Is O(N) better or O(NLogN) better?

#1

I am getting confused in this, if N is too large then the combined value of NLogN will be less than O(N) ?

#2

@raghav6 NlogN > N so O(N) is no doubt better than O(NlogN)

#3

But suppose our n=4 let’s consider it worst case, so log4=0.60 and nlogn = 4*0.60= 2.4 which is less than n=4 itself?

#4

@raghav6 You are considering the base 10. Consider it as 2 for these case.
1

#5

So, we consider base2 for example binary search , our array gets reduced to half.

#6

Yes for Binary search the time complexity in Log(n) not nlog(n). So it will be less than O(n).
But N*Log(N) is greater than O(N).

#7

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