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XOR of all even numbers from a given range
  • Difficulty Level : Expert
  • Last Updated : 05 Jan, 2021

Given two integers L and R, the task is to calculate Bitwise XOR of all even numbers in the range [L, R].

Examples:

Example: 
Input: L = 10, R = 20 
Output: 30 
Explanation: 
Bitwise XOR = 10 ^ 12 ^ 14 ^ 16 ^ 18 ^ 20 = 30 
Therefore, the required output is 30.

Example: 
Input: L = 15, R = 23 
Output:
Explanation: 
Bitwise XOR = 16 ^ 18 ^ 20 ^ 22 = 0 
Therefore, the required output is 0.

Naive Approach:The simplest approach to solve the problem is to traverse all even numbers in the range [L, R] and print the Bitwise XOR of all the even numbers. 



Time Complexity: O(R – L)
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized based on the following observations:

If N is an even number: 
2 ^ 4 … ^ (N) = 2 * (1 ^ 2 ^ … ^ (N / 2))

If N is an odd number: 
2 ^ 4 … ^ (N) = 2 * (1 ^ 2 ^ … ^ ((N – 1) / 2))

Follow the steps below to solve the problem:

Below is the implementation of the above approach:

C++

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// C++ Implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to calculate XOR of
// numbers in the range [1, n]
int bitwiseXorRange(int n)
{
 
    // If n is divisible by 4
    if (n % 4 == 0)
        return n;
 
    // If n mod 4 is equal to 1
    if (n % 4 == 1)
        return 1;
 
    // If n mod 4 is equal to 2
    if (n % 4 == 2)
        return n + 1;
 
    return 0;
}
 
// Function to find XOR of even
// numbers in the range [l, r]
int evenXorRange(int l, int r)
{
 
    // Stores XOR of even numbers
    // in the range [1, l - 1]
    int xor_l;
 
    // Stores XOR of even numbers
    // in the range [1, r]
    int xor_r;
 
    // Update xor_r
    xor_r
        = 2 * bitwiseXorRange(r / 2);
 
    // Update xor_l
    xor_l
        = 2 * bitwiseXorRange((l - 1) / 2);
 
    return xor_l ^ xor_r;
}
 
// Driver Code
int main()
{
    int l = 10;
    int r = 20;
    cout << evenXorRange(l, r);
 
    return 0;
}

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Java

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// Java Implementation of the above approach
class GFG
{
     
  // Function to calculate XOR of
  // numbers in the range [1, n]
  static int bitwiseXorRange(int n)
  {
 
    // If n is divisible by 4
    if (n % 4 == 0)
      return n;
 
    // If n mod 4 is equal to 1
    if (n % 4 == 1)
      return 1;
 
    // If n mod 4 is equal to 2
    if (n % 4 == 2)
      return n + 1;
 
    return 0;
  }
 
  // Function to find XOR of even
  // numbers in the range [l, r]
  static int evenXorRange(int l, int r)
  {
 
    // Stores XOR of even numbers
    // in the range [1, l - 1]
    int xor_l;
 
    // Stores XOR of even numbers
    // in the range [1, r]
    int xor_r;
 
    // Update xor_r
    xor_r
      = 2 * bitwiseXorRange(r / 2);
 
    // Update xor_l
    xor_l
      = 2 * bitwiseXorRange((l - 1) / 2);
 
    return xor_l ^ xor_r;
  }
 
  // Driver Code
  public static void main (String[] args)
  {
    int l = 10;
    int r = 20;
    System.out.print(evenXorRange(l, r));   
  }
}
 
// This code is contributed by AnkThon

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Python3

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# Python3 implementation of the above approach
 
# Function to calculate XOR of
# numbers in the range [1, n]
def bitwiseXorRange(n):
 
    # If n is divisible by 4
    if (n % 4 == 0):
        return n
 
    # If n mod 4 is equal to 1
    if (n % 4 == 1):
        return 1
 
    # If n mod 4 is equal to 2
    if (n % 4 == 2):
        return n + 1
 
    return 0
 
# Function to find XOR of even
# numbers in the range [l, r]
def evenXorRange(l, r):
 
    # Stores XOR of even numbers
    # in the range [1, l - 1]
    #xor_l
 
    # Stores XOR of even numbers
    # in the range [1, r]
    #xor_r
 
    # Update xor_r
    xor_r = 2 * bitwiseXorRange(r // 2)
 
    # Update xor_l
    xor_l = 2 * bitwiseXorRange((l - 1) // 2)
 
    return xor_l ^ xor_r
 
# Driver Code
if __name__ == '__main__':
     
    l = 10
    r = 20
     
    print(evenXorRange(l, r))
 
# This code is contributed by mohit kumar 29

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C#

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// C# Implementation of the above approach
using System;
class GFG {
     
  // Function to calculate XOR of
  // numbers in the range [1, n]
  static int bitwiseXorRange(int n)
  {
  
    // If n is divisible by 4
    if (n % 4 == 0)
      return n;
  
    // If n mod 4 is equal to 1
    if (n % 4 == 1)
      return 1;
  
    // If n mod 4 is equal to 2
    if (n % 4 == 2)
      return n + 1;
  
    return 0;
  }
  
  // Function to find XOR of even
  // numbers in the range [l, r]
  static int evenXorRange(int l, int r)
  {
  
    // Stores XOR of even numbers
    // in the range [1, l - 1]
    int xor_l;
  
    // Stores XOR of even numbers
    // in the range [1, r]
    int xor_r;
  
    // Update xor_r
    xor_r
      = 2 * bitwiseXorRange(r / 2);
  
    // Update xor_l
    xor_l
      = 2 * bitwiseXorRange((l - 1) / 2);
  
    return xor_l ^ xor_r;
  }
   
  // Driver code
  static void Main()
  {
    int l = 10;
    int r = 20;
    Console.Write(evenXorRange(l, r));   
  }
}
 
// This code is contributed by divyeshrabadiya07

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Output: 

30

 

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

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