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

Given an array **a**rr 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 = 2Output:2Explanation:There are 2 numbers whose digit count is multiple of 2 {12, 3444}.Input:arr[]={12, 345, 2, 68, 7896}, K = 3Output:1Explanation:There is 1 number whose digit count is multiple of 3 {345}.

**Approach:**

- Traverse the numbers in the array one by one
- Count the digits of every number in the array
- 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 <bits/stdc++.h>` `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

`<script>` `// JavaScript implementation of above approach` `// Function to find` `// digit count of numbers` `function` `digit_count(x)` `{` ` ` `let sum = 0;` ` ` `while` `(x) {` ` ` `sum++;` ` ` `x = x / 10;` ` ` `}` ` ` `return` `sum;` `}` `// Function to find the count of numbers` `function` `find_count(arr, k)` `{` ` ` `let ans = 0;` ` ` `for` `(let i of arr) {` ` ` `// Get the digit count of each element` ` ` `let 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` `let arr = [ 12, 345, 2, 68, 7896 ];` `let K = 2;` `document.write(find_count(arr, K));` `// This code is contributed by _saurabh_jaiswal` `</script>` |

**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. To complete your preparation from learning a language to DS Algo and many more, please refer **Complete Interview Preparation Course****.**

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