# Count the number of elements in an array which are divisible by k

• Difficulty Level : Basic
• Last Updated : 23 Sep, 2021

Given an array of integers. The task is to calculate the count of a number of elements which are divisible by a given number k.

Examples:

```Input: arr[] = { 2, 6, 7, 12, 14, 18 }, k = 3
Output: 3
Numbers which are divisible by k are { 6, 12, 18 }

Input: arr[] = { 2, 6, 7, 12, 14, 18 }, k = 2
Output: 5```

Method-1: Start traversing the array and check if the current element is divisible by K. If yes then increment the count. Print the count when all the elements get traversed.

Below is the implementation of the above approach:

## C++

 `// C++ program to Count the number of elements``// in an array which are divisible by k``#include ``using` `namespace` `std;` `// Function to count the elements``int` `CountTheElements(``int` `arr[], ``int` `n, ``int` `k)``{``    ``int` `counter = 0;` `    ``for` `(``int` `i = 0; i < n; i++) {``        ``if` `(arr[i] % k == 0)``            ``counter++;``    ``}` `    ``return` `counter;``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 2, 6, 7, 12, 14, 18 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``int` `k = 3;` `    ``cout << CountTheElements(arr, n, k);` `    ``return` `0;``}`

## Java

 `// Java program to Count the number of elements``// in an array which are divisible by k``import` `java.util.*;` `class` `Geeks {`  `// Function to count the elements``static` `int` `CountTheElements(``int` `arr[], ``int` `n, ``int` `k)``{``    ``int` `counter = ``0``;` `    ``for` `(``int` `i = ``0``; i < n; i++) {``        ``if` `(arr[i] % k == ``0``)``            ``counter++;``    ``}` `    ``return` `counter;``}` `// Driver code``public` `static` `void` `main(String args[])``{``    ``int` `arr[] = { ``2``, ``6``, ``7``, ``12``, ``14``, ``18` `};``    ``int` `n = arr.length;``    ``int` `k = ``3``;` `    ``System.out.println(CountTheElements(arr, n, k));``}``}` `// This code is contributed by ankita_saini`

## Python3

 `# Python 3 program to Count the``# number of elements in an array``# which are divisible by k` `# Function to count the elements``def` `CountTheElements(arr, n, k):``    ``counter ``=` `0` `    ``for` `i ``in` `range``(``0``, n, ``1``):``        ``if` `(arr[i] ``%` `k ``=``=` `0``):``            ``counter ``=` `counter``+``1``    ` `    ``return` `counter` `# Driver code``if` `__name__ ``=``=` `'__main__'``:``    ``arr ``=` `[``2``, ``6``, ``7``, ``12``, ``14``, ``18``];``    ``n ``=` `len``(arr)``    ``k ``=` `3` `    ``print``(CountTheElements(arr, n, k))` `# This code is contributed by``# Surendra_Gangwar`

## C#

 `// C# program to Count the number of elements``// in an array which are divisible by k``using` `System;` `class` `Geeks {`  `// Function to count the elements``static` `int` `CountTheElements(``int` `[]arr, ``int` `n, ``int` `k)``{``    ``int` `counter = 0;` `    ``for` `(``int` `i = 0; i < n; i++) {``        ``if` `(arr[i] % k == 0)``            ``counter++;``    ``}` `    ``return` `counter;``}` `// Driver code``public` `static` `void` `Main()``{``    ``int` `[]arr = { 2, 6, 7, 12, 14, 18 };``    ``int` `n = arr.Length;``    ``int` `k = 3;` `    ``Console.WriteLine(CountTheElements(arr, n, k));``}``}``//This code is contributed by inder_verma..`

## PHP

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## Javascript

 ``
Output:
`3`

Method-2: Another way to do that is to use inbuilt function – Count_if . This functions takes in pointers to beginning and ending of the container containing the elements and auxiliary function. It will return the count of elements for which the function will return true.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the above approach``#include ``using` `namespace` `std;``int` `main()``{``    ``vector<``int``> v{ 2, 6, 7, 12, 14, 18 };` `    ``// Count the number elements which when passed``    ``// through the function (3rd argument) returns true``    ``int` `res = count_if(v.begin(), v.end(),``                       ``[](``int` `i, ``int` `k = 3) { ``return` `i % k == 0; });``    ``cout << res;` `    ``return` `0;``}`

## Java

 `// Java implementation of the above approach``import` `java.util.*;` `class` `GFG``{``public` `static` `void` `main(String[] args)``{``   ``int` `[]v = { ``2``, ``6``, ``7``, ``12``, ``14``, ``18` `};` `    ``// Count the number elements which when passed``    ``// through the function (3rd argument) returns true``    ``int` `res = ``0``;``    ``for``(``int` `i = ``0``; i < v.length; i++) {``        ``if``(v[i] % ``3` `== ``0``)``            ``res++;``    ``}``     ``System.out.print(res);` `}``}` `// This code is contributed by gauravrajput1`

## Python3

 `# Java implementation of the above approach``v ``=` `[ ``2``, ``6``, ``7``, ``12``, ``14``, ``18` `];` `# Count the number elements which when passed``# through the function (3rd argument) returns true``res ``=` `0``for` `i ``in` `range``(``0``,``len``(v)):``    ``if``(v[i] ``%` `3` `=``=` `0``):``        ``res``+``=``1``    ` `print``(res)``# This code is contributed by shivanisinghss2110`

## C#

 `// C# implementation of the above approach``using` `System;` `class` `GFG``{``public` `static` `void` `Main(String[] args)``{``   ``int` `[]v = { 2, 6, 7, 12, 14, 18 };` `    ``// Count the number elements which when passed``    ``// through the function (3rd argument) returns true``    ``int` `res = 0;``    ``for``(``int` `i = 0; i < v.Length; i++) {``        ``if``(v[i] % 3 == 0)``            ``res++;``    ``}``     ``Console.Write(res);` `}``}` `// This code is contributed by shivanisinghss2110`

## Javascript

 ``
Output:
`3`

Time Complexity: O(N)

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