Print all distinct even and odd prefix Bitwise XORs of first N natural numbers

Given a positive integer N, the task is to print all the distinct even and odd values of prefix Bitwise XORs of first N natural numbers.

Examples:

Input: N = 6
Output:
Even: 0 4
Odd: 1 3 7
Explanation:
The prefix Bitwise XOR of the first 6 natural number si {1, 3, 0, 4, 1, 7}.
Even prefix Bitwise XORs are 0, 4.
Odd prefix Bitwise XORs are 1, 3, 7.

Input: N = 9
Output:
Even: 0 4 8
Odd: 1 3 7

Approach: The given problem can be solved based on the below observations:

• If the value of N modulo 4 is 0, then the value of Bitwise XOR of the first N natural numbers is N.
• If the value of N modulo 4 is 1, then the value of Bitwise XOR of the first N natural numbers is 1.
• If the value of N modulo 4 is 2, then the value of Bitwise XOR of the first N natural numbers is (N + 1).
• If the value of N modulo 4 is 3, then the value of Bitwise XOR of the first N natural numbers is 0.

Therefore, from the above principle, it can be said that Bitwise XOR as even numbers will always come as 0 or multiples of 4, and Bitwise XOR as odd numbers will always come as 1 or 1 less than multiples of 4.

Below is the implementation of the above approach.

C++

// C++ program for the above approach

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

// Print all distinct even & odd
// prefix Bitwise XORs from 1 to N
void evenOddBitwiseXOR(int N)
{

    cout << "Even: " << 0 << " ";

    // Print the even number
    for (int i = 4; i <= N; i = i + 4) {
        cout << i << " ";
    }

    cout << "\n";

    cout << "Odd: " << 1 << " ";

    // Print the odd number
    for (int i = 4; i <= N; i = i + 4) {
        cout << i - 1 << " ";
    }

    if (N % 4 == 2)
        cout << N + 1;
    else if (N % 4 == 3)
        cout << N;
}

// Driver Code
int main()
{
    int N = 6;
    evenOddBitwiseXOR(N);

    return 0;
}

Java

// Java approach for the above approach
class GFG{

// Print all distinct even & odd
// prefix Bitwise XORs from 1 to N
static void evenOddBitwiseXOR(int N)
{
    System.out.print("Even: " + 0 + " ");

    // Print the even number
    for(int i = 4; i <= N; i = i + 4)
    {
        System.out.print(i + " ");
    }

    System.out.print("\n");

    System.out.print("Odd: " + 1 + " ");

    // Print the odd number
    for(int i = 4; i <= N; i = i + 4)
    {
        System.out.print(i - 1 + " ");
    }

    if (N % 4 == 2)
        System.out.print(N + 1);
    else if (N % 4 == 3)
        System.out.print(N);
}

// Driver Code
public static void main(String[] args)
{
    int N = 6;
    evenOddBitwiseXOR(N);
}
}

// This code is contributed by abhinavjain194

Python3

# Python3 program for the above approach

# Print all distinct even & odd
# prefix Bitwise XORs from 1 to N
def evenOddBitwiseXOR(N):

    print("Even: ", 0, end = " ")

    # Print the even number
    for i in range(4, N + 1, 4):
        print(i, end = " ")

    print()

    print("Odd: ", 1, end = " ")

    # Print the odd number
    for i in range(4, N + 1, 4):
        print(i - 1, end = " ")

    if (N % 4 == 2):
        print(N + 1)
    elif (N % 4 == 3):
        print(N)

# Driver Code
N = 6

evenOddBitwiseXOR(N)

# This code is contributed by sanjoy_62

C#

// C# program for the above approach
using System;

class GFG{

// Print all distinct even & odd
// prefix Bitwise XORs from 1 to N
static void evenOddBitwiseXOR(int N)
{
    Console.Write("Even: " + 0 + " ");

    // Print the even number
    for(int i = 4; i <= N; i = i + 4)
    {
        Console.Write(i + " ");
    }

   Console.Write("\n");

    Console.Write("Odd: " + 1 + " ");

    // Print the odd number
    for(int i = 4; i <= N; i = i + 4)
    {
        Console.Write(i - 1 + " ");
    }

    if (N % 4 == 2)
        Console.Write(N + 1);
    else if (N % 4 == 3)
        Console.Write(N);
}

// Driver code
public static void Main()
{
    int N = 6;
    evenOddBitwiseXOR(N);
}
}

// This code is contributed by splevel62

Javascript

<script>

// Javascript implementation of the above approach

// Print all distinct even & odd
// prefix Bitwise XORs from 1 to N
function evenOddBitwiseXOR(N)
{
    document.write("Even: " + 0 + " ");

    // Print the even number
    for(let i = 4; i <= N; i = i + 4)
    {
        document.write(i + " ");
    }

    document.write("<br/>");

    document.write("Odd: " + 1 + " ");

    // Print the odd number
    for(let i = 4; i <= N; i = i + 4)
    {
       document.write(i - 1 + " ");
    }

    if (N % 4 == 2)
        document.write(N + 1);
    else if (N % 4 == 3)
        document.write(N);
}


  // Driver Code

    let N = 6;
    evenOddBitwiseXOR(N);

</script>

Output:

Even: 0 4 
Odd: 1 3 7

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