1];
 
    // Initialize the base cases
    arr[0] = arr[1] = 1;
 
    // Function Call to fill
    // the array, arr[]
    sieve(arr, n);
 
    // Store the required result
    int ans = 0;
 
    // Iterate over the range [2, N]
    for(int i = 2; i <= n; i++)
    {
         
        // Check if the i-th number has
        // exactly two distinct prime factor
        if (arr[i] == 2)
        {
 
            // If true, update the answer
            ans = (ans ^ i);
        }
    }
 
    // Print the result
    Console.WriteLine(ans);
}
 
// Driver code
public static void Main(String[] args)
{
     
    // Given Input
    int n = 20;
     
    // Function Call
    findXOR(n);
}
}
 
// This code is contributed by ankThon
Output: 
7

 

Time Complexity: O(N*log(logN))
Auxiliary Space: O(N)

 

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