Aquileo | XOR of Number Range [L, R]

Given two integers L and R, the task is to find the XOR of elements of the range [L, R].
Examples:

Input: L = 4, R = 8
Output: 8
Explanation: 4 ^ 5 ^ 6 ^ 7 ^ 8 = 8
Input: L = 2, R = 4
Output: 5
Explanation: 2 ^ 3 ^ 4 = 5

Try It Yourself

Table of Content

[Naive Approach] Simple Loop - O(n) Time and O(1) Space
[Efficient Approach] Using XOR Pattern - O(1) Time and O(1) Space

[Naive Approach] Simple Loop - O(n) Time and O(1) Space

Initialize answer as zero, Traverse all numbers from L to R and perform XOR of the numbers one by one with the answer.

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the XOR of elements
// from the range [l, r]
int findXOR(int l, int r)
{
    int ans = 0;
    for (int i = l; i <= r; i++) {
        ans = ans ^ i;
    }
    return ans;
}

// Driver code
int main()
{
    int l = 4, r = 8;

    cout << findXOR(l, r);

    return 0;
}

// this code is contributed by devendra solunke

Java

/*package whatever //do not write package name here */

import java.io.*;

class GFG {
    // Function to return the XOR of elements
    // from the range [l, r]
    public static int findXOR(int l, int r)
    {
        int ans = 0;
        for (int i = l; i <= r; i++) {
            ans = ans ^ i;
        }
        return ans;
    }

    // Driver code
    public static void main(String[] args)
    {
        int l = 4;
        int r = 8;

        System.out.println(findXOR(l, r));
    }
}
// this code is contributed by devendra solunke

Python

# Python3 implementation of the approach
from operator import xor

# Function to return the XOR of elements
# from the range [1, n]
def findXOR(l, r):
  ans = 0
  for i in range(l,r+1):
    ans = xor(ans,i)
  return ans

# Driver code
l = 4; r = 8;

print(findXOR(l, r));

# This code is contributed by Arpit Jain

// c# implementation of find xor between given range
using System;

public class GFG {

    // Function to return the XOR of elements
    // from the range [l, r]
    static int findXOR(int l, int r)
    {
        int ans = 0;
        for (int i = l; i <= r; i++) {
            ans = ans ^ i;
        }
        return ans;
    }

    // Driver code
    static void Main(String[] args)
    {
        int l = 4;
        int r = 8;

        Console.WriteLine(findXOR(l, r));
    }
}

// this code is contributed by devendra saunke

JavaScript

 // Javascript code to find xor of given range
 <script>
        var l = 4;
        var r = 8;
        var ans = 0;
        var(int i = l; i <= r; i++)
        {
          ans = ans ^ i;
        }
      document.write(ans);
 </script>
 // this code is contributed by devendra solunke

Output

[Efficient Approach] Using XOR Pattern - O(1) Time and O(1) Space

Use the repeating XOR pattern from 1 to n and compute the range XOR as: XOR(L to R) = XOR(1 to R) ^ XOR(1 to L-1).

How does this work? We mainly get the elements from 1 to L-1 appeared twice in x1 ^ x2 and if we do XOR of an element with itself, we get 0 and if we do XOR of 0 with an element x, we get x.

How do we find XOR of 1 to n Find the remainder of n by moduling it with 4.

If rem = 0, then XOR will be same as n.
If rem = 1, then XOR will be 1.
If rem = 2, then XOR will be n+1.
If rem = 3 ,then XOR will be 0.

Please refer XOR of 1 to n for working of this.

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the XOR of elements
// from the range [1, n]
int findXOR(int n)
{
    int mod = n % 4;

    // If n is a multiple of 4
    if (mod == 0)
        return n;

    // If n % 4 gives remainder 1
    else if (mod == 1)
        return 1;

    // If n % 4 gives remainder 2
    else if (mod == 2)
        return n + 1;

    // If n % 4 gives remainder 3
    else if (mod == 3)
        return 0;
}

// Function to return the XOR of elements
// from the range [l, r]
int findXOR(int l, int r)
{
    return (findXOR(l - 1) ^ findXOR(r));
}

// Driver code
int main()
{
    int l = 4, r = 8;

    cout << findXOR(l, r);

    return 0;
}