XOR of all even numbers from a given range
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)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!
