bug: #[ts(optional)] does not work with Option type alias

[gustavo-shigueo]

This is a rewording of #379 focusing on Option not working properly if spelled any other way rather than making optional work with any type

Describe the bug
Whenever the Option type is spelled out as anything other than the word Option by itself, the #[ts(optional)] attribute fails to
recognize it as an Option, this is because of the extract_option_argument function, which expects the type to have only one
component, and expects that component to be the word Option with one generic argument.

This means spelling Option as core::option::Option (which is quite important for macros) or std::option::Option or creating some type
alias type Foo<T> = Option<T> and using the type Foo will cause compiler errors when using #[ts(optional)]

To Reproduce
Steps to reproduce the behavior:

1. Create the following type:

type Opt<T> = Option<T>;

#[derive(TS)]
struct Foo {
  a: Option<T>, // Works
  b: core::option::Option<i32>, // Error
  c: std::option::Option<i32>, // Error
  d: Opt<i32>, // Error
}

Version
10.1

There is already a fix for this issue in #366, but this PR is a breaking change and still needs approval

[gustavo-shigueo]

Turns out this is the last remnant of #70, as type aliasing hasn't been an issue since all the way back in #233