Given an integer n
, return the number of trailing zeroes in n!
.
Note that n! = n * (n - 1) * (n - 2) * ... * 3 * 2 * 1
.
Example 1:
Input: n = 3
Output: 0
Explanation: 3! = 6, no trailing zero.
Example 2:
Input: n = 5
Output: 1
Explanation: 5! = 120, one trailing zero.
Example 3:
Input: n = 0
Output: 0
Constraints:
-
0 <= n <= 104
Follow up: Could you write a solution that works in logarithmic time complexity?
SOLUTION:
class Solution:
def trailingZeroes(self, n: int) -> int:
ctr = [0, 0]
for i in range(1, n + 1):
curr = i
while curr % 2 == 0:
ctr[0] += 1
curr = curr // 2
while curr % 5 == 0:
ctr[1] += 1
curr = curr // 5
return min(ctr)
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