algorithm - Sajha Mobile
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algorithm
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nepali8
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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
at_its_best
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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.
Kiddo
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The link posted above should help you. If that still doesn't help, post back and will try to clarify.
bairagiKancho
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if n<100, O(n^2) runs faster then O(nlogn) - should be for some value of k.

Hope it helps.

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