# Count of integers in an Array whose length is a multiple of K

Given an array arr of N elements and an integer K, the task is to count all the elements whose length is a multiple of K.

Examples:

```Input: arr[]={1, 12, 3444, 544, 9}, K = 2
Output: 2
Explanation:
There are 2 numbers whose digit count is multiple of 2 {12, 3444}.

Input: arr[]={12, 345, 2, 68, 7896}, K = 3
Output: 1
Explanation:
There is 1 number whose digit count is multiple of 3 {345}.
```

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

Approach:

1. Traverse the numbers in the array one by one
2. Count the digits of every number in the array
3. Check if it’s digit count is a multiple of K or not.

Below is the implementation of above approach:

## C++

 `// C++ implementation of above approach ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to find ` `// digit count of numbers ` `int` `digit_count(``int` `x) ` `{ ` `    ``int` `sum = 0; ` `    ``while` `(x) { ` `        ``sum++; ` `        ``x = x / 10; ` `    ``} ` `    ``return` `sum; ` `} ` ` `  `// Function to find the count of numbers ` `int` `find_count(vector<``int``> arr, ``int` `k) ` `{ ` ` `  `    ``int` `ans = 0; ` `    ``for` `(``int` `i : arr) { ` ` `  `        ``// Get the digit count of each element ` `        ``int` `x = digit_count(i); ` ` `  `        ``// Check if the digit count ` `        ``// is divisible by K ` `        ``if` `(x % k == 0) ` ` `  `            ``// Increment the count ` `            ``// of required numbers by 1 ` `            ``ans += 1; ` `    ``} ` ` `  `    ``return` `ans; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``vector<``int``> arr ` `        ``= { 12, 345, 2, 68, 7896 }; ` `    ``int` `K = 2; ` ` `  `    ``cout << find_count(arr, K); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of above approach ` ` `  `class` `GFG{ ` `  `  `// Function to find ` `// digit count of numbers ` `static` `int` `digit_count(``int` `x) ` `{ ` `    ``int` `sum = ``0``; ` `    ``while` `(x > ``0``) { ` `        ``sum++; ` `        ``x = x / ``10``; ` `    ``} ` `    ``return` `sum; ` `} ` `  `  `// Function to find the count of numbers ` `static` `int` `find_count(``int` `[]arr, ``int` `k) ` `{ ` `  `  `    ``int` `ans = ``0``; ` `    ``for` `(``int` `i : arr) { ` `  `  `        ``// Get the digit count of each element ` `        ``int` `x = digit_count(i); ` `  `  `        ``// Check if the digit count ` `        ``// is divisible by K ` `        ``if` `(x % k == ``0``) ` `  `  `            ``// Increment the count ` `            ``// of required numbers by 1 ` `            ``ans += ``1``; ` `    ``} ` `  `  `    ``return` `ans; ` `} ` `  `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `[]arr = { ``12``, ``345``, ``2``, ``68``, ``7896` `}; ` `    ``int` `K = ``2``; ` `  `  `    ``System.out.print(find_count(arr, K)); ` `  `  `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

## Python3

 `# Python3 implementation of above approach ` ` `  `# Function to find ` `# digit count of numbers ` `def` `digit_count(x): ` `    ``sum` `=` `0` `    ``while` `(x): ` `        ``sum` `+``=` `1` `        ``x ``=` `x ``/``/` `10` `    ``return` `sum` ` `  `# Function to find the count of numbers ` `def` `find_count(arr,k): ` `    ``ans ``=` `0` `    ``for` `i ``in` `arr: ` `        ``# Get the digit count of each element ` `        ``x ``=` `digit_count(i) ` ` `  `        ``# Check if the digit count ` `        ``# is divisible by K ` `        ``if` `(x ``%` `k ``=``=` `0``): ` `            ``# Increment the count ` `            ``# of required numbers by 1 ` `            ``ans ``+``=` `1` ` `  `    ``return` `ans ` ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``arr  ``=`  `[``12``, ``345``, ``2``, ``68``, ``7896``] ` `    ``K ``=` `2` ` `  `    ``print``(find_count(arr, K)) ` ` `  `# This code is contributed by Surendra_Gangwar `

## C#

 `// C# implementation of above approach ` ` `  `using` `System; ` ` `  `public` `class` `GFG{ ` ` `  `// Function to find ` `// digit count of numbers ` `static` `int` `digit_count(``int` `x) ` `{ ` `    ``int` `sum = 0; ` `    ``while` `(x > 0) { ` `        ``sum++; ` `        ``x = x / 10; ` `    ``} ` `    ``return` `sum; ` `} ` ` `  `// Function to find the count of numbers ` `static` `int` `find_count(``int` `[]arr, ``int` `k) ` `{ ` ` `  `    ``int` `ans = 0; ` `    ``foreach` `(``int` `i ``in` `arr) { ` ` `  `        ``// Get the digit count of each element ` `        ``int` `x = digit_count(i); ` ` `  `        ``// Check if the digit count ` `        ``// is divisible by K ` `        ``if` `(x % k == 0) ` ` `  `            ``// Increment the count ` `            ``// of required numbers by 1 ` `            ``ans += 1; ` `    ``} ` ` `  `    ``return` `ans; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``int` `[]arr = { 12, 345, 2, 68, 7896 }; ` `    ``int` `K = 2; ` ` `  `    ``Console.Write(find_count(arr, K)); ` ` `  `} ` `} ` ` `  `// This code is contributed by Rajput-Ji `

Output:

```3
```

Time complexity:- O(N*M), where N is the size of array, and M is the digit count of the largest number in the array.
Space complexity:- O(1)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

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.

Article Tags :
Practice Tags :

Be the First to upvote.

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.