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Yeah, they're not particularly hard and not intended to be; the goal is to just see if someone is a competent coder, not a genius. I still only ended up giving a LH or higher recommendation to ~20% of people.
In some ways going for the best and brightest would be disadvantageous; they'd get bored wiring protos and updating config files all day. The main things selected for are competence, willingness to do some bare minimum of work, and compliance/desire not to rock the boat too much. Which probably makes sense.
Are the people applying and getting hired at major tech companies really that bad at basic algorithmic thinking? Conceptually, that stuff was at the level of a quiz in AP CS in junior year in high school.
Maybe I should crank out an online course or something...
did they post the follow up questions elsewhere? All I see as a question is the one where you just compare the first and last element in the array. I imagine the actual questions are much harder.
The example follow-ups were "write a function to verify that an array satisfies this constraint; "invert" an array that initially satisfies this constraint such that it decreases and then increases; or sort an array that initially satisfies this constraint.". It might not have been the most elegant solutions, but that's the kind of stuff I was expected to accomplish in C++ in 1-2 45 minutes classes in 2002.
Seems pretty doable, I assume efficiency counts?
verify() only really exists as an O(n) operation unless someone is trying to be dumb.
sort() can be done inefficiently if someone does not know much about sorting, but there is an O(n) solution waiting to be found; it can even be done in-place.
Sort is just pull off one side, compare to other side, put the smaller on stack, check next on that side compare to other side put smaller on stack repeat right? Probably a way to do less compares. For in place I'm not sure on the cost of shifts for just comparing the next element you're iterating through from left to opposite end.
Yes, if you have two sorted lists, you can combine them into one sorted list just by comparing the next element in both and putting the smaller in the list. (Merge Sort implementations rely on this.)
I thought through my in-place implementation some more and I am becoming skeptical of it. It might be possible in place but I think it means having three lists to compare (the two original and a third you build up as you march along the line). Same big-O technically but might require twice as many compares.
It's definitely possible to do in O(n) in-place, but I don't think there's a workable solution that involves only processing each element once (but I think there is one that involves processing each element at most twice).
That said I expect
return sorted(nums)is probably as fast as or maybe even faster than the clever solution in practice.More options
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