Submitted 7 months ago by reddthat@reddthat.com to memes@reddthat.com
https://reddthat.com/pictrs/image/769d2aaf-8539-4b1c-be53-dbc9fe102a11.jpeg
Question: Assuming you have two integers, x and y, with y bigger than x. Sum all the numbers from x to y.
The imprecision of the question itself bugs me, but I think the Geordi meme above has it right for the most reasonable reading of the question text.
Usually interview questions are designed to be imprecise. It is a great way to see how the candidate handles it. Do they ask follow-up questions? Or do they just assume the details?
I’d be perfectly happy for a candidate to do the second answer
I’m not a programmer, just an occasional hobbyist. Can you fill me in on why the second way is better? Is it because it’s asking for “all numbers between x and y” instead of “all integers between x and y”? I probably would have done it the first way assuming they meant integers, maybe incorrectly.
The really short version is that the first solution (with the for loop) can take “a long(er) time” depending on the values of x and y.
Say x is negative one million and y is positive one million. If the loop takes, say, a microsecond per iteration, the result would take roughly 2 seconds.
The second option is one decrement operation, one increment operation, two multiplication operations, one subtraction operation, and one division operation. No matter the values of x and y, the latter option will always take exactly the same amount of time.
There’s a concept of “Big-O notation”. If the amount of time an algorithm takes increases linearly as some value in the problem (in this case, the value of y minus x), we say that’s a “O(n)” runtime. That’s the case for the first solution I posted above. If the algorithm takes the same amount of time no matter what parameters are given, we say it’s “O(1)”. That’s the case for the second option above. (If it takes the square of some parameter, we say it’s “O(n^2)”. Exponential? “O(2^n)”. Etc.)
(And, I’m handwaving quite a bit of rigor and details here. But hopefully you get the basic idea.)
99% of the time, if you’ve got a choice between a O(n) algorithm and an O(1) algorithm, you’re better off to go with the O(1) algorithm. It is entirely possible that an O(n) algorithm could run faster than O(1) in some special cases, but those special cases are almost always when n is small and the runtime of either one is going to be negligible. If n is large, you can get some very large time savings.
(And again, I’m leaving out details, but there’s a lot online about Big-O notation or “asymptotic runtime bound”.)
The different Big-O classifications tell you how “quickly” the runtime increases based on the parameter. O(n^n) grows quicker than O(2^n) which grows quicker than O(n^2) (or even O(n^c) where c is any constant) which grows quicker than O(n*log(n)) which grows quicker than O(n) which grows quicker than O(log(n)) which grows quicker than O(1). And in general, picking an option that’s alter in this list is almost always better than picking one earlier.
Now, why the latter works. The sum of integers from 1 through n is the same as n*(n+1)/2. The sum of integers from x to y is just the sum of integers from 1 to y minus the sum of integers from 1 to x-1. y*(y+1)/2 - (x-1)*(x-1+1)/2 = y*(y+1)/2 - x*(x-1)/2 = (y*(y+1)-x*(x-1))/2.
Oh, and to your specific question, both of the solutions I posted in the meme assume that the question meant specifically all “integers” between x and y inclusive. So, no, doesn’t have to do with whether we read “numbers” there to mean integers or something else. Theoretically, the two solutions I posted should always give the same answer (so long as y >= x, which is a given in the question; perhaps also so long as no integer overflow/underflow occurs… I haven’t thought that one through fully yet. Heh.) The only reason the second is preferable is because it’s “quicker” and its runtime doesn’t depend on the value of the parameters. I guess one could also argue the latter, being a one-liner, is more terse, which may also be desirable.
Damn, I took a C class about 20 years ago, and other than that I’ve done various little bash scripts for myself over the decades, and I’m pretty sure I would do better than those people.
That read made me feel better about myself, thank you for that. The imposter syndrome demon tries to pester us all from time to time.
Hiring has been so broken at the companies I’ve worked at. A lot of my coworkers decided code tests are bad, I think because it makes them uncomfortable to watch a candidate squirm. It makes me uncomfortable too, and live coding is stressful. But It also tells you a lot about their thinking.
Each company I worked for who eschewed code tests hired exceptionally bad programmers at times. I’m dealing with that right now. Sometimes I feel like I am going crazy because my coworkers tend not to notice how poorly these new folks interviewed or perform on the job. I was always the hardest interviewer and I honestly felt I went easy on people 90% of the time.
Most of our interviews weren’t as awful as these in the post but once we video chatted with a guy who was clearly googling and reading answers to us. You could see the reflection of his monitor in his glasses. It was hilarious but painful AF at the same time.
In today’s episode of “Things That Never Happened”:
Like, I’m not a good coder and struggle to get a job but uh…
i_dont_want_to@lemmy.blahaj.zone 7 months ago
“it has logic errors too” was my very favorite story