So you implemented bubble sort, well done

So you implemented bubble sort, well done.
Now for some theoretical computer science:

Is 3^n-O(2^n)=O(3^n)?
Is 3^n-O(2^n)=\Omega(3^n)?

Sorry, but these questions are a waste of my time. Can I please speak to one of the developers on the team I'll be working with? I have several opportunities I am perusing at the moment so I'd like to see if this workplace is a good fit for me.

List arr = new ArrayList();
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I forgot what goes in the middle, took this class years ago and dont write my own sort functions

well said hehehe

Thanks for coming, we'll call you.

Sorry, we are actually trying to diversify out development team. Thanks for the 5 hour interview.

>So you implemented bubble sort, well done.
I didn't, I ripped it off someone else's code by running a github search.
>Now for some theoretical computer science:
>Is 3^n-O(2^n)=O(3^n)?
>Is 3^n-O(2^n)=\Omega(3^n)?
yes and no.

Thanks for the free plane tickets. I really enjoyed Disneyworld

man i just want to code

We have an open position in our department in India if you're interested, we'll let you know.

>I really enjoyed Disneyworld
>Disney
>Disney giving you a free pass to DW
They are a money hungry, lawsuit company and have the luxury of being able to raise prices every year without backlash.

You can reapply in 6 months

>Simply saying yes and no is a "waste of time"

Sorry but that's wrong. We thank you for your interest in our company and wish you good luck in your job search

Sup Forums is so afraid of things it never did, job interviews, women, operating systems.

Joke's on you, I'm the owner, you're all fired for failing to hire me.

>Omega
What kind of Mickey Mouse company is this?

f(c)=Omega(g(x)) iff g(x)=O(f(x))

>iff
you're fired too, you can't code if you have dyslexia

>Is 3^n-O(2^n)=O(3^n)?
Yes.

>Is 3^n-O(2^n)=\Omega(3^n)?
No.

This is not a misspelling. "Iff" is shorthand for "if and only if", and is standard among mathematicians.

confirmed for retard

A simple way to think of Omega is like a lower bound. Obviously 3^n - O(2^n ) < 3^n. So its not at least 3^n, it could be less, meaning the answer to the second question is "no".

lel the answer to both questions is Yes

>Is 3^n-O(2^n)=O(3^n)?
Yes
3^n-O(2^n)=\Omega(3^n) if and only if 3^n=O(3^n-O(2^n)), which is true.

Just as n^3=O(n^3-n^2)

You plebs don't know complexity theory, go back to your gaymen

>Is 3^n-O(2^n)=O(3^n)?
Yes.
>Is 3^n-O(2^n)=\Omega(3^n)?
Yes.

>go into an interview recently
>ready for any technical question they throw at me
>midway through the interview
>"so tell us about yourself"
>completely freeze up

They didn't call back

Your question is malformed. Big O and Omega represent sets, so 3^n - O(2^n) isn't well-defined.

Did you mean "Is 3^n - 2^n in O(3^n)"?

It can represent a set or it can represent a variable, it really doesn't matter what it represents

Not for triggered mathematicians