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I do not know how to get exact Asymptotic Complexity as following question asks me to solve.

Here is Question:

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Everything I learned from class is that, if there exists "for loop" that runs N times, or recursively call "recursive method" N times, then either one is O(N). This is practically all the base knowledge I know of.

Question 1) What answer says about "N(N+1)/2", is this same as "O(N(N+1)/2)"?

Also, the answer says followng "On iteration i, both add and trimToSize require that i array elements be copied to a new array.", I think it says this because it calls both "list.add()" and "list.trimToSize()" N times.

From my perspective, this looks more like just O(N)... because since it adds ith element N times using "list.add()" and also array-copies element N times through "list.trimToSize()". Thus, N + N <= 2N so O(N) where witness C=2 and k=1, not O(N(N+1)/2). However, my answer is wrong. Therefore,

Question 2) I just do not know how to obtain N(N+1)/2 as answer says.

Thank you very much if you can answer!

online.0227
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From my perspective, this looks more like just O(N)... because since it adds ith element N times using "list.add()" and also array-copies element N times through "list.trimToSize()". Thus, N + N <= 2N so O(N) where witness C=2 and k=1, not O(N(N+1)/2).

However, my answer is wrong.

Yes. It is.

You are treating each call to trimToSize having a constant cost. That is not correct. If you look at what trimToSize does, you will that each call is copying all of the elements (so far) to a new array. That is NOT a constant cost

  • In the 1st call to trimToSize the array length is 1, so you copy 1 element.
  • In the 2nd call to trimToSize the array length is 2, so you copy 2 elements.
  • ...
  • In the Nth call to trimToSize the array length is N, so you copy N elements.

Add that all up and you get N(N+1)/2 for all of the trimToSize calls.

(And, in fact, given the way that the code is using the ArrayList, each time that add is called, the list's backing array will need to be reallocated. So the same calculation applies there too.)

Note that O(N(N+1)/2) is the same complexity class as O(N^2). You can prove this from first principles.)

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Stephen C
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  • ah now i got it, thank you very much! so O(N) from running "list.add()" N times, and also O(N(N+1)/2) running time from "list.trimToSize()" method. So in total by adding them, "N + N(N+1)/2". But you said O(N(N+1)/2) can be O(N^2).. so it can be replaced to "N + N^2". To calculate big oh from it, "N + N^2 < N^2 + N^2 = 2N^2" thus O(N^2) with witness of C=2 and K=1, which is same as O(N(N+1)/2). Hope I got it... Thank you. – online.0227 Oct 20 '16 at 02:35