N log n? Is plain n possible?

Hi all,

So I am somewhat confused with this lecture, and I have a few questions to clear it up.

  1. Is “n log n” just reordering it like you talked about? (turning boom into bmoo and then checking bmoo) I have no clue what “log” does because I’m just starting Alg. 2 this school year.

  2. I don’t understand how plain “n” could be possible. Is it possible, or is it not being possible a point that Ben is trying to make?

I’m not sure, but I think that “n log n” is the best solution here for isogram checking. I’ll look at the next video to see if it reveals anything to me.

Aidan M.

Edit: There was a small part I missed (that it was “n log n plus n” at around 8 mins), I understand now, thanks.

Edit 2: I also had trouble with understanding what you meant by O(n), I thought about having an array or something of the alphabet on file to compare with.

log is short for logarythm,

I believe that (n log n) does not take into consideration the additional check for pairs. Just N is the checking for pairs. Just N therefore is not a valid check since its just comparing neighbor values and can not properly check for isograms. (n log n) + n would have to be the fastest sort to actually take into account the value checking for an isogram.

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