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

• Last Updated : 27 May, 2021

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}.```

Approach:

1. Traverse the numbers in the array one by one
2. Count the digits of every number in the array
3. Check if its 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`

## Javascript

 ``

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)

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