# Bitwise OR( | ) of all even number from 1 to N

Given a number N, the task is to find the bitwise OR( | ) of all even number from 1 to N.

Examples:

Input: 2
Output: 2

Input: 10
Output: 14
Explanation: 2 | 4 | 6 | 8 | 10 = 14

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Naive Approach: Initialize result as 2.Iterate loop from 4 to n (for all even number) and update result by finding bitwise or ( | ).

Below is the implementation of the approach:

## C++

 `// C++ implementation of the above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the bitwise OR ` `// of all the even numbers upto N ` `int` `bitwiseOrTillN(``int` `n) ` `{ ` `    ``// Initialize result as 2 ` `    ``int` `result = 2; ` ` `  `    ``for` `(``int` `i = 4; i <= n; i = i + 2) { ` `        ``result = result | i; ` `    ``} ` `    ``return` `result; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 10; ` `    ``cout << bitwiseOrTillN(n); ` `    ``return` `0; ` `} `

## Java

 `// Java implementation of the above approach  ` `class` `GFG ` `{ ` `     `  `    ``// Function to return the bitwise OR  ` `    ``// of all the even numbers upto N  ` `    ``static` `int` `bitwiseOrTillN(``int` `n)  ` `    ``{  ` `        ``// Initialize result as 2  ` `        ``int` `result = ``2``;  ` `     `  `        ``for` `(``int` `i = ``4``; i <= n; i = i + ``2``) ` `        ``{  ` `            ``result = result | i;  ` `        ``}  ` `        ``return` `result;  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``static` `public` `void` `main (String args[]) ` `    ``{  ` `        ``int` `n = ``10``;  ` `        ``System.out.println(bitwiseOrTillN(n));  ` `    ``}  ` `} ` ` `  `// This code is contributed by AnkitRai01 `

## Python3

 `# Python 3 implementation of the above approach ` ` `  `# Function to return the bitwise OR ` `# of all the even numbers upto N ` `def` `bitwiseOrTillN ( n ): ` `     `  `    ``# Initialize result as 2 ` `    ``result ``=` `2``; ` ` `  `    ``for` `i ``in` `range``(``4``, n ``+` `1``, ``2``) : ` `        ``result ``=` `result | i ` `     `  `    ``return` `result ` ` `  `# Driver code ` `n ``=` `10``; ` `print``(bitwiseOrTillN(n)); ` ` `  `# This code is contributed by ANKITKUMAR34 `

## C#

 `// C# implementation of the above approach  ` `using` `System; ` ` `  `class` `GFG ` `{ ` `     `  `    ``// Function to return the bitwise OR  ` `    ``// of all the even numbers upto N  ` `    ``static` `int` `bitwiseOrTillN(``int` `n)  ` `    ``{  ` `        ``// Initialize result as 2  ` `        ``int` `result = 2;  ` `     `  `        ``for` `(``int` `i = 4; i <= n; i = i + 2) ` `        ``{  ` `            ``result = result | i;  ` `        ``}  ` `        ``return` `result;  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``static` `public` `void` `Main () ` `    ``{  ` `        ``int` `n = 10;  ` `        ``Console.WriteLine(bitwiseOrTillN(n));  ` `    ``}  ` `} ` ` `  `// This code is contributed by AnkitRai01 `

Output:

```14
```

Efficient Approach: Compute the total number of bits in N. In bitwise OR, the rightmost bit will be 0 and all other bits will be 1. Therefore, return pow(2, total no. of bits)-2. It will give the equivalent value in decimal of bitwise OR.

Below is the implementation of the approach:

## C++

 `// C++ implementation of the above approach ` `#include ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the bitwise OR ` `// of all even numbers upto N ` `int` `bitwiseOrTillN(``int` `n) ` `{ ` `    ``// For value less than 2 ` `    ``if` `(n < 2) ` `        ``return` `0; ` ` `  `    ``// Count total number of bits in bitwise or ` `    ``// all bits will be set except last bit ` `    ``int` `bitCount = log2(n) + 1; ` ` `  `    ``// Compute 2 to the power bitCount and subtract 2 ` `    ``return` `pow``(2, bitCount) - 2; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 10; ` `    ``cout << bitwiseOrTillN(n); ` `    ``return` `0; ` `} `

## Java

 `// Java implementation of the above approach  ` `class` `GFG  ` `{ ` `     `  `    ``// Function to return the bitwise OR  ` `    ``// of all even numbers upto N  ` `    ``static` `int` `bitwiseOrTillN(``int` `n)  ` `    ``{  ` `        ``// For value less than 2  ` `        ``if` `(n < ``2``)  ` `            ``return` `0``;  ` `     `  `        ``// Count total number of bits in bitwise or  ` `        ``// all bits will be set except last bit  ` `        ``int` `bitCount = (``int``)(Math.log(n)/Math.log(``2``)) + ``1``;  ` `     `  `        ``// Compute 2 to the power bitCount and subtract 2  ` `        ``return` `(``int``)Math.pow(``2``, bitCount) - ``2``;  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `main (String[] args) ` `    ``{  ` `        ``int` `n = ``10``;  ` `        ``System.out.println(bitwiseOrTillN(n));  ` `    ``}  ` `} ` ` `  `// This code is contributed by AnkitRai01 `

## Python3

 `# Python3 implementation of the above approach  ` `from` `math ``import` `log2 ` ` `  `# Function to return the bitwise OR  ` `# of all even numbers upto N  ` `def` `bitwiseOrTillN(n) :  ` ` `  `    ``# For value less than 2  ` `    ``if` `(n < ``2``) : ` `        ``return` `0``;  ` ` `  `    ``# Count total number of bits in bitwise or  ` `    ``# all bits will be set except last bit  ` `    ``bitCount ``=` `int``(log2(n)) ``+` `1``;  ` ` `  `    ``# Compute 2 to the power bitCount and subtract 2  ` `    ``return` `pow``(``2``, bitCount) ``-` `2``;  ` ` `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"` `:  ` ` `  `    ``n ``=` `10``;  ` `    ``print``(bitwiseOrTillN(n));  ` ` `  `# This code is contributed by AnkitRai01 `

## C#

 `// C# implementation of the above approach  ` `using` `System; ` ` `  `class` `GFG  ` `{ ` `     `  `    ``// Function to return the bitwise OR  ` `    ``// of all even numbers upto N  ` `    ``static` `int` `bitwiseOrTillN(``int` `n)  ` `    ``{  ` `        ``// For value less than 2  ` `        ``if` `(n < 2)  ` `            ``return` `0;  ` `     `  `        ``// Count total number of bits in bitwise or  ` `        ``// all bits will be set except last bit  ` `        ``int` `bitCount = (``int``)(Math.Log(n)/Math.Log(2)) + 1;  ` `     `  `        ``// Compute 2 to the power bitCount and subtract 2  ` `        ``return` `(``int``)Math.Pow(2, bitCount) - 2;  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `Main() ` `    ``{  ` `        ``int` `n = 10;  ` `        ``Console.WriteLine(bitwiseOrTillN(n));  ` `    ``}  ` `} ` ` `  `// This code is contributed by AnkitRai01 `

Output:

```14
```

My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Improved By : ANKITKUMAR34, AnkitRai01