Factorial Trailing Zeroes LeetCode Solution

Problem – Factorial Trailing Zeroes

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?

Factorial Trailing Zeroes LeetCode Solution in Java

    return n == 0 ? 0 : n / 5 + trailingZeroes(n / 5);

Factorial Trailing Zeroes LeetCode Solution in C++

    return n == 0 ? 0 : n / 5 + trailingZeroes(n / 5);

Factorial Trailing Zeroes LeetCode Solution in Python

    return 0 if n == 0 else n / 5 + self.trailingZeroes(n / 5)
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