Skip to content
Related Articles

Related Articles

Calculate XOR from 1 to n.
  • Difficulty Level : Medium
  • Last Updated : 03 Oct, 2018

Given a number n, the task is to find the XOR from 1 to n.

Examples :

Input : n = 6
Output : 7
// 1 ^ 2 ^ 3 ^ 4 ^ 5 ^ 6  = 7

Input : n = 7
Output : 0
// 1 ^ 2 ^ 3 ^ 4 ^ 5 ^ 6 ^ 7 = 0

Method 1 (Naive Approach):
1- Initialize result as 0.
1- Traverse all numbers from 1 to n.
2- Do XOR of numbers one by one with result.
3- At the end, return result.

Method 2 (Efficient method) :
1- Find the remainder of n by moduling it with 4.
2- If rem = 0, then xor will be same as n.
3- If rem = 1, then xor will be 1.
4- If rem = 2, then xor will be n+1.
5- If rem = 3 ,then xor will be 0.

C/C++




// C++ program to find XOR of numbers
// from 1 to n.
#include <bits/stdc++.h>
using namespace std;
  
// Function to calculate xor
long computeXOR(const int n)
{
    // Modulus operator are expensive on most of the 
    // computers. n & 3 will be equivalent to n % 4.   
  
    switch(n & 3) // n % 4 
    {
    case 0: return n;     // if n is multiple of 4
    case 1: return 1;     // If n % 4 gives remainder 1  
    case 2: return n + 1; // If n % 4 gives remainder 2    
    case 3: return 0;     // If n % 4 gives remainder 3  
    }
}
  
// Driver code
int main()
{
    // your code goes here
    int n = 5;
    cout <<    computeXOR(n);
    return 0;
}


Java




// Java program to find XOR of numbers
// from 1 to n.
  
class GFG 
{
    // Method to calculate xor
    static int computeXOR(int n)
    {
        // If n is a multiple of 4
        if (n % 4 == 0)
            return n;
       
        // If n%4 gives remainder 1
        if (n % 4 == 1)
            return 1;
       
        // If n%4 gives remainder 2
        if (n % 4 == 2)
            return n + 1;
       
        // If n%4 gives remainder 3
        return 0;
    }
      
    // Driver method
    public static void main (String[] args)
    {
         int n = 5;
         System.out.println(computeXOR(n));
    }
}


Python 3




# Python 3 Program to find 
# XOR of numbers from 1 to n. 
  
# Function to calculate xor 
def computeXOR(n) :
  
    # Modulus operator are expensive 
    # on most of the computers. n & 3 
    # will be equivalent to n % 4.
  
    # if n is multiple of 4 
    if n % 4 == 0 :
        return n
  
    # If n % 4 gives remainder 1
    if n % 4 == 1 :
        return 1
  
    # If n%4 gives remainder 2 
    if n % 4 == 2 :
        return n + 1
  
    # If n%4 gives remainder 3
    return 0
  
# Driver Code
if __name__ == "__main__" :
  
    n = 5
  
    # function calling
    print(computeXOR(n))
          
# This code is contributed by ANKITRAI1


C#




// C# program to find XOR 
// of numbers from 1 to n.
using System;
  
class GFG
{
      
    // Method to calculate xor
    static int computeXOR(int n)
    {
        // If n is a multiple of 4
        if (n % 4 == 0)
            return n;
      
        // If n%4 gives remainder 1
        if (n % 4 == 1)
            return 1;
      
        // If n%4 gives remainder 2
        if (n % 4 == 2)
            return n + 1;
      
        // If n%4 gives remainder 3
        return 0;
    }
      
    // Driver Code
    static public void Main ()
    {
        int n = 5;
        Console.WriteLine(computeXOR(n));
    }
}
  
// This code is contributed by ajit


PHP




<?php
// PHP program to find XOR 
// of numbers from 1 to n.
  
// Function to calculate xor
function computeXOR($n)
{
    // Modulus operator are expensive 
    // on most of the computers. n & 3
    // will be equivalent to n % 4. 
  
    switch($n & 3) // n % 4 
    {
    // 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; 
    }
}
  
// Driver code
$n = 5;
echo computeXOR($n);
  
// This code is contributed by aj_36
?>



Output :



1

How does this work?
When we do XOR of numbers, we get 0 as XOR value just before a multiple of 4. This keeps repeating before every multiple of 4.

Number Binary-Repr  XOR-from-1-to-n
1         1           [0001]
2        10           [0011]
3        11           [0000]  <----- We get a 0
4       100           [0100]  <----- Equals to n
5       101           [0001]
6       110           [0111]
7       111           [0000]  <----- We get 0
8      1000           [1000]  <----- Equals to n
9      1001           [0001]
10     1010           [1011]
11     1011           [0000] <------ We get 0
12     1100           [1100] <------ Equals to n

This article is contributed by Sahil Chhabra. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up
Recommended Articles
Page :