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Aug 1, 2021 at 4:37 history edited Emily L. CC BY-SA 4.0
Typos
Jul 31, 2021 at 19:52 comment added Emily L. @KellyBundy well technically yes I was trying to make the point that it needs to be considered part of the big-O behaviour as OP simplified kO(n) to O(n) implying to me that they though k was constant. I'll update soon
Jul 31, 2021 at 15:55 comment added pacmaninbw Nice to see you are still providing good answers!
Jul 31, 2021 at 15:23 comment added D_S_X @KellyBundy It says f.O(g) = O(fg) so i guess they're the same right?
Jul 31, 2021 at 15:16 vote accept D_S_X
Jul 31, 2021 at 14:16 comment added Kelly Bundy "not kO(n), it's actually O(kn)" - Aren't they the same?
Jul 31, 2021 at 12:01 history edited Emily L. CC BY-SA 4.0
added 92 characters in body
Jul 31, 2021 at 11:53 comment added Emily L. @D_S_X I updated my answer, ptal
Jul 31, 2021 at 11:52 history edited Emily L. CC BY-SA 4.0
added 710 characters in body
Jul 31, 2021 at 10:33 vote accept D_S_X
Jul 31, 2021 at 10:33
Jul 31, 2021 at 10:30 comment added D_S_X Thanks, yeah i understand why my solution is performing so. But in case k = n, won't heap solution O(Nlogk) take O(NlogN) which is same as a simple sort and fetch kth element solution? Why is heap preferred over sorting solution?
Jul 31, 2021 at 10:23 history answered Emily L. CC BY-SA 4.0