LeetCode 793, Hard. Topics: Math, Binary Search. View on LeetCode.
Generate this problem as a practice environment: tested reference solution, 28 parametrized pytest cases, and a playground notebook:
Problem
Let <code>f(x)</code> be the number of zeroes at the end of <code>x!</code>. Recall that <code>x! = 1 * 2 * 3 * … * x</code> and by convention, <code>0! = 1</code>.
<ul>
<li>For example, <code>f(3) = 0</code> because <code>3! = 6</code> has no zeroes at the end, while <code>f(11) = 2</code> because <code>11! = 39916800</code> has two zeroes at the end.</li>
</ul>
<p>Given an integer <code>k</code>, return <em>the number of non-negative integers</em> <code>x</code> <em>have the property that</em> <code>f(x) = k</code>.</p>
Examples
Constraints
Solution
Reference implementation from solution.py on GitHub, full suite in test_solution.py:
Complexity
Last modified on September 7, 2026