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Find XOR of numbers from the range [L, R]

  • Difficulty Level : Basic
  • Last Updated : 21 Oct, 2021

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

 

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.
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

 




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