Count the factors of K present in the given Array

Given an array arr[] and an integer K, the task is to calculate the count the factors of K present in the array.
Examples:

Input: arr[] = {1, 2, 4, 5, 6}, K = 6 
Output: 3 
Explanation:
There are three numbers present in the array those are factors of K = 6 – {1, 2, 6}
Input: arr[] = {1, 2, 12, 24}, K = 20 
Output: 2 
Explanation:
There are two numbers present in the array those are factors of K = 20 – {1, 2} 

Naive Approach: A simple solution for this problem is to find all the factors of K and then for each factor iterate over the array and check that it is present in the array or not. If yes then increment the count of factors by 1.
Efficient Approach: The idea is to instead of finding all factors of the number K iterate over the array and check for each element that it is the factor of K, or not with the help of the modulo operator. If yes then increment the count of factors of K.
Below is the implementation of the above approach:

C++

// C++ implementation to find the count
// of factors of K present in array

#include <iostream>
using namespace std;

// Function to find the count
// of factors of K present in array
int calcCount(int arr[], int n, int k)
{
    int count = 0;

    // Loop to consider every
    // element of array
    for (int i = 0; i < n; i++) {
        if (k % arr[i] == 0)
            count++;
    }

    return count;
}

// Driver Code
int main()
{
    int arr[] = { 1, 2, 4, 5, 6 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 6;

    // Function Call
    cout << calcCount(arr, n, k);

    return 0;
}

Java

// Java implementation to find the count
// of factors of K present in array
class GFG{

// Function to find the count
// of factors of K present in array
static int calcCount(int arr[], int n, int k)
{
    int count = 0;

    // Loop to consider every
    // element of array
    for(int i = 0; i < n; i++)
    {
       if (k % arr[i] == 0)
           count++;
    }
    return count;
}

// Driver Code
public static void main(String[] args)
{
    int arr[] = { 1, 2, 4, 5, 6 };
    int n = arr.length;
    int k = 6;

    // Function Call
    System.out.print(calcCount(arr, n, k));
}
}

// This code is contributed by gauravrajput1

Python3

# Python3 implementation to find the count
# of factors of K present in array

# Function to find the count
# of factors of K present in array
def calcCount(arr, n, k):

    count = 0

    # Loop to consider every
    # element of array
    for i in range(0, n):
        if (k % arr[i] == 0):
            count = count + 1

    return count

# Driver Code
arr = [ 1, 2, 4, 5, 6 ]
n = len(arr)
k = 6

# Function Call
print(calcCount(arr, n, k))

# This code is contributed by PratikBasu

C#

// C# implementation to find the count
// of factors of K present in array
using System;

class GFG{

// Function to find the count
// of factors of K present in array
static int calcCount(int []arr, int n, int k)
{
    int count = 0;

    // Loop to consider every
    // element of array
    for(int i = 0; i < n; i++)
    {
       if (k % arr[i] == 0)
           count++;
    }
    return count;
}

// Driver Code
public static void Main(String[] args)
{
    int []arr = { 1, 2, 4, 5, 6 };
    int n = arr.Length;
    int k = 6;

    // Function Call
    Console.Write(calcCount(arr, n, k));
}
}

// This code is contributed by Amit Katiyar

Javascript

<script>

// Javascript implementation to find the count
// of factors of K present in array

// Function to find the count
// of factors of K present in array
function calcCount(arr, n, k)
{
    var count = 0;

    // Loop to consider every
    // element of array
    for (var i = 0; i < n; i++) {
        if (k % arr[i] == 0)
            count++;
    }

    return count;
}

// Driver Code
var arr = [ 1, 2, 4, 5, 6 ];
var n = arr.length;
var k = 6;

// Function Call
document.write( calcCount(arr, n, k));

</script>

Output:

3

Time Complexity: O(N)

Auxiliary Space: O(1), since no extra space has been taken.