no-uninferred-type-parameter: check type parameters with constraint

[ajafff]

TypeScript uses the constraint of a type parameter if it cannot be inferred and has no default.
no-uninferred-type-parameter currently only checks for the implicit constraint ({} or unknown as of ts3.5).

It should be able to tell if a type parameter with constraint cannot be inferred. That will also fix false-positives where the inferred type equals the implicit constraint.
This is a bit more involved:

• type parameter has no default
• inferred type equals constraint
• there's nothing the type parameter could be inferred from
  • optional parameter not provided
  • optional property not present or no property matching index signature
  • there might be some edge cases involving nested mapped types: https://typescript-play.js.org/#code/CYUwxgNghgTiAEAzArgOzAFwJYHtVNQB4AVAPgAooAueAJXBxmEIHIoX4AfeFgIxYA0dBk0IBnDDCyoA5kLKkAlDQBuOLMADcAKG2JU5AN7V4hsDQCMAXyG8ah4JatXFmoA