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Calculate XOR from 1 to n.

  • Difficulty Level : Medium
  • Last Updated : 30 Aug, 2021

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++ program to find XOR of numbers
// from 1 to n.
#include<bits/stdc++.h>
using namespace std;
 
// Method to calculate xor
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
int main()
{
  int n = 5;
  cout<<computeXOR(n);
}
 
 
// This code is contributed by rutvik_56.

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
?>

Javascript




<script>
 
// JavaScript 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.
 
    // 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
    if(n % 4 == 3)
        return 0;    
      
}
 
// Driver code
 
    // your code goes here
    let n = 5;
    document.write(computeXOR(n));
 
// This code is constributed by Surbhi Tyagi.
 
</script>

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 write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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