[Solved] Given an integer n, return the number of trailing zeroes in n!. Note that n! = n * (n – 1) * (n – 2) * … * 3 * 2 * 1.

Question

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?

Python Solution

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

Started my Data Science journey in my 2nd year of college and since then continuously into it because of the magical powers of ML and continuously doing projects in almost every domain of AI like ML, DL, CV, NLP.

Articles: 520

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