The Parking Lot Problem: A Successor to FizzBuzz?

In one of my side projects, I encountered a pair of interesting problems, the first of which which I call the Parking Lot problem, and the sister problem the Parking Load problem. They are but toy problems, but to anyone outside of the programming field, they sound quite complicated!

I’ve provided “Expected Output” sections for you to try out your own solutions against, so don’t be too aggressive with the scrollbar if you want to quiz yourself!

Problem Statement

Imagine an infinite parking lot, where each parking space is given a natural number, starting with 0, 1, 2, 3, … to infinity. The entire lot is a single row of spaces — so that space 0 is the one closest to the entrance. There are a finite number of cars randomly parked in the lot, as people come and go as they wish. A new car has just arrived outside the lot, and the driver asks you, “Where is the closest open parking space?”

The only information you have is the taken spaces list (T), which lists all the occupied parking spaces. This list is sometimes sorted — sometimes, it is not!

Expected Output

Here is a set of expected inputs and outputs for our problem:

Input -> Output
[] -> 0
[0] -> 1
[0, 1] -> 2
[1] -> 0
[1,2,3] -> 0
[0,1,2,3] -> 4
[0,1,2,3,1000] -> 4

. You may notice that the input list is sorted (for readability), but recall that T may or may not be sorted! Try your own solution against these expected outputs!

Some Preliminary Observations

The most important thing here is T. If T is empty, then our parking lot is empty, so the answer is obviously 0. In all other cases, we have to figure out the lowest number that is absent from T. Essentially, we can reduce the problem as follows:

• Find the lowest number not present in the set of integers T, where the lowest possible number is 0.

I used the metaphor of the parking spaces because it is more memorable, and more importantly, more familiar to the general public.

Mathematical Approach

Conceptually, the problem is very simple. Consider the infinite set N, which includes all natural numbers, starting from 0. Then obviously, we can start with N itself, and remove every number in N that is also present in T. Then, when we are done (T is empty), we can look at N and retrieve the lowest number.

Sorting

Intuitively, we are really only interested in the first consecutive group of cars parked next to each other. Essentially, what you do is walk along from the very first parking space, 0, and see if it is occupied. If it is occupied, you move on to the next space — if it is not occupied, you can stop because you found your answer!

There is no way to solve the problem without first sorting T.¹ You can solve the problem without sorting only in the edge case where T’s lowest number is greater than 0 — i.e., if the car most closely parked to the entrance is not on 0, then you do not need to bother sorting T, because the answer is simply 0. And, finding the lowest number in a given set is always a \(O(n)\) operation because you need to loop through the entire set just once, keeping track of the lowest number found.

In the more common case (as seen in real life), the parking space closest to the driver (the most desirable parking space) is already taken; and what’s more, the group of spaces closest to the driver is already taken. In this case, we have to proceed from space 0, and incrementally observe every subsequent space, until we arrive at an empty parking space. The insight here is that we are treating the multitude of consecutively parked cars as one giant car parked across multiple spaces, and are just trying to find the total number of spaces this car occupies.

Ruby

I think it is time for some actual code. While I could explain how the Ruby version works, I feel that I have already expounded upon the problem enough above, so that the code here should make intuitive sense.

def get_parking_space(t)
	t_sorted = t.sort

	i = 0
	while i < t_sorted.size
		if i < t_sorted[i]
			break
		end
		i += 1
	end

	return i
end

Haskell

The Haskell version is a direct translation of the Ruby version, as an iterative approach works just fine. The only drawback is that we use Int instead of the arbitrarily large Integer type for succinctness (we can use Integer, but some built-in functions like sort only work on Int).

import Data.List

getParkingSpace :: [Int] -> Int
getParkingSpace ts
  | null ts = 0
  | otherwise = loop 0 $ sort ts
  where
  loop i xs
    | i == length xs = i
    | i < xs !! i = i
    | otherwise = loop (i + 1) xs

Functional solution

Michele Alzetta kindly sent me another solution, which has a much stronger “Haskell” flavor. I’ve made some slight adjustments; here it is:

import Data.List
import Data.Maybe

getParkingSpace' :: [Int] -> Int
getParkingSpace' ts = fromMaybe
  (length ts)
  (elemIndex False . zipWith (==) [0..] $ sort ts)

. The advantage of this version is that it only evaluates length ts (the input list) as a last resort, thanks to fromMaybe. It also uses standard Prelude functions like elemIndex and zipWith for easier understanding. Thanks Michele!

Low-Level Interlude

Did you realize that the basic concept here is the same as finding the least significant bit (LSB)? I.e., if the parking lot is one giant computer word, and a 1 bit represents an available parking space, then we are simply trying to find the LSB (the index of the bit being the same thing as the parking space number). This is such a common scenario, that there are native hardware instructions for this on most all CPUs. The canonical name for this operation is find first set or find first one.² In Linux, you can do man ffs to learn about how to use it in your C program.

Of course, the parking lot in our problem is of infinite size, which makes the size of T arbitrarily large; alas, we cannot use ffs here.

The Parking Load Problem

An interesting related problem is what I call the Parking Load problem. The scenario is the same as in the Parking Lot problem, but with the following twist: if the last (most far away) parked car determines the bounds of the parking lot (i.e., it is no longer considered infinitely large), then how many parking spaces are available?

Interestingly, this problem is easier to solve than the first problem. This seems paradoxical — surely, finding the first available parking space is easier than counting every single available space! But it is true — this problem is indeed easier — because we can solve the problem without sorting.

If we talk in terms of our “Low-Level Interlude,” this is essentially the same as saying, “Count the number of 1 bits.” There are very clever ways of counting bits, but that is not our concern, and so I will present a naive solution in Ruby.

Expected Output

Like in the previous problem, below are some expected outputs for you to test your own version against.

Input -> Output
[] -> "N/A"
[0] -> 0
[0,1] -> 0
[0,1,3] -> 1
[999] -> 999
[1,5,999] -> 997

We return “N/A” for the empty case because this condition does not make sense under the terms of the problem, which defines the bounds of the parking lot based on the car farthest away; if there are no cars to begin with, the problem cannot be posed.

Ruby

Without further ado, here is the Ruby solution.

def get_parking_spaces(t)
	if t.empty?
		return "N/A"
	else
		return (t.max - t.size) + 1
	end
end

Haskell

The Haskell version is not much different.

getParkingSpaces :: [Int] -> Either String Int
getParkingSpaces t
  | null t = Left "N/A"
  | otherwise = Right $ (maximum t) - (length t) + 1

A Successor to FizzBuzz?

For some reason, I get a strong feeling that these problems are much more interesting than FizzBuzz. The fact that we talk about an infinitely large parking lot will probably throw a lot of newbies and naive thinkers off the right track. You have to be especially careful about edge cases, such as the empty list in the second Haskell version. I believe that good coders have a keen sense of edge cases, because correct algorithms must withstand them without blowing up. Just review the Haskell solutions and notice all of the edge cases that we have to look out for!

You might even get some crazy answers that consider at length related nonessential tangents and Big-O notation, but fail to realize just how simple, at least conceptually, the problem becomes once you sort T. And, I like these problems more than FizzBuzz because there are so many interesting points about it. For example, you can ask a simple related question: if you were the keeper of this parking lot, what kind of data structure would you use to keep track of the taken parking spaces? And of course, there are very conspicuous low-level analogues that more experienced coders can relate to.

I hope you enjoyed reading about these problems. Maybe you can quiz your friend about it, and see how they respond!

Happy hacking!

Update: January 2015

I applied for a job at Periscope, and during the process Tom O’Neill pointed out that the Parking Lot Problem can be solved without sorting using sparse bitsets.³ The idea is pretty simple — each bit represents a parking space, and if it is set to 1, we treat it as occupied, and if it is 0, we treat it as empty. If there are 1,000,000 parking spaces, then we would need 1,000,000 bits. But this is where “sparseness” comes in; if there are a ton of 0’s — say, 50,000 of them — in one spot, we can easily encode that information in fewer bits; the traditional approach is to use run-length encoding (RLE) to do this. E.g., you could write the number “50,000”, and then enclose it in special OP-codes that you reserve to declare that the “50,000” here means something special (in our case, that these 50,000 bits are all zeroes).

Another approach is to use an octree (a tree that always has 8 child nodes), or “Bzets” as coined by the late Robert Uzgalis, who passed away in 2012. An excellent set of slides are available online if you Google them, but I’ve also uploaded them to my site here. Essentially, the idea is to have a tree with 3 possible values for the child nodes: 0 1 T. A 0 means that you can stop recursing down the tree, as all child nodes, recursively are all set to 0. A 1, similarly, means you can stop because all child nodes are set to 1. A T just means that you have to walk down to the child nodes, because if you gather up all the child nodes’ bits, they are mixed with 0 s and 1 s. You can imagine how compact some representations can become. In fact, Uzgalis calls bzets as having logarithmic compression! A project on Github exists that seems to be from one of the 3 UCLA student teams charged with implementing a version of bzets.

Copyright (C) 2013-2023 Linus Arver. All rights reserved.