Bitwise XOR of all odd numbers from a given range

Given an integer N, the task is to find the Bitwise XOR of all odd numbers in the range [1, N].

Examples:

Input: 11
Output: 2
Explanation: Bitwise XOR of all odd numbers up to 11 = 1 ^ 3 ^ 5 ^ 7 ^ 9 ^ 11 = 2.

Input: 10
Output: 9
Explanation: Bitwise XOR of all odd numbers up to 10 = 1 ^ 3 ^ 5 ^ 7 ^ 9 = 9.

Naive Approach: The simplest approach to solve the problem is to iterate over the range [1, N] and for every value, check if it is odd or not. Calculate Bitwise XOR of every odd element found and print it as the required result. 
Time Complexity: O(N)
Auxiliary Space: O(1)

Efficient Approach: To optimize the above approach, the idea is to use the following formula to calculate Bitwise XOR of all odd numbers less than or equal to N:

Let f(N) = 2 ^ 4 ^ 6 ^ … ^ (N − 2) ^ n and g(n) = 1 ^ 2 ^ 3 ^ … ^ (N − 2) / 2 ^ (N / 2) 
=> f(N) = 2 * g(N)

Now, let k(N) = 1 ^ 2 ^ 3 ^ … ^ (N − 2) ^ N

XOR of all odd numbers less than or equal to N = k(N) ^ f(N) [Since all even numbers cancel their own bits leaving only odd numbers]. 
Substituting the value of f(N) = 2 * g(N), XOR of all odd numbers up to N = k(N) ^ (2 * g(N))

Follow the steps below to solve the problem:

• Store XOR of numbers in the range[1, N] in a variable, say K.
• If N is odd, store the XOR of numbers in the range[1, (N – 1) / 2] in a variable, say g. Otherwise, store XOR of numbers in the range[1, N / 2] in g.
• Using the derived formula, update result to K ^ (2 * g).
• After completing the above steps, print the value of the result as the answer.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to calculate Bitwise
// XOR of odd numbers in the range [1, N]
int findXOR(int n)
{
    // N & 3 is equivalent to n % 4
    switch (n & 3) {

    // If n is multiple of 4
    case 0:
        return n;

    // If n % 4 gives remainder 1
    case 1:
        return 1;

    // If n % 4 gives remainder 2
    case 2:
        return n + 1;

    // If n % 4 gives remainder 3
    case 3:
        return 0;
    }
}

// Function to find the XOR of odd
// numbers less than or equal to N
void findOddXOR(int n)
{

    // If number is even
    if (n % 2 == 0)

        // Print the answer
        cout << ((findXOR(n))
                 ^ (2 * findXOR(n / 2)));

    // If number is odd
    else

        // Print the answer
        cout << ((findXOR(n))
                 ^ (2 * findXOR((n - 1) / 2)));
}

// Driver Code
int main()
{
    int N = 11;

    // Function Call
    findOddXOR(N);

    return 0;
}

Java

// Java program for the above approach
import java.util.*;
class GFG
{

  // Function to calculate Bitwise
  // XOR of odd numbers in the range [1, N]
  static int findXOR(int n)
  {

    // N & 3 is equivalent to n % 4
    switch (n & 3)
    {

        // If n is multiple of 4
      case 0:
        return n;

        // If n % 4 gives remainder 1
      case 1:
        return 1;

        // If n % 4 gives remainder 2
      case 2:
        return n + 1;
    }
    // If n % 4 gives remainder 3
    return 0;

  }

  // Function to find the XOR of odd
  // numbers less than or equal to N
  static void findOddXOR(int n)
  {

    // If number is even
    if (n % 2 == 0)

      // Print the answer
      System.out.print(((findXOR(n))
                        ^ (2 * findXOR(n / 2))));

    // If number is odd
    else

      // Print the answer
      System.out.print(((findXOR(n))
                        ^ (2 * findXOR((n - 1) / 2))));
  }

  // Driver Code
  public static void main(String[] args)
  {
    int N = 11;

    // Function Call
    findOddXOR(N);
  }
}

// This code is contributed by shikhasingrajput

Python3

# Python program for the above approach

# Function to calculate Bitwise
# XOR of odd numbers in the range [1, N]
def findXOR(n):

    # N & 3 is equivalent to n % 4
    if (n % 4 == 0):

        # If n is multiple of 4
        return n;
    elif (n % 4 == 1):

        # If n % 4 gives remainder 1
        return 1;

    # If n % 4 gives remainder 2
    elif (n % 4 == 2):
        return n + 1;

    # If n % 4 gives remainder 3
    elif (n % 4 == 3):
        return 0;

# Function to find the XOR of odd
# numbers less than or equal to N
def findOddXOR(n):

    # If number is even
    if (n % 2 == 0):

        # Print the answer
        print(((findXOR(n)) ^ (2 * findXOR(n // 2))));

    # If number is odd
    else:

        # Print the answer
        print(((findXOR(n)) ^ (2 * findXOR((n - 1) // 2))));

# Driver Code
if __name__ == '__main__':
    N = 11;

    # Function Call
    findOddXOR(N);

# This code is contributed by 29AjayKumar

C#

// C# program for the above approach
using System;
public class GFG
{

  // Function to calculate Bitwise
  // XOR of odd numbers in the range [1, N]
  static int findXOR(int n)
  {

    // N & 3 is equivalent to n % 4
    switch (n & 3)
    {

        // If n is multiple of 4
      case 0:
        return n;

        // If n % 4 gives remainder 1
      case 1:
        return 1;

        // If n % 4 gives remainder 2
      case 2:
        return n + 1;
    }
    // If n % 4 gives remainder 3
    return 0;

  }

  // Function to find the XOR of odd
  // numbers less than or equal to N
  static void findOddXOR(int n)
  {

    // If number is even
    if (n % 2 == 0)

      // Print the answer
      Console.Write(((findXOR(n))
                     ^ (2 * findXOR(n / 2))));

    // If number is odd
    else

      // Print the answer
      Console.Write(((findXOR(n))
                     ^ (2 * findXOR((n - 1) / 2))));
  }

  // Driver Code
  public static void Main(String[] args)
  {
    int N = 11;

    // Function Call
    findOddXOR(N);
  }
}

// This code is contributed by shikhasingrajput

Javascript

<script>
// JavaScript program for the above approach

// Function to calculate Bitwise
// XOR of odd numbers in the range [1, N]
function findXOR(n)
{
    // N & 3 is equivalent to n % 4
    switch (n & 3) {

    // If n is multiple of 4
    case 0:
        return n;

    // If n % 4 gives remainder 1
    case 1:
        return 1;

    // If n % 4 gives remainder 2
    case 2:
        return n + 1;

    // If n % 4 gives remainder 3
    case 3:
        return 0;
    }
}

// Function to find the XOR of odd
// numbers less than or equal to N
function findOddXOR(n)
{

    // If number is even
    if (n % 2 == 0)

        // Print the answer
        document.write((findXOR(n))
                ^ (2 * findXOR(n / 2)));

    // If number is odd
    else

        // Print the answer
        document.write((findXOR(n))
                ^ (2 * findXOR((n - 1) / 2)));
}

// Driver Code
    let N = 11;

    // Function Call
    findOddXOR(N);

// This code is contributed by Surbhi Tyagi.

</script>

Output:

2

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