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

  • Last Updated : 17 May, 2021

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)

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