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Let f(x)
be the number of zeroes at the end of x!
. Recall that x! = 1 * 2 * 3 * ... * x
and by convention, 0! = 1
.
- For example,
f(3) = 0
because 3! = 6
has no zeroes at the end, while f(11) = 2
because 11! = 39916800
has two zeroes at the end.
Given an integer k
, return the number of non-negative integers x
have the property that f(x) = k
.
Example 1:
Input: k = 0
Output: 5
Explanation: 0!, 1!, 2!, 3!, and 4! end with k = 0 zeroes.
Example 2:
Input: k = 5
Output: 0
Explanation: There is no x such that x! ends in k = 5 zeroes.
Example 3:
Input: k = 3
Output: 5
Constraints:
[Math]
[Binary Search]
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