C++     Math     Medium    

Problem Statement:

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:

You should notice that the number of zeros is bounded by the number of factors of 5 and its powers because 2 will have always more. So we just count the number of factors of each power of 5 from 1 to n using a simple divide operation.

 
int trailingZeroes(int n) 
{
    return n/3125 + n/625 + n/125 + n/25 + n/5;
}