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Looking at the big O of functions, If f1(N)=O(NlogN) and f2(N)=O(log N), then what is "big...

Looking at the big O of functions,

If f1(N)=O(NlogN) and f2(N)=O(log N), then what is "big O" of f1 +f2?

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Answer #1

From the question:

f1(N)=O(NlogN)

Here the function grows logarithemically as well as sequentially.

f2(N)=O(log N)

Here the function grows only logarithemically.

f1+f2 is the sun of O(NlogN) + O(log N).

In the above O(NlogN) > O(logN)

So we can justify that f1 has highest complexity.

In Big O addition calculation highest complexity is the overall complexity.

The overall complexity is O(NlogN).

Therefore f1 +f2 = O(NlogN).

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