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 = 7Input : 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));
Output
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
?>
Output
1
Time Complexity: O(1)
Auxiliary Space: O(1)
