Is there a minimal extension of $L$ that is not a forcing extension?

Question

It's well known that Sacks forcing constructs a real of minimal constructability degree, i.e. a real $x$ such that for any $y\in L(x) \setminus L$, $L(y) = L(x)$. It's also well known that certain objects, such as $0^\sharp$, can never be created by a forcing extension.

Given these two facts there is a natural question:

Can there exist a real $x\notin L$ such that any inner model $W$ with $L(x) \supseteq W \supseteq L$ either $W=L$ or $W= L(x)$ but such that $L(x)$ is not a (set) forcing extension of $L$?

Obviously we could also ask the question regarding larger sets than reals or class forcing extensions.

Answer 1

Yes, this is possible and follows from Sy Friedman's paper Minimal coding (you may better look at Fine structure and class forcing).

It follows from the results of the above paper that there is an $L$-definable class forcing for producing a real $R$ which is minimal over $L$ but is not set-generic over $L$.