172. Factorial Trailing Zeroes 

Description

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?

Solutions

Solution 1: Mathematics

The problem is actually asking how many factors of $5$ are there in $[1,n]$.

Let's take $130$ as an example for analysis:

Divide by $5$ for the first time, get $26$, indicating that there are $26$ numbers containing the factor $5$;
Divide by $5$ for the second time, get $5$, indicating that there are $5$ numbers containing the factor $5^2$;
Divide by $5$ for the third time, get $1$, indicating that there is $1$ number containing the factor $5^3$;
Sum up to get the count of all factors of $5$ in $[1,n]$.

The time complexity is $O(\log n)$, and the space complexity is $O(1)$.

Python3

class Solution:
    def trailingZeroes(self, n: int) -> int:
        ans = 0
        while n:
            n //= 5
            ans += n
        return ans

Java

class Solution {
    public int trailingZeroes(int n) {
        int ans = 0;
        while (n > 0) {
            n /= 5;
            ans += n;
        }
        return ans;
    }
}

C++

class Solution {
public:
    int trailingZeroes(int n) {
        int ans = 0;
        while (n) {
            n /= 5;
            ans += n;
        }
        return ans;
    }
};

Go

func trailingZeroes(n int) int {
	ans := 0
	for n > 0 {
		n /= 5
		ans += n
	}
	return ans
}

TypeScript

function trailingZeroes(n: number): number {
    let ans = 0;
    while (n > 0) {
        n = Math.floor(n / 5);
        ans += n;
    }
    return ans;
}