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

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:  
Traverse the numbers in the array one by oneCount the digits of every number in the arrayCheck if its digit count is a multiple of K or not.
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++
Java
Python3
C#
Javascript
// C++ implementation of above approach #include <bits/stdc++.h>using namespace std; // Function to find// digit count of numbersint digit_count(int x){    int sum = 0;    while (x) {        sum++;        x = x / 10;    }    return sum;} // Function to find the count of numbersint 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 codeint main(){    vector<int> arr        = { 12, 345, 2, 68, 7896 };    int K = 2;     cout << find_count(arr, K);     return 0;}
// Java implementation of above approach class GFG{  // Function to find// digit count of numbersstatic int digit_count(int x){    int sum = 0;    while (x > 0) {        sum++;        x = x / 10;    }    return sum;}  // Function to find the count of numbersstatic 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 codepublic 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 implementation of above approach # Function to find# digit count of numbersdef digit_count(x):    sum = 0    while (x):        sum += 1        x = x // 10    return sum # Function to find the count of numbersdef 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 codeif __name__ == '__main__':    arr  =  [12, 345, 2, 68, 7896]    K = 2     print(find_count(arr, K)) # This code is contributed by Surendra_Gangwar
// C# implementation of above approach using System; public class GFG{ // Function to find// digit count of numbersstatic int digit_count(int x){    int sum = 0;    while (x > 0) {        sum++;        x = x / 10;    }    return sum;} // Function to find the count of numbersstatic 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 codepublic 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
<script> // JavaScript implementation of above approach // Function to find// digit count of numbersfunction digit_count(x){    let sum = 0;    while (x) {        sum++;        x = x / 10;    }    return sum;} // Function to find the count of numbersfunction 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>
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) 
SURENDRA_GANGWAR
29AjayKumar
Rajput-Ji
_saurabh_jaiswal
divisibility
number-digits
Arrays
Arrays
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