Factorial Trailing Zeroes
Brute force solution is to calculate and keep dividing by and count number of 's.
# Number of 5's
Factor a number: . We can see only has 1 trailing zero b/c the number of pair of and is 1.
For a factorial, the number of
2's is definitely more than that of
Thus, the number of
5's is the number of trailing zeros.
def trailingZeroes(self, n: int) -> int: cnt = 0 while n>0: cnt += n//5 n //= 5 # this also works # n //= 5 # cnt += n return cnt