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 –Â
Python3
from math import ceil, sqrt  # function to count the factors of an arraydef 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 = 6print(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)
