- Bit Magic, Bitwise-XOR, Mathematical

Count of integers K in range [0, N] such that (K XOR K+1) equals (K+2 XOR K+3)

Count of integers K in range [0, N] such that (K XOR K+1) equals (K+2 XOR K+3)Given an integer N, the task is to print the count of all the non-negative integers K less than or equal to N, such that bitwise XOR of K and K+1 equals bitwise XOR of K+2 and K+3.Examples:Input: N = 3Output: 2Explanation: The numbers satisfying the conditions are:K = 0, the bitwise XOR of 0 and 1 is equal to 1 and the bitwise xor of 2 and 3 is also equal to 1.K = 2, the bitwise XOR of 2 and 3 is equal to 1 and the bitwise xor of 4 and 5 is also equal to 1.Therefore, there are 2 numbers satisfying the condition.Input: 4Output: 3Naive Approach: The simplest approach is to iterate over the range [0, N] and check if the current number satisfies the condition or not. If it satisfies, increment the count by 1. After checking all the numbers, print the value of the count.Time Complexity: O(N)Auxiliary Space: O(1)Efficient Approach: The above approach can be optimized by the observation that all the even numbers in the range [0, N] satisfy the given condition.Below is the implementation of the above approach:C++#include using namespace std;  int countXor(int N){              int cnt = N / 2 + 1;          return cnt;}  int main(){        int N = 4;          cout