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Notes -
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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