172_factorial_trailing_zeroes.md

172 - Factorial Trailing Zeroes

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Given an integer n, return the number of trailing zeroes in n!.

Follow up: Could you write a solution that works in logarithmic time complexity?

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:

• 1 <= n <= 104

Solution

// 23/04/2020
int trailingZeroes(int n) {
    long mul = 5;
    int res = 0;
    int tmp = 1;
    while(tmp){
        tmp = n / mul;
        res += tmp;
        mul *= 5;
    }
    return res;
}