Preimage Size of Factorial Zeroes Function

Problem

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 <= 10^9

Solution

class Solution {
    public int preimageSizeFZF(int K) {
        int count = 0;
        long r = search_rank(K);
        count = r==-1?0:5;
        return count;
    }

    private long search_rank(long K) {
        long lo = 4*K;
        long hi = 5*K;
        while (lo <= hi) {
            long mid = lo + (hi - lo) / 2;
            long y = f(mid);
            if (K < y)
                hi = mid - 1;
            else if (K > y)
                lo = mid + 1;
            else
                return mid;
        }
        return -1;
    }

    private long f(long x) {
        long count = 0;
        for (long i = 5; i <= x; i *= 5)
            count += x / i;
        return count;
    }
}

Explain:

nope.

Complexity:

Time complexity : O(n).
Space complexity : O(n).