# LeetCode – LFU Cache [HARD]

## Introduction

The Least Frequently Used (LFU) Cache problem brings together a number of Computer Science techniques and principles to solve a very “Computer Science’y” problem.

In summary, the problem asks you to build a cache that could mimic the behavior found in hardware caches (i.e. CPU caches) as well as software based caches (i.e. Redis, memcached, etc.). The cache is composed of key:value pair objects.

The problem is essentially structured into 3 areas:

1. Intialising a cache of a given size
2. Supporting the addition of items to the cache such that if the cache is full, the least frequently accessed item is removed (to make space) and the new item added.
1. If the item already exists in the cache, its value is updated and the frequency of that item being accessed is increased by 1
3. Supporting the retrieval of items from the cache such that each retrieval updates the frequency at which the retrieved item is accessed, as well as when it was last accessed (and importantly in relation to when it was last accessed, when compared with items accessed at the same frequency)

My O(1) JavaScript solution was 99.31% faster (although this measurement isn’t particularly fair as the environment is not the same for each run (i.e. it’s a shared environment)) and 100% more memory efficient than all other accepted JavaScript solutions.

## Approaches

In the problem description, it mentions that a solution of complexity O(1) is desirable – meaning the retrieval or addition of an item takes the same number of steps, irrelevant of the item or the size of the cache (from a computational perspective – obviously larger software caches may perform slower where hardware caching, virtual memory and paging come into play).

Ignoring the O(1) requirement, I identified two possible solutions:

• Iterative approach (not O(1))

### Iterative Approach

The initial approach that comes to mind is to split the items into buckets based on frequency, with the buckets containing an array of items. The item at the top of the array is the most recently accessed item for that frequency. This is demonstrated in the diagram below:

The issue with this solution is that in order to determine whether an item is in the bucket, you must loop through potentially all the items in the bucket. This prevents the solution from achieving O(1) and is more O(n).

Other issues I encountered with this approach around around maintaining index pointers into the array and the complexity this introduced when items are removed in particular. Essentially, I found a loop was always required (although I’d love to see an O(1) solution for this approach if it exists!).

In order to achieve O(1), a solution must be designed such that, in simple terms, there are no loops or recursion (which are valid approaches to solving the problem). Whereas the previous solution relied on buckets and loops to look through all the items, an O(1) solution must therefore utilise a different data structure(s).

Those data structures are:

The solution is highlighted in the diagram below:

1. The first doubly linked-list creates a set of buckets that contain items based on the frequency at which they’re accessed
2. The second doubly linked-list (the root of which is the ‘data’ segment of the first doubly linked-list – the frequency bucket) stores the items in order of last accessed
1. This is particularly useful when wanting to delete the least frequently accessed item

This data structure on its own does not meet our O(1) requirement; if we wanted to determine whether an item exists, we’d have to traverse each bucket and the items within each bucket (a complexity of essentially O(n)). The solution is simple, maintain a dictionary which points to the items within the doubly linked-list structure.

There are additional minor details which won’t be expanded upon in this post, but I invite you to critique the solution on GitHub.

## Analysis

This problem was definitely less of a thought experiment and more of a problem you may come across during a career as a Software Engineer. My takeaways are as follows:

• Understand when a solution will benefit from a more rigid approach – apart from scribbling a bit-part solution on paper, I started development straight away. The doubly linked-list solution naturally developed and therefore I hadn’t coded for it to be reusable or use a reusable library. This resulted in some code duplication and generally increased the difficulty of debugging the solution. Obviously if I was to productionise a solution such as this, I would refactor to use a library implementation of a doubly linked-list that was well tested.
• Having a basic understanding of computational complexity is important for all software engineers – regardless as to whether you’re developing for embedded systems or consumer websites. Problems can be solved in any number of ways, but its not usually until scaled out (not least in the enterprise) that the way the problem was solved becomes important. By understanding the impact of your solution, you’ll save on some pain further down the road.
• All software boils down to data structures and algorithms. You may have been taught about them at University and now subconsciously use them in your job; or you may have never studied them but as a Software Engineer find yourself using them naturally. Having an understanding of the common data structures (linked lists, stacks, dictionaries, etc.) and how to operate upon them can increase the speed at which you develop reliable, quality code (especially if using libraries). Data Structures and Algorithms are not just topics to be studied for exams – keep them at the front of your mind.

## Conclusion

I’ve solved many problems on other ‘coding’ challenge sites such as Project Euler – these problems can often seem too theoretical to take any practical lessons from. It’s been enjoyable to solve a problem that is Computer Science focused – whilst I didn’t learn as much as I may do when compared with a Project Euler problem, it has reinforced some Computer Science theory and the importance of it in your day-to-day job as a Software Engineer.

# LeetCode – Max Points on a Line

## Introduction

Max Points on a Line is the third least accepted (against submitted solutions) problem on LeetCode; it is also the second hardest problem in the ‘hard’ category. At first glance, the problem seems simple – from a set of points, determine the longest straight line that can be formed from those points and return the number of points in the subset. However, as the saying goes, if you assume it makes an ass out of u and me.

This is my write-up detailing the approach I took to solving the problem and the issues I encountered along the way. Lessons learnt can always be taken from one context and applied to another – hopefully you find this useful.

## Approaches

I explored two options when it comes to solving this problem, both explained below. An ‘agricultural’ approach which I quickly abandoned, and a ‘formulaic’ approach that did work but encountered some interesting issues.

Whilst I won’t be posting my solution in this article, you can find my implementation on GitHub (linked in the Social Banner above).

### Agricultural

Having forgotten some high school mathematics – my brain instantly went trying to understand how an algorithm based on scanning and or recursion / iteration could solve the problem. Could I determine the ‘step’ between two points and then for each remaining point, see if I continue the step pattern whether or not the point in question would be reached.

The algorithm quickly became complex – you would obviously need to check the step in both positive and negative directions (reducing performance), you may want to start finding the point in the line that would be closest to the point your analysing to slightly improve the performance, etc.

Good problem solving doesn’t mean coming to the correct solution first time – it’s also recognising when an approach isn’t working and looking for another way. y = mx + b.

### Formulaic

The straight line equation in its well known form will determine y for a points associated x where the slope (m) and y-intercept (b) of the line are known.

My initial approach was to take two points in an existing line (any two points can be the start of a line) and determine their slope m = ( (y1 – y2) / (x1 – x2) ). The remaining variable to determine was the y-intercept which with the formula rearranged stated b = y – mx. I knew the slope from the previous calculation and I know x and y from an existing point so I could determine b, the y-intercept. The final step was to determine if for the given x of the point being analysed, given the y-intercept and the defined slope, did the resulting y == y of the point being analysed.

The approach worked – I went from passing 50% of the test cases using the agricultural approach to passing approx. 90%. But what was causing the final 10% to fail? Precision.

In day-to-day life as a Software Engineer, precision doesn’t usually impact you unless you’re working on financial systems, sensor based system, and other similarly complex, numerically focussed applications.

My solution was based on JavaScript which by default represents all floating point numbers as 64-bit floats. This means a number with a fractional component can be represented up to 17 decimal places, and even then may not be accurate. Both of these limitations caused issues.

The following test case enabled me to identify the issue:

```[[0,0],[94911151,94911150],[94911152,94911151]]
```

Whilst the final two points are very close to each other, when on a slope with 0,0, all three do not form a straight line. If we try to determine the slope of the line containing the first two points, we would calculate -94911150/-94911151 – the answer according to the JavaScript engine is 0.9999999894638303. To determine the slope between the first and third point, we would do -94911151/-94911152 – the answer according to the JavaScript engine is 0.9999999894638303. Brilliant! All three points are on a straight line! But they’re not…

Use a high precision calculator and you’ll see that -94911150/-94911151 = 0.999999989463830230022181482131641201991112719726684170124541003617161907 and -94911151/-94911152 = 0.999999989463830341033053734296682016882483946670460811601991723796588202.

You can see around digits 16/17, the answer begins to vary… the slope is not the same. However, due to JavaScripts (although this problem isn’t limited to native JavaScript) precision limit and accuracy issues (i.e. rounding) mean that for this problem, I am unable to correctly to determine the slope. So, how do you solve the problem?

Initially I knew I had a precision error, but I wasn’t sure whether it was determining the slope, y-intercept, and y value for the cross-comparison. My first thought was do I need to determine the y-intercept and the y value? Surely if two points are on the same slope as two other points on the line, we’re good. Having removed the other elements, I noticed the precision issue above.

My immediate thought was that by using 0,0 as a comparison point, we were working with an answer for the slope that was extremely precise. If another point was used, the result may not require the same precision. If points 2 and 3 form the input were used to determine the slope, the result would be 1 (-1/-1). As 1 != 0.9999999894638303 – we pass the test case. This solution worked for all test cases and the ultimate acceptance of the solution. However, it’s obvious to see for test cases that result in a similar precision no matter what points are selected, the solution would encounter problems.

## Analysis

LeetCode provides some great solution analytics – at the time of solving the problem, my solution execution time was 79% faster than other JavaScript solutions. My assumption is that that these solutions may have taken more agricultural approaches.

There are a few key takeaways from working through this solution:

1. All programming languages have limitations (of which some can be solved with third-party libraries – in this case, several BigNumber libraries). Whilst they may not impact you day-to-day, if you’re encountering ‘strange’ issues, they’re good to have in the back of your mind. Worse, often the runtime won’t throw any errors to tell you it has automatically rounded a result or is not accurate past a certain point.
2. Formal Specification techniques are used in critical systems to at runtime ensure code is behaving according to some mathematical constraints. Whilst formal specifications may not be necessary for all projects, there’s nothing wrong with having code to check the validity of results. For example, if adding two decimals both with two digits following the decimal point, the result won’t have more than two digits following the decimal point. If it does, you know something has gone wrong.
3. Understand what the minimum required steps are to solve the problem – my initial formulaic approach trying to understand the y value was adding complexity to the solution and making it harder to debug. I only needed to calculate the slope to solve this problem.
4. Another bug I encountered was due to not fully understanding the specification – my assumption (the second time I assumed wrong in solving this problem!) was that the points had to be consecutive according to the step function. This wasn’t the case and was not clearly stipulated in the problem description.

## Conslusion

This problem was great fun to solve. Day-to-day in the workplace, my clients are using software in ways were these sorts of problems do not manifest themselves. But they exists – it’s important as a Software Engineer to be aware of them and to challenge yourself in ways that you may not be challenged in your day job (not that it lacks its own unique set of challenges!).

Apart from the usual cliches such as testing is only as good as the test cases and not fully understanding requirements is the downfall of many IT projects, I will take two ideas from this exercise:

1. Testing can be built into your code – if the result of performing an action is that n number of things should be in a list (without knowing the contents), add some code to check that n things are in the list. If not, an issue must have occurred (either data or code)
2. Understand what the minimum required steps are to solve the problem – initially you may just want to solve a problem, with the result not being the most elegant solution. Even if you don’t encounter issues, refactoring is important. It results in less instructions, potentially leading to less bugs in future; for those bugs that do occur, the solution will be easier to debug.