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Apr 28, 2018 at 12:21 vote accept Kevin
Apr 25, 2018 at 12:27 comment added Joseph O'Rourke I think you mean $O(n \log n)$ rather than $\Omega(n \log n)$: the latter is a lowerbound, but you seek an algorithm no worse than the upperbound $O(n \log n)$.
Apr 25, 2018 at 12:25 answer added Joseph O'Rourke timeline score: 5
Apr 25, 2018 at 11:06 history edited Kevin CC BY-SA 3.0
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Apr 25, 2018 at 8:57 review First posts
Apr 25, 2018 at 10:06
Apr 25, 2018 at 8:56 history asked Kevin CC BY-SA 3.0