If an algorithm has running time of $\Theta(n^2)$, is it possible to have a best-case running time of $\Omega(n)$? Or is the fastest running time only $c n^2$ for some constant factor $c$?
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There are two possible interpretations for "running time of an algorithm". The first interpretation is the worst-case running time, depending on some parameters such as the input length $n$. The second interpretation is that the running time is some bound that holds for all inputs with the given parameters. When we give a big $O$ estimate, both interpretations amount to the same thing: if an algorithm has a running time of $O(n^2)$, then this is both a worst-case bound and a bound which is valid for all inputs of length $n$.
What is then the interpretation of $\Theta(n^2)$ in your case? It's hard to say without more details, and could depend on the context. If it's a worst-case bound then the meaning is "there is a family of inputs for which the algorithm runs in $\Omega(n^2)$ time". If it's a global bound then the meaning is "the algorithm runs in $\Omega(n^2)$ time on all inputs". Out of these two possible interpretations, the first seems slightly more likely, but others may hold different opinions.
David Richerby
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Yuval Filmus
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Normally, it's the worst-case running time that is given but one should never say just "running time" since it could be anything – best-case, worst-case, average case according to some probability distribution, amortized, ...
If the worst-case running time is $\Theta(n^2)$ all you can say about the best case is that it can't be worse than the worst case, so the best case is $O(n^2)$. So, yes, the best case could be $\Omega(n)$.
David Richerby
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