# 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 = 3
Output: 2
Explanation:
The numbers satisfying the conditions are:

1. 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.
2. 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: 4
Output: 3

Naive 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++

 `// C++ program for the above approach` `#include ` `using` `namespace` `std;`   `// Function to count all the integers` `// less than N satisfying the given` `// condition` `int` `countXor(``int` `N)` `{`   `    ``// Store the count of even` `    ``// numbers less than N+1` `    ``int` `cnt = N / 2 + 1;`   `    ``// Return the count` `    ``return` `cnt;` `}`   `// Driver Code` `int` `main()` `{` `    ``// Given Input` `    ``int` `N = 4;`   `    ``// Function Call` `    ``cout << countXor(N);`   `    ``return` `0;` `}`

## Java

 `// Java Program for the above approach` `import` `java.io.*;`   `class` `GFG` `{` `  `  `    ``// Function to count all the integers` `    ``// less than N satisfying the given` `    ``// condition` `    ``static` `int` `countXor(``int` `N)` `    ``{`   `        ``// Store the count of even` `        ``// numbers less than N+1` `        ``int` `cnt = (``int``) N / ``2` `+ ``1``;`   `        ``// Return the count` `        ``return` `cnt;` `    ``}`   `    ``// Driver Code` `    ``public` `static` `void` `main(String[] args)` `    ``{`   `        ``// Given Input` `        ``int` `N = ``4``;`   `        ``// Function Call` `        ``System.out.println(countXor(N));`   `    ``}` `}`   `// This code is contributed by Potta Lokesh`

## Python3

 `# Python 3 program for the above approach`   `# Function to count all the integers` `# less than N satisfying the given` `# condition` `def` `countXor(N):` `  `  `    ``# Store the count of even` `    ``# numbers less than N+1` `    ``cnt ``=` `N ``/``/` `2` `+` `1`   `    ``# Return the count` `    ``return` `cnt`   `# Driver Code` `if` `__name__ ``=``=` `'__main__'``:` `  `  `    ``# Given Input` `    ``N ``=` `4`   `    ``# Function Call` `    ``print``(countXor(N))` `    `  `    ``# This code is contributed by SUTENDRA_GANGWAR.`

## C#

 `// C# program for the above approach` `using` `System;`   `class` `GFG{` `    `  `// Function to count all the integers` `// less than N satisfying the given` `// condition` `static` `int` `countXor(``int` `N)` `{` `    `  `    ``// Store the count of even` `    ``// numbers less than N+1` `    ``int` `cnt = (``int``)N / 2 + 1;`   `    ``// Return the count` `    ``return` `cnt;` `}`   `// Driver code` `static` `void` `Main()` `{` `    `  `    ``// Given Input` `    ``int` `N = 4;`   `    ``// Function Call` `    ``Console.WriteLine(countXor(N));` `}` `}`   `// This code is contributed by abhinavjain194`

## Javascript

 ``

Output

`3`

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

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