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Namaste!!!!! if n<100,   O(n^2) runs faster then O(nlogn) if n>=100  O(nlogn) is better, kripaya garera yeso explain garidinu hola  thanks in advance

nepali8 · Sep 11, 2012 7:05 PM · 13,850 views

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 Did you google it? Is this what you are looking for  etutorials.org/Programming/Java+performance+tuning/Chapter+9.+Sorting/9.3+Better+Than+Onlogn+Sorting/  I tried to help.

at_its_best · Sep 11, 2012 9:30 PM

The link posted above should help you. If that still doesn't help, post back and will try to clarify.

Kiddo · Sep 12, 2012 10:43 AM

if n<100, O(n^2) runs faster then O(nlogn) - should be for some value of k. Hope it helps.

bairagiKancho · Sep 12, 2012 11:51 AM

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