$Question:$
$Approach:$
A is reduced to B . Here reduction is done at polynomial time.
Here B is solved in polynomial time. So A should also be in polynomial time.
Now A can not be harder than B, So I think A can be $O(n^3)$ or $O(n^2)$. But logically if I reduced A to B and if B is $O(n^3)$ then it makes no sense for A to be $O(n^2)$, else why would I reduce it to higher complexity? So A is $O(n^3)$.
But my doubt is, we say while reduction that A can not be harder than B then how can we decide whether it is $O(n^2)$ or $O(n^3)$.
Or is this argument only valid for P, NP classes of problem when I say A can not be harder than B or B must be at least as hard as A?
Answer given is "A is $O(n^3)$", but why can't it be $O(n^2)$ as "A can not be harder than B" but can be of equal complexity? Or is reducibility just an argument for complexity classes?
Explain if possible how reduction actually works, and why I couldn't apply it to my problem.
