# Bitwise XOR of first N natural numbers that are product of two distinct Prime Numbers

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 ` `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`

## Javascript

 ``

Output:

`7`

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

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