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. 

Below is the implementation of above approach :

C++

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

// Function to find the XOR of odd numbers
// less than or equal to N using the naive approach
void findOddXOR(int n)
{
    int xorVal = 0;
    for (int i = 1; i <= n; i++) {
        if (i % 2 != 0) {
            xorVal ^= i;
        }
    }
    // Print the answer
    cout << xorVal;
}

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

    // Function Call
    findOddXOR(N);

    return 0;
}

Output

2

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)