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