Valid Perfect Square - Leetcode 367 - Binary Search (Python)

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Опубликовано:

18 май 2024

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Комментарии : 14

@Jonas-gm4my 17 часов назад

The nth perfect square is the sum of the first n odd number, cant we just subtract odd numbers while the number is bigger or equal to zero and then check whether we are at 0 at the end?

@brycejohansen7114 23 дня назад

You could use Newton's Method which is O(log log n) because it converses quadratically.

@GregHogg 23 дня назад

Yeah very true!

@LelouchVDK 23 дня назад

I might be dumb, but putting it on the internet anyway. But wouldn't any square of an integer larger than the half of the original number already be bigger than the original number, meaning you could set the initial range to be from 1 to num // 2. Either way amazing question with an awesome solution from you

@GregHogg 23 дня назад

Yes you could probably optimize the end range quite a bit. However because it's log and dividing by two each time, you'll only spare one or two iterations max anyway. Excellent observation. Thanks so much!