# Calculate XOR from 1 to n.

• 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 ``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``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<

## 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

 ``

## Javascript

 ``

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           
2        10           
3        11             <----- We get a 0
4       100             <----- Equals to n
5       101           
6       110           
7       111             <----- We get 0
8      1000             <----- Equals to n
9      1001           
10     1010           
11     1011            <------ We get 0
12     1100            <------ Equals to n```

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