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 = 7Input: 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++ 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 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 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# 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 |
<script> // JavaScript program for the above approach // Function to count prime factors // using Sieve of Eratosthenes function sieve( arr, n)
{ // Iterate in the [2, N]
for ( var i = 2; i <= n; i++) {
// If the current number is prime
if (arr[i] == 0)
// Iterate over all multiples of i
for ( var 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 function findXOR( n)
{ // arr[i]: Stores the number of
// distinct prime factors of i
var arr = new Array(n + 1);
arr.fill(0);
// Initialize the base cases
arr[0] = arr[1] = 1;
// Function Call to fill
// the array, arr[]
sieve(arr, n);
// Store the required result
var ans = 0;
// Iterate over the range [2, N]
for ( var 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
document.write( ans);
} n = 20; // Function Call findXOR(n); // This code is contributed by SoumikMondal </script> |
7
Time Complexity: O(N*log(logN))
Auxiliary Space: O(N)