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Calculate XOR from 1 to n.

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  • Difficulty Level : Medium
  • Last Updated : 24 Jun, 2022

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 the result as 0. 
1- Traverse all numbers from 1 to n. 
2- Do XOR of numbers one by one with results. 
3- At the end, return the result.

C++




// C++ program to find XOR of numbers
// from 1 to n.
#include <bits/stdc++.h>
using namespace std;
int computeXOR(int n)
{
    if (n == 0)
        return 0; // base case
    int uni = 0;
    for (int i = 1; i <= n; i++) {
        uni = uni ^ i; // calculate XOR
    }
    return uni;
}
int main()
{
    int n = 7;
    int result = computeXOR(n);
    cout << result << endl;
    return 0;
}
/* This code is contributed by Rishab Dugar */


Java




// Given a number n, find the XOR from 1 to n for given n number
import java.io.*;
 
public class GFG {
    public static void main(String[] args) {
        int n = 7;
        int ans = computeXor(n);
        System.out.println(ans);
    }
    static int computeXor(int n){
        if(n == 0) return 0; // base case
        int uni = 0;
        for (int i = 1; i <= n; i++) {
 
            uni = uni^i; // calculate XOR
        }
        return uni;
    }
 
}
/* This code is contributed by devendra salunke */


Python3




#defining a function computeXOR
def computeXOR(n):
    uni = 0
    if n==0:
        return 0 #base case
    for i in range(1,n+1):
        uni = uni ^ i
    return uni
 
n = 7
ans = computeXOR(n) #calling the function
print(ans)
 
#This code is contributed by Gayatri Deshmukh


C#




// C# program  that finds the XOR
// from 1 to n for a given number n
using System;
 
public class GFG {
    static int computeXor(int n)
    {
        if (n == 0)
            return 0; // base case
        int uni = 0;
        for (int i = 1; i <= n; i++) {
 
            uni = uni ^ i; // calculate XOR
        }
        return uni;
    }
 
    // Driver code
    public static void Main(string[] args)
    {
        int n = 7;
 
        // Function call
        int ans = computeXor(n);
        Console.WriteLine(ans);
    }
}
 
// This code is contributed by phasing17


Javascript




// JavaScript that for a number n
// finds the XOR from 1 to n for given n number
function computeXor(n){
    if(n == 0)
        return 0; // base case
    var uni = 0;
    for (var i = 1; i <= n; i++)
        uni = uni^i; // calculate XOR
 
    return uni;
}
 
// Driver Code
var n = 7;
 
// Function Call
var ans = computeXor(n);
console.log(ans);
 
// This code is contributed by phasing17


Output

0

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

Time Complexity: O(1)

Auxiliary Space: O(1)

How does this work? 
When we do XOR of numbers, we get 0 as the 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.
Please write comments if you find anything incorrect, or if you want to share more information about the topic discussed above.
 


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