Calculate XOR from 1 to n.

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++

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// 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;
}

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Java

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// 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));
    }
}

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Python 3

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# 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

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C#

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// 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

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PHP

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<?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
?>

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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.

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