# Bitwise AND of all even number up to N

Given an integer N, the task is to find bitwise and (&) of all even number from 1 to N.

Examples:

Input: 2
Output: 2

Input :10
Output :0
Explanation: Bitwise and of 2, 4, 6, 8 and 10 is 0.

## 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 and (&).

Below is the implementation of the approach:

## C++

 `// C++ implementation of the above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the bitwise & ` `// of all the even numbers upto N ` `int` `bitwiseAndTillN(``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 = 2; ` `    ``cout << bitwiseAndTillN(n); ` `    ``return` `0; ` `} `

## Java

 `// Java implementation of the above approach  ` `class` `GFG  ` `{ ` `     `  `    ``// Function to return the bitwise &  ` `    ``// of all the even numbers upto N  ` `    ``static` `int` `bitwiseAndTillN(``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  ` `    ``public` `static` `void` `main (String[] args)  ` `    ``{  ` `        ``int` `n = ``2``;  ` `        ``System.out.println(bitwiseAndTillN(n));  ` `    ``}  ` `} ` ` `  `// This code is contributed by AnkitRai01 `

## Python3

 `# Python3 implementation of the above approach  ` ` `  `# Function to return the bitwise &  ` `# of all the even numbers upto N  ` `def` `bitwiseAndTillN(n) :  ` ` `  `    ``# Initialize result as 2  ` `    ``result ``=` `2``;  ` ` `  `    ``for` `i ``in` `range``(``4``, n ``+` `1``, ``2``) :  ` `        ``result ``=` `result & i;  ` `     `  `    ``return` `result;  ` ` `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"` `: ` `     `  `    ``n ``=` `2``;  ` `    ``print``(bitwiseAndTillN(n));  ` ` `  `# This code is contributed by AnkitRai01 `

## C#

 `// C# implementation of the above approach  ` `using` `System; ` ` `  `class` `GFG  ` `{ ` `     `  `    ``// Function to return the bitwise &  ` `    ``// of all the even numbers upto N  ` `    ``static` `int` `bitwiseAndTillN(``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  ` `    ``public` `static` `void` `Main()  ` `    ``{  ` `        ``int` `n = 2;  ` `        ``Console.WriteLine(bitwiseAndTillN(n));  ` `    ``}  ` `} ` ` `  `// This code is contributed by AnkitRai01 `

Output:

```2
```

Efficient approach: Efficient approach is to return 2 for N less than 4 and return 0 for all N>=4 because bitwise and of 2 and 4 is 0 and bitwise and of 0 with any number is 0.

Below is the implementation of the approach:

## C++

 `// C++ imlpementation of the above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the bitwise & ` `// of all the numbers upto N ` `int` `bitwiseAndTillN(``int` `n) ` `{ ` `    ``if` `(n < 4) ` `        ``return` `2; ` `    ``else` `        ``return` `0; ` `} ` ` `  `int` `main() ` `{ ` `    ``int` `n = 2; ` `    ``cout << bitwiseAndTillN(n); ` `    ``return` `0; ` `} `

## Java

 `// Java imlpementation of the above approach ` `class` `GFG ` `{ ` `     `  `    ``// Function to return the bitwise & ` `    ``// of all the numbers upto N ` `    ``static` `int` `bitwiseAndTillN(``int` `n) ` `    ``{ ` `        ``if` `(n < ``4``) ` `            ``return` `2``; ` `        ``else` `            ``return` `0``; ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `main (String[] args)  ` `    ``{ ` `        ``int` `n = ``2``; ` `        ``System.out.println(bitwiseAndTillN(n)); ` `    ``} ` `} ` ` `  `// This code is contributed by AnkitRai01 `

## Python3

 `# Python3 imlpementation of the above approach ` ` `  `# Function to return the bitwise & ` `# of all the numbers upto N ` `def` `bitwiseAndTillN( n): ` `    ``if` `(n < ``4``): ` `        ``return` `2` `    ``else``: ` `        ``return` `0` ` `  `# Driver code ` `n ``=` `2` `print``(bitwiseAndTillN(n)) ` ` `  `# This code is contributed by ANKITKUMAR34 `

## C#

 `// C# imlpementation of the above approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` `     `  `    ``// Function to return the bitwise & ` `    ``// of all the numbers upto N ` `    ``static` `int` `bitwiseAndTillN(``int` `n) ` `    ``{ ` `        ``if` `(n < 4) ` `            ``return` `2; ` `        ``else` `            ``return` `0; ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `Main()  ` `    ``{ ` `        ``int` `n = 2; ` `        ``Console.WriteLine(bitwiseAndTillN(n)); ` `    ``} ` `} ` ` `  `// This code is contributed by AnkitRai01 `

Output:

```2
```

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