Python | Count the array elements with factors less than or equal to the factors of given x

Given an array, the task is to count the elements of array whose factors are less than the given number x. Examples:

Input: arr = [2, 12, 4, 6], x = 6 Output: 2 factors of x = 6 is [1, 2, 3] factors of arr[0] = 2 is [1] factors of arr[1] = 12 is [1, 2, 3, 4] factors of arr[2] = 4 is [1, 2] factors of arr[3] = 6 is [1, 2, 3] so only arr[0] and arr[2] are the answer.

Approach: Find out factors of all the elements and that of given xand after that compare all of them if factors of elements are less than that of factors of x, then increment the count. Below is the implementation of above problem – 

Python3

from math import ceil, sqrt

# function to count the factors of an array
def factorscount(x):
    count = 0
    for i in range(1,ceil(sqrt(x))):
        if x%i==i:
            count+=1
        else:
            count+=2
    return count

def Totalcount(arr, x):

    # count of factors of  given x
    count_fac = factorscount(x)

    # store the count of each factors
    arr_fac = [factorscount(i) for i in arr]

    ans = 0

    for i in arr_fac:
        # if factors count of element of array is
        #small than that of given number
        if i<count_fac:
            ans+=1

    return ans


# Driver code

arr = [2,12,4,6]
x = 6
print(Totalcount(arr, x))

Output:

2

Time Complexity: O(?x+n?k) where n is the size of the array, x is the given number, and k is the maximum element in the array.
Auxiliary Space: O(n)