# Solving Leetcode Interviews in Seconds with AI: Preimage Size of Factorial Zeroes Function


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "793" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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:  0 <= k <= 109  

	# Explanation
	Here's the breakdown of the solution:

*   **Understanding Trailing Zeroes:** The number of trailing zeroes in x! is determined by the number of times 5 appears as a factor in the numbers from 1 to x. This is because the number of factors of 2 will always be greater than or equal to the number of factors of 5, and a factor of 2 and a factor of 5 combine to create a factor of 10, which contributes to a trailing zero.
*   **Monotonicity:** The function f(x) is monotonically increasing (non-decreasing). This means that as x increases, the number of trailing zeroes can only increase or stay the same. This property allows us to use binary search to efficiently find the smallest x such that f(x) = k.
*   **Finding the Range:**  Since f(x) is monotonic, if we find the smallest x such that f(x) = k, then the next 5 values of x will also have f(x) = k, until we reach an x where a new factor of 5 appears in the factorial. Thus, we can find the smallest `x1` such that `f(x1) = k`, and the smallest `x2` such that `f(x2) = k+1`, and the answer will be `x2 - x1`.

*   **Time & Space Complexity:** O(log<sup>2</sup>k) time complexity, O(1) space complexity. The outer binary search for k and k+1 is O(log k).  The `zeros` function which does the calculation for the number of trailing zeroes given a particular `x` is O(log x). Since x is in the ballpark of k, the `zeros` is O(log k) as well.  Thus O(log k) * O(log k) = O(log<sup>2</sup>k). Constant space is used.

	
	# Code
	```python
	def trailingZeroes(n):
    count = 0
    i = 5
    while n // i >= 1:
        count += n // i
        i *= 5
    return count

def find_k(k):
    low = 0
    high = 5 * k  # Upper bound for x (since every 5th number contributes a factor of 5)

    while low <= high:
        mid = (low + high) // 2
        zeroes = trailingZeroes(mid)

        if zeroes < k:
            low = mid + 1
        elif zeroes > k:
            high = mid - 1
        else:
            # Found a potential solution. Need to find the smallest one.
            high = mid -1
    return low

class Solution:
    def preimageSizeFZF(self, k: int) -> int:
        x1 = find_k(k)
        x2 = find_k(k+1)

        return x2-x1
	```