# Maximum possible elements which are divisible by 2

Given an integer array arr of size N. The task is to find the maximum possible elements in the array which are divisible by 2 after modifying the array. One can perform below operation an arbitrary number of times(possibly zero times).

Replace any two elements in the array with their sum.

Examples:

Input : arr = [1, 2, 3, 1, 3]
Output : 3
After adding elements at index 0 and 2, and index 3 and 4, array becomes arr=[4, 2, 4].

Input : arr = [1, 2, 3, 4, 5]
Output : 3
After adding 1 and 3, array becomes arr=[4, 2, 4, 5].

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

Approach :
First, observation is that we don’t need to modify elements which are divisible by 2(i.e., even numbers). Then we left with odd numbers. Addition of two numbers will give an even number which is divisible by 2.

So finally, the result will be:

count_even + count_odd/2.

Below is the implementation of the above approach:

## CPP

 `// CPP program to find maximum possible ` `// elements which divisible by 2 ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find maximum possible ` `// elements which divisible by 2 ` `int` `Divisible(``int` `arr[], ``int` `n) ` `{ ` `    ``// To store count of even numbers ` `    ``int` `count_even = 0; ` ` `  `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``if` `(arr[i] % 2 == 0) ` `            ``count_even++; ` ` `  `    ``// All even numbers and half of odd numbers ` `    ``return` `count_even + (n - count_even) / 2; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 1, 2, 3, 4, 5 }; ` ` `  `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]); ` ` `  `    ``// Function call ` `    ``cout << Divisible(arr, n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program to find maximum possible  ` `// elements which divisible by 2  ` `class` `GFG  ` `{ ` ` `  `    ``// Function to find maximum possible  ` `    ``// elements which divisible by 2  ` `    ``static` `int` `Divisible(``int` `arr[], ``int` `n)  ` `    ``{  ` `        ``// To store count of even numbers  ` `        ``int` `count_even = ``0``;  ` `     `  `        ``for` `(``int` `i = ``0``; i < n; i++)  ` `            ``if` `(arr[i] % ``2` `== ``0``)  ` `                ``count_even++;  ` `     `  `        ``// All even numbers and half of odd numbers  ` `        ``return` `count_even + (n - count_even) / ``2``;  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `main (String[] args)  ` `    ``{ ` `        ``int` `arr[] = { ``1``, ``2``, ``3``, ``4``, ``5` `};  ` `     `  `        ``int` `n = arr.length;  ` `     `  `        ``// Function call  ` `        ``System.out.println(Divisible(arr, n));  ` `    ``} ` `} ` ` `  `// This code is contributed by AnkitRai01 `

## Python3

 `# Python3 program to find maximum possible ` `# elements which divisible by 2 ` ` `  `# Function to find maximum possible ` `# elements which divisible by 2 ` `def` `Divisible(arr, n): ` `    ``# To store count of even numbers ` `    ``count_even ``=` `0` ` `  `    ``for` `i ``in` `range``(n): ` `        ``if` `(arr[i] ``%` `2` `=``=` `0``): ` `            ``count_even``+``=``1` ` `  `    ``# All even numbers and half of odd numbers ` `    ``return` `count_even ``+` `(n ``-` `count_even) ``/``/` `2` ` `  `# Driver code ` ` `  `arr``=``[``1``, ``2``, ``3``, ``4``, ``5``] ` ` `  `n ``=` `len``(arr) ` ` `  `# Function call ` `print``(Divisible(arr, n)) ` ` `  `# This code is contribute by mohit kumar 29 `

## C#

 `// C# program to find maximum possible  ` `// elements which divisible by 2  ` `using` `System; ` ` `  `class` `GFG ` `{ ` `     `  `    ``// Function to find maximum possible  ` `    ``// elements which divisible by 2  ` `    ``static` `int` `Divisible(``int` `[]arr, ``int` `n)  ` `    ``{  ` `        ``// To store count of even numbers  ` `        ``int` `count_even = 0;  ` `     `  `        ``for` `(``int` `i = 0; i < n; i++)  ` `            ``if` `(arr[i] % 2 == 0)  ` `                ``count_even++;  ` `     `  `        ``// All even numbers and half of odd numbers  ` `        ``return` `count_even + (n - count_even) / 2;  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``static` `public` `void` `Main () ` `    ``{ ` `         `  `        ``int` `[]arr = { 1, 2, 3, 4, 5 };  ` `        ``int` `n = arr.Length;  ` `         `  `        ``// Function call  ` `        ``Console.Write(Divisible(arr, n));  ` `    ``} ` `} ` ` `  `// This code is contributed by ajit. `

Output:

```3
```

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.