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

 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.

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

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

Hope it helps.