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Jul 24, 2023 at 15:10 answer added LD50 timeline score: 1
Jul 24, 2023 at 13:35 answer added LD50 timeline score: 1
May 23, 2017 at 12:37 history edited CommunityBot
replaced http://stackoverflow.com/ with https://stackoverflow.com/
Sep 9, 2016 at 20:28 comment added KWillets Oops, my brain is elsewhere today. N^2 in the second term.
Sep 9, 2016 at 19:53 comment added KWillets It will be $O(N \log N + N)$ in that case, if you obtain the LCP's during the sort.
Sep 9, 2016 at 18:46 comment added D.W. This algorithm is correct, but depending on what $N$ is intended to represent, the running time is not $O(N \lg N)$. Consider a set of $N$ strings, where each string starts with $N$ a's. Then the running time is more like $O(N^2 \lg N)$, because comparing a pair of strings takes $O(N)$ time rather than $O(1)$ time.
Sep 9, 2016 at 18:44 history edited D.W. CC BY-SA 3.0
deleted 43 characters in body; edited title
Sep 9, 2016 at 17:18 vote accept Ahmed Fasih
Sep 9, 2016 at 17:04 answer added KWillets timeline score: 2
Sep 9, 2016 at 16:51 history edited Ahmed Fasih CC BY-SA 3.0
speed test
Sep 9, 2016 at 16:12 history asked Ahmed Fasih CC BY-SA 3.0