human
I found an interesting problem on https://leetcode.com/problems/preimage-size-of-factorial-zeroes-function/:
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 2 zeroes at the end. Given K, find how many 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.

Note:

       K will be an integer in the range [0, 10^9].

I tried it in Python, but could not do it. Can you solve it?

 ... Give me a Python code to find how many non-negative integers x have the property that f(x) = K?
assistant
#Here is Python code to find how many non-negative integers x have the property that f(x) = K.

class Solution:
     def canTransform(self, start, end):
         """
         :type start: str
         :type end: str
         :rtype: bool
         """
         def countX(s, char='R'):
             if char == 'L':
                 s = s[::-1]
             countX = 0   
             ans = []
             for c in s:
                 if c == char:
                     ans.append(countX)
                 elif c == 'X':
                     countX += 1
             return ans
 
         startLR = "".join(c for c in start if c != 'X')
         endLR   = "".join(c for c in end   if c != 'X')
         startR, startL = countX(start, 'R'), countX(start, 'L')
         endR, endL     = countX(end, 'R'), countX(end, 'L')
         
         return (startLR == endLR) and \
                all([startR[i] <= endR[i] for i in range(len(startR))]) and \
                all([startL[i] <= endL[i] for i in range(len(startL))])