"""
Solution for -> Number of trailing zeros of N!.
Created by Egor Kostan.
GitHub: https://github.com/ikostan
"""
[docs]
def zeros(n) -> int:
"""
Zeros function.
A program that calculate the number of trailing
zeros in a factorial of a given number.
N! = 1 * 2 * 3 * ... * N
For more info, see: http://mathworld.wolfram.com/Factorial.html
A simple way is to calculate floor(n/5). For example, 7! has
one 5, 10! has two 5s. It is done yet, there is one more thing
to consider. Numbers like 25, 125, etc have more than one 5.
For example if we consider 28!, we get one extra 5 and number of 0s become 6.
Handling this is simple, first divide n by 5 and remove all single 5s,
then divide by 25 to remove extra 5s and so on.
Following is the summarized formula for counting trailing 0s.
Trailing 0s in n! = Count of 5s in prime factors of n!
= floor(n/5) + floor(n/25) + floor(n/125) + ....
:param n: int
:return: int
"""
# Initialize result
count: int = 0
i: int = 5
while (n / i) >= 1:
# update Count
count += int(n / i)
i *= 5
return int(count)