Skip to content

16.05. Factorial Zeros

Description

Write an algorithm which computes the number of trailing zeros in n factorial.

Example 1:


Input: 3

Output: 0

Explanation: 3! = 6, no trailing zero.

Example 2:


Input: 5

Output: 1

Explanation: 5! = 120, one trailing zero.

Note: Your solution should be in logarithmic time complexity.

Solutions

Solution 1: Mathematics

The problem is actually asking for the number of factors of $5$ in $[1,n]$.

Let's take $130$ as an example:

  1. Divide $130$ by $5$ for the first time, and get $26$, which means there are $26$ numbers containing a factor of $5$.
  2. Divide $26$ by $5$ for the second time, and get $5$, which means there are $5$ numbers containing a factor of $5^2$.
  3. Divide $5$ by $5$ for the third time, and get $1$, which means there is $1$ number containing a factor of $5^3$.
  4. Add up all the counts to get the total number of factors of $5$ in $[1,n]$.

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

1
2
3
4
5
6
7
class Solution:
    def trailingZeroes(self, n: int) -> int:
        ans = 0
        while n:
            n //= 5
            ans += n
        return ans

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
class Solution {
    public int trailingZeroes(int n) {
        int ans = 0;
        while (n > 0) {
            n /= 5;
            ans += n;
        }
        return ans;
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
class Solution {
public:
    int trailingZeroes(int n) {
        int ans = 0;
        while (n) {
            n /= 5;
            ans += n;
        }
        return ans;
    }
};

1
2
3
4
5
6
7
8
func trailingZeroes(n int) int {
    ans := 0
    for n > 0 {
        n /= 5
        ans += n
    }
    return ans
}

1
2
3
4
5
6
7
8
function trailingZeroes(n: number): number {
    let ans = 0;
    while (n) {
        n = Math.floor(n / 5);
        ans += n;
    }
    return ans;
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
class Solution {
    func trailingZeroes(_ n: Int) -> Int {
        var count = 0
        var number = n
        while number > 0 {
            number /= 5
            count += number
        }
        return count
    }
}

Comments