type-challenges-solutions/en/medium-flatten about type-challenges-solutions HOT 4 OPEN

There are two issues with this solution:

Given the following test case...

// @ts-expect-error
type error = Flatten<undefined>

• The T extends [] check is redundant. T is an empty array iff. T is an array and T extends [infer H, ...infer R] is false.

type Flatten<T extends unknown[]> = 
  T extends [infer H, ...infer Rest]
    ? H extends unknown[]
      ? [...Flatten<H>, ...Flatten<Rest>]
      : [H, ...Flatten<Rest>]
    : []

• The solution uses T ambiguously -- for both the input array type and the inferred "tail" array type. Here's a slightly different formulation that relies on those two types being defined independently.

type Flatten<T extends any[]> = // T[0] cannot extend unknown
  T extends [infer H, ...infer Rest]
    ? H extends unknown[]
      ? [...Flatten<T[0]>, ...Flatten<Rest>]
      : [T[0], ...Flatten<Rest>]
    : []

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My solution is similar with @MajorLift one, except I'm using Flattern 1 instantiation less :P

type Flatten<T extends unknown[]> = T extends [infer Head, ...infer Tail]
  ? Head extends unknown[]
    ? Flatten<[...Head, ...Tail]>
    : [Head, ...Flatten<Tail>]
  : [];

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@albert-luta Nice! It seems that your solution would result in less optimal recursion depth, as the Flatten call will attempt to process the entire input array (with Head one level flattened) on a single stack. Depending on how deeply nested the elements of T are, this could result in a significant performance hit.

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I believe this initial example could be improved

type Flatten<T> = T extends []
  ? []
  : T extends [infer H, ...infer T]
  ? [H, T]
  : [T];

• : T extends [infer H, ...infer T] -> : T extends [infer H, ...infer K]
T could be replaced with K. Otherwise T after extends is "shadowing" the original T
• ? [H, T] -> ? [H, ...K]
  ** T could be replaced with K as explained in the previous point
  ** There should be ... before K. Otherwise K is a tuple

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