Find XOR of numbers from the 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 
4 ^ 5 ^ 6 ^ 7 ^ 8 = 8

Input: L = 3, R = 7 
Output: 3  

Naive Approach: Initialize answer as zero, Traverse all numbers from L to R and perform XOR of the numbers one by one with the answer. This would take O(N) time.

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

Python3

# 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#

// 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

8

Time complexity: O(N)
Auxiliary Space: O(1)

Efficient Approach: By following the approach discussed here, we can find the XOR of elements from the range [1, N] in O(1) time. 
Using this approach, we have to find xor of elements from the range [1, L – 1] and from the range [1, R] and then xor the respective answers again to get the xor of the elements from the range [L, R]. This is because every element from the range [1, L – 1] will get XORed twice in the result resulting in a 0 which when XORed with the elements of the range [L, R] will give the result.
Below is the implementation of the above approach: 

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;
}

Java

// Java implementation of the approach
class GFG
{
    // Function to return the XOR of elements
    // from the range [1, n]
    static 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;
        return 0;
    }

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

    // Driver code
    public static void main(String[] args)
    {

        int l = 4, r = 8;

            System.out.println(findXOR(l, r));
    }
}

// This code contributed by Rajput-Ji

Python3

# Python3 implementation of the approach
from operator import xor

# Function to return the XOR of elements
# from the range [1, n]
def findXOR(n):
    mod = n % 4;

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

    # If n % 4 gives remainder 1
    elif (mod == 1):
        return 1;

    # If n % 4 gives remainder 2
    elif (mod == 2):
        return n + 1;

    # If n % 4 gives remainder 3
    elif (mod == 3):
        return 0;

# Function to return the XOR of elements
# from the range [l, r]
def findXORFun(l, r):
    return (xor(findXOR(l - 1) , findXOR(r)));

# Driver code
l = 4; r = 8;

print(findXORFun(l, r));

# This code is contributed by PrinciRaj1992

C#

// C# implementation of the approach
using System;

class GFG
{
    // Function to return the XOR of elements
    // from the range [1, n]
    static 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;
        return 0;
    }

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

    // Driver code
    public static void Main()
    {

        int l = 4, r = 8;

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

// This code is contributed by AnkitRai01

Javascript

<script>

    // Javascript implementation of the approach

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

    let l = 4, r = 8;
    document.write(findXOR(l, r));

</script>

Output

8

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