# 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}.```

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 the 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*log10M), where N is the size of array, and M is the largest number in the array.
Space complexity:- O(1)

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