长度为K的倍数的数组中的整数计数

给定一个数组，其中rr由N个元素和一个整数K组成，任务是计算长度为K的倍数的所有元素。

例子：

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}.

方法：

1. 遍历数组中的数字
2. 计算数组中每个数字的位数
3. 检查其位数是否为K的倍数。

下面是上述方法的实现：

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 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 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

输出：

3

时间复杂度： -O(N * M) ，其中N是数组的大小，M是数组中最大数字的位数。
空间复杂度： -O(1)