Given two integers L and R, the task is to calculate Bitwise XOR of all even numbers in the range [L, R].
Examples:
Example:
Input: L = 10, R = 20
Output: 30
Explanation:
Bitwise XOR = 10 ^ 12 ^ 14 ^ 16 ^ 18 ^ 20 = 30
Therefore, the required output is 30.Example:
Input: L = 15, R = 23
Output: 0
Explanation:
Bitwise XOR = 16 ^ 18 ^ 20 ^ 22 = 0
Therefore, the required output is 0.
Naive Approach:The simplest approach to solve the problem is to traverse all even numbers in the range [L, R] and print the Bitwise XOR of all the even numbers.
Time Complexity: O(R – L)
Auxiliary Space: O(1)
Efficient Approach: The above approach can be optimized based on the following observations:
If N is an even number:
2 ^ 4 … ^ (N) = 2 * (1 ^ 2 ^ … ^ (N / 2))If N is an odd number:
2 ^ 4 … ^ (N) = 2 * (1 ^ 2 ^ … ^ ((N – 1) / 2))
Follow the steps below to solve the problem:
- Find the bitwise XOR of all the numbers from the range [1, (R) / 2] and store it in a variable, say xor_r.
- Find the bitwise XOR of all the numbers from the range [1, (L – 1) / 2] and store it in a variable, say xor_l.
- Finally, print the value of xor_r ^ xor_l.
Below is the implementation of the above approach:
C++
// C++ Implementation of the above approach#include <bits/stdc++.h>using namespace std;// Function to calculate XOR of// numbers in the range [1, n]int bitwiseXorRange(int n){ // If n is divisible by 4 if (n % 4 == 0) return n; // If n mod 4 is equal to 1 if (n % 4 == 1) return 1; // If n mod 4 is equal to 2 if (n % 4 == 2) return n + 1; return 0;}// Function to find XOR of even// numbers in the range [l, r]int evenXorRange(int l, int r){ // Stores XOR of even numbers // in the range [1, l - 1] int xor_l; // Stores XOR of even numbers // in the range [1, r] int xor_r; // Update xor_r xor_r = 2 * bitwiseXorRange(r / 2); // Update xor_l xor_l = 2 * bitwiseXorRange((l - 1) / 2); return xor_l ^ xor_r;}// Driver Codeint main(){ int l = 10; int r = 20; cout << evenXorRange(l, r); return 0;} |
Java
// Java Implementation of the above approachclass GFG { // Function to calculate XOR of // numbers in the range [1, n] static int bitwiseXorRange(int n) { // If n is divisible by 4 if (n % 4 == 0) return n; // If n mod 4 is equal to 1 if (n % 4 == 1) return 1; // If n mod 4 is equal to 2 if (n % 4 == 2) return n + 1; return 0; } // Function to find XOR of even // numbers in the range [l, r] static int evenXorRange(int l, int r) { // Stores XOR of even numbers // in the range [1, l - 1] int xor_l; // Stores XOR of even numbers // in the range [1, r] int xor_r; // Update xor_r xor_r = 2 * bitwiseXorRange(r / 2); // Update xor_l xor_l = 2 * bitwiseXorRange((l - 1) / 2); return xor_l ^ xor_r; } // Driver Code public static void main (String[] args) { int l = 10; int r = 20; System.out.print(evenXorRange(l, r)); }}// This code is contributed by AnkThon |
Python3
# Python3 implementation of the above approach# Function to calculate XOR of# numbers in the range [1, n]def bitwiseXorRange(n): # If n is divisible by 4 if (n % 4 == 0): return n # If n mod 4 is equal to 1 if (n % 4 == 1): return 1 # If n mod 4 is equal to 2 if (n % 4 == 2): return n + 1 return 0# Function to find XOR of even# numbers in the range [l, r]def evenXorRange(l, r): # Stores XOR of even numbers # in the range [1, l - 1] #xor_l # Stores XOR of even numbers # in the range [1, r] #xor_r # Update xor_r xor_r = 2 * bitwiseXorRange(r // 2) # Update xor_l xor_l = 2 * bitwiseXorRange((l - 1) // 2) return xor_l ^ xor_r# Driver Codeif __name__ == '__main__': l = 10 r = 20 print(evenXorRange(l, r))# This code is contributed by mohit kumar 29 |
C#
// C# Implementation of the above approachusing System;class GFG { // Function to calculate XOR of // numbers in the range [1, n] static int bitwiseXorRange(int n) { // If n is divisible by 4 if (n % 4 == 0) return n; // If n mod 4 is equal to 1 if (n % 4 == 1) return 1; // If n mod 4 is equal to 2 if (n % 4 == 2) return n + 1; return 0; } // Function to find XOR of even // numbers in the range [l, r] static int evenXorRange(int l, int r) { // Stores XOR of even numbers // in the range [1, l - 1] int xor_l; // Stores XOR of even numbers // in the range [1, r] int xor_r; // Update xor_r xor_r = 2 * bitwiseXorRange(r / 2); // Update xor_l xor_l = 2 * bitwiseXorRange((l - 1) / 2); return xor_l ^ xor_r; } // Driver code static void Main() { int l = 10; int r = 20; Console.Write(evenXorRange(l, r)); }}// This code is contributed by divyeshrabadiya07 |
Javascript
<script>// JavaScript Implementation of the above approach// Function to calculate XOR of// numbers in the range [1, n]function bitwiseXorRange(n){ // If n is divisible by 4 if (n % 4 == 0) return n; // If n mod 4 is equal to 1 if (n % 4 == 1) return 1; // If n mod 4 is equal to 2 if (n % 4 == 2) return n + 1; return 0;}// Function to find XOR of even// numbers in the range [l, r]function evenXorRange(l, r){ // Stores XOR of even numbers // in the range [1, l - 1] let xor_l; // Stores XOR of even numbers // in the range [1, r] let xor_r; // Update xor_r xor_r = 2 * bitwiseXorRange(Math.floor(r / 2)); // Update xor_l xor_l = 2 * bitwiseXorRange(Math.floor((l - 1) / 2)); return xor_l ^ xor_r;}// Driver Code let l = 10; let r = 20; document.write(evenXorRange(l, r));// This code is contributed by Surbhi Tyagi.</script> |
Output:
30
Time Complexity: O(1).
Auxiliary Space: O(1)
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