Coding interview, haskell example

Recently I've been looking for a job and as usual in the IT field, I got lots of tests and programming quizz (more on this in another post). Since I wanted to flex my haskell muscles, I tried to solve one of the problem I got with it.

Count harmonic slices

Here's the problem: Given an array A of integer count the number of harmonic slices in this array. An harmonic array is an array where all consecutive elements differ only by the same constant. For example [-1, 1, 3] is harmonic as well as [-1, -1, -1] but [-1, 1, 1] is not. If the array's length is less than 2, it's defined as non harmonic. An harmonic slice is a sub array: A.slice(i, j).

So for example, the array A = [-1, 1, 3, 3, 3, 2, 1, 0] has 5 harmonic slices defined by the indices: (0, 2) (2, 4) (4, 6) (4, 7) (5, 7).

Classical solution

I'll not spoil the solution here, but it fairly easy to come up with a O(n) time complexity, and O(1) space complexity. And btw, the platform used for the test did not allow haskell anyway.

Haskell solution

I first started to replicate the classical solution using Haskell. While it would have lead to a more efficient algorithm, it was not very educational in learning Haskell. So I switched to a very declarative approach:

Get all potential slices

First, given a list of Int, return all the potential slices for this list. This is easily achieved with list comprehension:

Check if a slice is harmonic

This one had me thinking a lot. I tried various approach using recursion or foldl trying to replicate a for loop approach I would've used in other languages but with limited success. A declarative approach is actually easier. The trick is to zip the list with itself (minus the first element) to be able to compare an element with the next one.

Putting it all together

The end result is given by the length of the filtered list:

countHarmonicSlices :: [Int] -> Int
countHarmonicSlices = length $ filter isHarmonic $ subLists

Which gives in ghci:

λ: countHarmonicSlices [-1, 1, 3, 3, 3, 2, 1, 0]
5

yeah!

Final thoughts

Reasoning about space complexity in Haskell is very hard because of lazyness. For the same reason, time complexity is also harder to evaluate. I would tend to think though that this algorithm is vastly less efficient and most surely not O(n) since I'm actually evaluating all possible slices.

I'm not convinced "regular" Haskell is a good tool for this kind of task, but that wasn't the point of the exercise anyway.