Silly job interview questions in Haskell
60 comments
·May 23, 2025sshine
I live-coded Haskell a couple of times at job interviews for non-Haskell positions where you were free to choose.
It’s fun because the interviewer often doesn’t grok the paradigm. In one case, I was asked to make a password generator, and I went with QuickCheck’s Gen, which lets me write a random generator of things without explicitly defining a single function (they’re all monads).
So you go from somehow describing the thing you want to make to “I’m done.” :-D
18172828286177
Did you get the job?
sshine
Yeah.
I also refused to solve a part of the task because it would lead to bad password policy. Instead I property-tested the generator.
djtango
I remember I did an Advent of Code day 1 in Haskell a while back.
The meat of the code was a one liner. But opening a file, throwing away the first line and parsing a CSV took a lot more head scratching, especially if you've only ever done puzzles in Haskell.
Makes the hard things easy and the easy things hard
ngruhn
Same. Until I discovered those parser combinator libraries which are THE way to do parsing in Haskell. It makes parsing so pleasant you'll never need a regex again. A parser for that CSV file probably looks like this:
line = decimal `sepBy` char ','
file = line `sepBy` newline
djtango
Thanks for sharing! This was in a pre-LLM era and I couldn't be bothered to research the right way to do this. Those parser combinators are excellent, so clean
ReDress
> Silly job interview questions in Haskell.
Well, it's not clear to me why we're terming these as silly. They mostly seem simple, instead, to me.
lynx97
Well, that fizlle implementation is not a good example of a great answer. It checks for the two conditions twice. At least I would use a where clause to define fizz and buzz.
internet_points
which is nice with the laziness because you only eval what you need, e.g.
fizzle :: Int -> String
fizzle n
| fizz && baz = "Fizz Baz!" -- if this branch is taken, buzz is never evaluated
| fizz && buzz = "Fizz Buzz!"
| fizz = "Fizz!"
| buzz = "Buzz!"
| otherwise = show n
where
fizz = n `mod` 3 == 0
buzz = n `mod` 5 == 0
baz = n `mod` 7 == 0thdhhghgbhy
Typical Haskell coding post with no mention of order or efficiency of the algorithm solutions. The thing is, even in basic coding quizes, if the interviewer is worth their salt, they will ask about how to improve the efficiency of the naive solutions first put forward, like the palindrome answer in this post.
What is the order of sumNToTotal?
dmurray
It does discuss performance and points out some places where the Haskell implementation makes the natural naive solutions performant.
> While it's true that sort has an O(nlogn) performance profile, one interesting thing here is that sorting is lazy in Haskell! This means that if our two strings are unequal, they will only be sorted far enough to determine inequality!
I don't understand exactly how this works - don't all O(n log n) sorting algorithms still have to do some work on the other elements of the list before pulling out the smallest one? If you just scanned and pulled the smallest one to the front each time that is O(n^2). And there's an implementation detail here that differences at the start of the sorted string will be caught more quickly than at the end, so real-world behaviour is dependent on your choice of sort order.
Agreed that the performance analysis is quite superficial and that some of the solutions like sum NToTotal must be very sub-optimal.
kqr
The matchesSum check is constant in the length of the input list. The filter adds an iteration over all results of (combinations n xs). If the complexity of (combinations n) is T(n) in xs, then by analogy of its recursive definition we have two recursive calls that are T(n-1), we have an fmap that is O(1) and a concatenation that is O(1). That sets up the recurrence T(n) = 2T(n-1) making the (combinations n) function O(n) in xs.
Thus the whole thing is O(n²). It's not like this is particularly hard for Haskell. If anything, it's often easier because we can set up the recurrence by analogy with the definition.
iamevn
can you not narrow it down further to O(k * nCr(n, k)) (n=size of input, k=combination size) since it does k conses for each of the nCr(n, k) combinations?
(the final filter sums k elements nCr(n, k) times as well)
benmmurphy
how is combinations O(n)? for n choose 3 generating combinations is going to approach n^3 since the formula is basically (n * n * n) / constant and for sumToAny generating all the combinations is 2^n.
Paracompact
The execution order, or the order of the output list? The latter should be self-evident from the function body (first elements get collected first), and the former holds no tricks either since filters are evaluated from the head first, and concat (<>) allows lazy access to the first list's elements.
This is just a beginner's Haskell post. If an interviewer really meant to test optimization knowledge, I would hope they wouldn't ask me to optimize one O(n) utility function into another.
null
iamevn
with N as the length of the subsets and X as the length of input, `combinations n xs` basically walks through list xs and at each element, conses that element onto each combination of size n-1 of later elements in the list then appends those resulting lists together. this is on the order of N * nCr(X, N) aka O(NX!/(N!(X-N)!)) which dominates the overall runtime. the final filter walks through each combination and does O(N) work at each element to find the ones that sum to the desired amount, again O(NX!/(N!(X-N)!))
null
Sourabhsss1
Words for the functions are easy to understand.
rhdjsjebshjffn
[flagged]
90s_dev
For one thing, it's hard to know what to spend it on. What should take priority? Obviously your own basic needs, for starters. And then of course the needs of your immediate dependents. But then what? Do you try to increase your quality of life? Theirs? At what point do you draw the line and begin to help others altruistically? Are two coins worth more than billions in practice? Besides, what even is quality of life? It's a very difficult and heavy problem. One I'm glad I don't have.
zwnow
Altruism? In rich folk? Usually they convince themselves on thinking they need stuff they don't actually need. Employees can't afford everyday life? Not their issue.
fkfyshroglk
> it's hard to know what to spend it on.
Probably not Haskell developers, but then again, I don't have enough money for my opinions to be worth anything.
> One I'm glad I don't have.
you should probably put less thought into it if you want to be happy & not grapple with such difficult thoughts.
> At what point do you draw the line and begin to help others altruistically?
At the point you have money, IMO. Otherwise what is the point of having money?
pensatoio
Fifteen years into my career, and I'm finally realizing that "expressive" languages are practically unreadable.
tikhonj
unfamiliar languages are practically unreadable
rhdjsjebshjffn
I don't see how that's any better for a haskell shop. i got some empathy but you chose a rough life.
yakshaving_jgt
The trick is to hire people who are familiar with the language.
rrradical
As a haskell programmer, all of that was easily readable.
sgarland
As a non-Haskell programmer, all of that was easily readable. Straightforward words for functions makes that pretty easy.
RHSeeger
There were a couple of places that took me a couple reads to figure out, like the fact that `(x:)` was "prepend". But overall, I followed the code pretty well. From the context of someone that wrote a small amount of Haskell a decade ago.
90s_dev
I honestly just skimmed the code and assumed it probably does what I would guess it does. It seemed to make sense and be straightforward... assuming my guesses were right, I guess?
CamperBob2
As a C programmer, that's the worst FizzBuzz implementation ever. You're not supposed to special-case % 15, just
bool m3 = !(i % 3);
bool m5 = !(i % 5);
if (m3) printf("Fizz");
if (m5) printf("Buzz");
if (m3 || m5) printf("\n");
sponnath
Not all programming languages are obvious derivatives of C. Haskell is pretty readable once you spend some time getting to know the syntax.
90s_dev
The semantics are very different. It's much closer to a mathematical proof than a C program with different syntax, in terms of its lazy evaluation.
Darmani
I prefer to think of Haskell-like lazy evaluation as constructing a dataflow graph. The expression `map f (sort xs)` constructs a dataflow graph that streams each output of the sort function to `f`, and then printing the result begins running that job. Through that lens, the Haskell program is more like constructing a Spark pipeline. But you can also think of it as just sorting a list then transforming each element with a function. It only makes a difference in resource costs or when there's potential nontermination involved, unless you use unsafe effects (e.g.: unsafePerformIO).
Is there a way to think of proofs as being lazy? Yes, but it's not what you think. It's an idea in proof theory called polarization. Some parts of a proof can be called positive or negative. Positive parts roughly correspond to strict, and negative roughly correspond to lazy.
To explain a bit more: Suppose you want to prove all chess games terminate. You start by proving "There is no move in chess that increases the number of pieces on the board." This is a lemma with type `forall m: ChessMove, forall b: BoardState, numPieces b >= numPieces (applyMove m b)`. Suppose you now want to prove that, throughout a game of chess, the amount of material is decreasing. You would do this by inducting over the first lemma, which is essentially the same as using it in a recursive function that takes in a board state and a series of moves, and outputs a proof that the final state does not have more material than the initial state. This is compact, but intrinsically computational. But now you can imagine unrolling that recursive function and getting a different proof that the amount of material is always decreasing: simply write out every possible chess game and check. This is called "cut elimination."
So you can see there's a sense in which every component of a proof is "executable," and you can see whether it executes in a strict or lazy manner. Implications ("If A, then B") are lazy. Conjuctions ("A and B") can be either strict or lazy, depending on how they're used. I'm at the edge of my depth here and can't explain more -- in honesty, I never truly grokked proof polarization.
Conversely, in programming languages, it's not strictly accurate to say that the C program is strict and the Haskell program is lazy. In C, function definitions and macro expansions are lazy. You can have the BAR() macro create a #error, and yet FOO(BAR()) need not create a compile error. In Haskell, bang patterns, primitives like Int#, and the `seq` operator are all strict.
So it's not the case that proofs are lazy and C is strict and Haskell is lazy so it's more like a proof. It's not even accurate to say that C is strict and Haskell is lazy. Within a proof, and within a C and a Haskell program, you can find lazy parts and strict parts.
Always fun to see this kind of Rosetta Code thing, feel like no matter the topic there's something to glean from reading!
That said: if I ever gave the palindrome question to a candidate (I haven't) and they pulled out a `reverse` library func/keyword we'd be encouraging them to do more.