Calculate XOR from 1 to n.

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++
Java
Python3
C#
Javascript
// 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 */
// Given a number n, find the XOR from 1 to n for given n numberimport 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 */
#defining a function computeXORdef computeXOR(n):    uni = 0    if n==0:        return 0 #base case    for i in range(1,n+1):        uni = uni ^ i    return uni n = 7ans = computeXOR(n) #calling the functionprint(ans) #This code is contributed by Gayatri Deshmukh
// C# program  that finds the XOR// from 1 to n for a given number nusing 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 that for a number n// finds the XOR from 1 to n for given n numberfunction 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 Codevar n = 7; // Function Callvar ans = computeXor(n);console.log(ans); // This code is contributed by phasing17
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++
Java
Python 3
C#
PHP
Javascript
// C++ program to find XOR of numbers// from 1 to n.#include<bits/stdc++.h>using namespace std; // Method to calculate xorint 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 methodint 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 xordef 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 Codeif __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
<?php// PHP program to find XOR// of numbers from 1 to n. // Function to calculate xorfunction 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?>
<script> // JavaScript program to find XOR of numbers// from 1 to n. // Function to calculate xorfunction 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>
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
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