# 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: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 <bits/stdc++.h>` `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

`<script>` ` ` ` ` `// JavaScript program for the above approach` ` ` `// Function to count all the integers` ` ` `// less than N satisfying the given` ` ` `// condition` ` ` `function` `countXor(N) {` ` ` `// Store the count of even` ` ` `// numbers less than N+1` ` ` `let cnt = Math.floor(N / 2) + 1;` ` ` `// Return the count` ` ` `return` `cnt;` ` ` `}` ` ` `// Driver Code` ` ` `// Given Input` ` ` `let N = 4;` ` ` `// Function Call` ` ` `document.write(countXor(N));` ` ` `// This code is contributed by Potta Lokesh` `</script>` |

**Output**

3

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

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