XOR of 1 to n Numbers

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

Naive Approach - O(n) Time

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++ program to find XOR of numbers
// from 1 to n.
#include <bits/stdc++.h>
using namespace std;

int computeXOR(int n)
{
    int res = 0;
    for (int i = 1; i <= n; i++) {
        res = res ^ i; 
    }
    return res;
}
int main()
{
    int n = 7;
    cout << computeXOR(n) << endl;
    return 0;
}

// Given a number n, find the XOR from 1 to n for given n number
import java.io.*;

public class GfG {
    static int computeXor(int n){
        int res = 0;
        for (int i = 1; i <= n; i++) {
            res = res^i; 
        }
        return res;
    }
    public static void main(String[] args) {
        int n = 7;
        System.out.println(computeXor(n));
    }
}

#defining a function computeXOR
def computeXOR(n):
    res = 0
    for i in range(1,n+1):
        res = res ^ i
    return res

n = 7
print(computeXOR(n))

// 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)
    {
        int res = 0;
        for (int i = 1; i <= n; i++) {
            res = res ^ i; // calculate XOR
        }
        return res;
    }

    // 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 that for a number n
// finds the XOR from 1 to n for given n number
function computeXor(n){
    var res = 0;
    for (let i = 1; i <= n; i++)
        res = res^i; // calculate XOR

    return res;
}

// Driver Code
let n = 7;
console.log(computeXor(n));

0

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

Expected Approach - O(1) Time

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.
 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  

// 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 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 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# 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

<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>

<?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
?>

1

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