[Typescript] 123. Hard - Intersection

Implement the type version of Lodash.intersection with a little difference. Intersection takes an Array T containing several arrays or any type element including the union type, and returns a new union containing all intersection elements.

type Res = Intersection<[[1, 2], [2, 3], [2, 2]]>; // expected to be 2
type Res1 = Intersection<[[1, 2, 3], [2, 3, 4], [2, 2, 3]]>; // expected to be 2 | 3
type Res2 = Intersection<[[1, 2], [3, 4], [5, 6]]>; // expected to be never
type Res3 = Intersection<[[1, 2, 3], [2, 3, 4], 3]>; // expected to be 3
type Res4 = Intersection<[[1, 2, 3], 2 | 3 | 4, 2 | 3]>; // expected to be 2 | 3
type Res5 = Intersection<[[1, 2, 3], 2, 3]>; // expected to be never

 

/* _____________ Your Code Here _____________ */
type ToUnion<T> = T extends any[] ? T[number] : T
type Intersection<T> = T extends [infer F, ...infer RT]
  ? ToUnion<F> & Intersection<RT>
  : unknown

/* _____________ Test Cases _____________ */
import type { Equal, Expect } from '@type-challenges/utils'

type cases = [
  Expect<Equal<Intersection<[[1, 2], [2, 3], [2, 2]]>, 2>>,
  Expect<Equal<Intersection<[[1, 2, 3], [2, 3, 4], [2, 2, 3]]>, 2 | 3>>,
  Expect<Equal<Intersection<[[1, 2], [3, 4], [5, 6]]>, never>>,
  Expect<Equal<Intersection<[[1, 2, 3], [2, 3, 4], 3]>, 3>>,
  Expect<Equal<Intersection<[[1, 2, 3], 2 | 3 | 4, 2 | 3]>, 2 | 3>>,
  Expect<Equal<Intersection<[[1, 2, 3], 2, 3]>, never>>,
]