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Jan 27, 2022 at 18:12 comment added qwr For top k numbers, you can maintain a k-size max-heap. But that is probably not faster in practice for small k.
Jan 27, 2022 at 10:41 comment added Matthieu M. @Teepeemm: For a generic version of selecting the k-th largest, OP's approach is O(n * k); depending how large k is related to log n, sorting may offer better complexity. Also... this approach has the advantage of being very easy to understand, which is a clear advantage.
Jan 27, 2022 at 5:21 history edited Sᴀᴍ Onᴇᴌᴀ CC BY-SA 4.0
add line after last fence so the fence doesn't appear within code
Jan 27, 2022 at 2:51 comment added Teepeemm But sorting is an n log n operation. OP's approach is O(n).
Jan 27, 2022 at 2:35 history answered Bens Steves CC BY-SA 4.0