# Count number of even and odd elements in an array

• Difficulty Level : Easy
• Last Updated : 20 Jul, 2022

For the given array of integers, count even and odd elements.

Examples:

```Input:
int arr = {2, 3, 4, 5, 6}
Output:
Number of even elements = 3
Number of odd elements = 2

Input:
int arr = {22, 32, 42, 52, 62}
Output:
Number of even elements = 5
Number of odd elements = 0```

Solution: We can also check if a number is odd or even

• By doing AND of 1 and that digit, if the result comes out to be 1 then the number is odd otherwise even.
• By its divisibility by 2. A number is said to be odd if it is not divisible by 2, otherwise its even.

Here, we will check if a number is odd, then we will increment the odd counter otherwise we will increment the even counter.

Below is the implementation of the above approach:

## C++

 `// CPP program to count number of even``// and odd elements in an array``#include ``using` `namespace` `std;` `void` `CountingEvenOdd(``int` `arr[], ``int` `arr_size)``{``    ``int` `even_count = 0;``    ``int` `odd_count = 0;` `    ``// loop to read all the values in the array``    ``for` `(``int` `i = 0; i < arr_size; i++) {``        ` `          ``// checking if a number is completely``        ``// divisible by 2``        ``if` `(arr[i] & 1 == 1)``            ``odd_count++;``        ``else``            ``even_count++;``    ``}` `    ``cout << ``"Number of even elements = "` `<< even_count``         ``<< ``"\nNumber of odd elements = "` `<< odd_count;``}` `// Driver Code``int` `main()``{``    ``int` `arr[] = { 2, 3, 4, 5, 6 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);``      ` `      ``// Function Call``    ``CountingEvenOdd(arr, n);``}`

## Java

 `// JAVA program to count number of even``// and odd elements in an array``import` `java.io.*;` `class` `GFG {` `    ``static` `void` `CountingEvenOdd(``int` `arr[], ``int` `arr_size)``    ``{``        ``int` `even_count = ``0``;``        ``int` `odd_count = ``0``;` `        ``// loop to read all the values in``        ``// the array``        ``for` `(``int` `i = ``0``; i < arr_size; i++) {``            ` `              ``// checking if a number is``            ``// completely divisible by 2``            ``if` `((arr[i] & ``1``) == ``1``)``                ``odd_count++;``            ``else``                ``even_count++;``        ``}` `        ``System.out.println(``"Number of even"``                           ``+ ``" elements = "` `+ even_count``                           ``+ ``" Number of odd elements = "``                           ``+ odd_count);``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `arr[] = { ``2``, ``3``, ``4``, ``5``, ``6` `};``        ``int` `n = arr.length;``          ` `          ``// Function Call``        ``CountingEvenOdd(arr, n);``    ``}``}` `// This code is Contributed by anuj_67.`

## Python3

 `# Python3 program to count number of``# even and odd elements in an array`  `def` `CountingEvenOdd(arr, arr_size):``    ``even_count ``=` `0``    ``odd_count ``=` `0` `    ``# loop to read all the values``    ``# in the array``    ``for` `i ``in` `range``(arr_size):` `        ``# checking if a number is``        ``# completely divisible by 2``        ``if` `(arr[i] & ``1` `=``=` `1``):``            ``odd_count ``+``=` `1``        ``else``:``            ``even_count ``+``=` `1` `    ``print``(``"Number of even elements = "``,``          ``even_count)``    ``print``(``"Number of odd elements = "``,``          ``odd_count)`  `# Driver Code``arr ``=` `[``2``, ``3``, ``4``, ``5``, ``6``]``n ``=` `len``(arr)` `# Function Call``CountingEvenOdd(arr, n)` `# This code is contributed by sahishelangia`

## C#

 `// C# program to count number of even``// and odd elements in an array``using` `System;` `class` `GFG {` `    ``static` `void` `CountingEvenOdd(``int``[] arr, ``int` `arr_size)``    ``{``        ``int` `even_count = 0;``        ``int` `odd_count = 0;` `        ``// loop to read all the values in``        ``// the array``        ``for` `(``int` `i = 0; i < arr_size; i++) {``            ` `              ``// checking if a number is``            ``// completely divisible by 2``            ``if` `((arr[i] & 1) == 1)``                ``odd_count++;``            ``else``                ``even_count++;``        ``}` `        ``Console.WriteLine(``"Number of even"``                          ``+ ``" elements = "` `+ even_count``                          ``+ ``" Number of odd elements = "``                          ``+ odd_count);``    ``}` `    ``// Driver Code``    ``public` `static` `void` `Main()``    ``{``        ``int``[] arr = { 2, 3, 4, 5, 6 };``        ``int` `n = arr.Length;``      ` `          ``// Function Call``        ``CountingEvenOdd(arr, n);``    ``}``}` `// This code is Contributed by anuj_67.`

## PHP

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## Javascript

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Output

```Number of even elements = 3
Number of odd elements = 2```

Time Complexity: O(n)

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