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Timeline for Sever-sort an array

Current License: CC BY-SA 3.0

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Jan 29, 2024 at 19:51 answer added bigyihsuan timeline score: 0
Jan 29, 2024 at 17:20 answer added noodle person timeline score: 0
Jan 29, 2024 at 16:28 answer added Xcali timeline score: 0
Jan 29, 2024 at 6:29 answer added l4m2 timeline score: 0
Jan 29, 2024 at 1:59 answer added pacman256 timeline score: 1
Feb 11, 2021 at 22:57 answer added Dominic van Essen timeline score: 1
Feb 11, 2021 at 22:54 answer added Leo timeline score: 3
Feb 11, 2021 at 21:55 answer added coltim timeline score: 1
Feb 11, 2021 at 19:50 answer added Giuseppe timeline score: 1
Feb 11, 2021 at 19:37 answer added caird coinheringaahing timeline score: 2
Dec 28, 2016 at 14:50 answer added NikoNyrh timeline score: 0
Dec 24, 2016 at 0:32 answer added AvahW timeline score: 0
Dec 23, 2016 at 4:31 answer added rahnema1 timeline score: 1
Dec 23, 2016 at 1:10 answer added Sergei Patiakin timeline score: 0
Dec 22, 2016 at 19:28 answer added Martin Ender timeline score: 1
Dec 22, 2016 at 18:47 answer added Billywob timeline score: 1
Dec 22, 2016 at 17:54 comment added FlipTack Here are the average number of severs for random list sizes from 1-20. Each one was run 10000 times, with the total severs counted, and then an average calculated. Note that the "random lists" were filled with uniform random elements 1-100.
Dec 22, 2016 at 16:35 comment added Nathan Merrill It's definitely O(n^2). Rotating a sorted array by 1 causes causes the O(n^2) lower bound ([2,3,4,5,6,1]). However the upper bound is also O(n^2) (assuming a good implementation), as it is strictly better than bubble sort (as it swaps two numbers + possibly more)
Dec 22, 2016 at 14:27 comment added ETHproductions Reversing is O(n), but reversing can be built right into the algorithm (that's what my JS answer does); since each iteration loops over each item in the array once, a single iteration is O(n). (I think...)
Dec 22, 2016 at 13:23 answer added nimi timeline score: 1
Dec 22, 2016 at 11:58 answer added Noodle9 timeline score: 0
Dec 22, 2016 at 11:27 comment added Jasen @WheatWizard reversing an array doesn't require room for a copy of the array, only room for a single element. and is O(n). swap first and last elements then swap second and second last elements etc, when you get to the middle stop.
Dec 22, 2016 at 8:59 answer added Adám timeline score: 2
Dec 22, 2016 at 8:17 answer added Fatalize timeline score: 3
Dec 22, 2016 at 3:26 answer added Conor O'Brien timeline score: 0
Dec 22, 2016 at 3:06 answer added smls timeline score: 0
Dec 22, 2016 at 1:23 answer added manonthemat timeline score: 4
Dec 22, 2016 at 0:18 history tweeted twitter.com/StackCodeGolf/status/811727485960327168
Dec 22, 2016 at 0:04 answer added lynn timeline score: 11
Dec 22, 2016 at 0:02 answer added Dennis timeline score: 15
Dec 21, 2016 at 23:59 answer added LegionMammal978 timeline score: 7
Dec 21, 2016 at 23:33 answer added Maltysen timeline score: 0
Dec 21, 2016 at 23:16 answer added Arnauld timeline score: 11
Dec 21, 2016 at 23:03 answer added mbomb007 timeline score: 4
Dec 21, 2016 at 22:59 answer added ETHproductions timeline score: 19
Dec 21, 2016 at 22:45 comment added DJMcMayhem That sounds right to me, however it's worth pointing out that reversing an array isn't a very efficient operation, so it's a slow O(n^2)
Dec 21, 2016 at 22:40 answer added FlipTack timeline score: 5
Dec 21, 2016 at 22:39 answer added user62802 timeline score: 4
Dec 21, 2016 at 22:23 comment added ETHproductions @mbomb007 I don't understand big-o notation very well, but I think a single iteration is O(n). Multiply that by worst-case n iterations and you get O(n^2) (worst-case; best-case would be O(n), I think, for a single iteration).
Dec 21, 2016 at 22:19 answer added Blue timeline score: 4
Dec 21, 2016 at 22:08 comment added mbomb007 What's the big-o of this sorting method?
Dec 21, 2016 at 22:07 answer added Emigna timeline score: 5
Dec 21, 2016 at 22:03 answer added Luis Mendo timeline score: 8
Dec 21, 2016 at 21:46 history asked ETHproductions CC BY-SA 3.0