00296-medium-permutation

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type Permutation<T, K = T> = [T] extends [never] ? [] 
  : K extends K ? [K, ...Permutation<Exclude<T, K>>] : never

Solution by eunsukimme #35205 

type Permutation<T, K=T> =
    [T] extends [never]
      ? []
      : K extends K
        ? [K, ...Permutation<Exclude<T, K>>]
        : never

Solution by devshinthant #34579 

type Permutation<T, K=T> = [T] extends [never] ? [] : K extends infer R ? [R, ...Permutation<Exclude<T, R>>] : never

Solution by ouzexi #34006 

// Algorithm
function permute(arr: number[]): number[][] {
  var result : number[][] = []
  function dfs(a: number[]) {
    if (a.length === arr.length) {
      result.push(a)
    } else {
      for (const n of arr) {
        if (a.includes(n)) {
          continue
        }
        dfs([...a, n])
      }
    }
  }
  dfs([])
  return result
}

// Type
type Permutation<T, K=T> =
[T] extends [never]
  ? [] // 空元素空排列，递归出口
  : K extends infer U // 遍历T
    ? [U, ...Permutation<Exclude<T, U>>] // 递归得到除了U以外元素的排列
    : never // never happened

// Permutation<A | B | C>
// =>
// [A, ...Permutation<B | C>]
// [A, ...[B, ...Permutation<C>]] [A, ...[C, ...Permutation<B>]]
// [A, ...[B, ...[C, []]]]] [A, ...[C, ...[B, []]]]]
// [A, B, C] [A, C, B]

// [B, ...Permutation<A | C>] 
// [B, ...[A, ...Permutation<C>]] [B, ...[C, ...Permutation<A>]]
// [B, ...[A, ...[C, []]]]] [B, ...[C, ...[A, []]]]]
// [B, A, C] [B, C, A]

// [C, ...Permutation<A | B>]
// [C, ...[A, ...Permutation<B>]] [C, ...[B, ...Permutation<A>]]
// [C, ...[A, ...[B, []]]]] [C, ...[B, ...[A, []]]]]
// [C, A, B] [C, B, A]

Solution by ScriptBloom #33909

type Permutation<T,U=T> = [T] extends [never] ? [] : T extends any ? [T,...Permutation<Exclude<U,T>>] : [];

Solution by chihiro365yb #33857

Unlike the most upvoted answer, I don't want to use K extends K in the second loop (because it feels somewhat counterintuitive). For the permutation solution, it's natural to iterate over T itself. Therefore, I present the following solution (which has the same effect as K extends K).

----- Above is the translation from Chinese by ChatGPT XD -----

和票数最多的答案不同，二层循环我不想使用 K extends K（因为它有点反直觉），对于 permutation 解法，我们很自然地想到对 T 本身做遍历，因此我给出如下的解法（实际效果和 K extends K 相同）

// your answers
type Permutation<T extends string | number | symbol, A extends any[] = []> = [
  T
] extends [never]
  ? A
  : { [key in T]: Permutation<Exclude<T, key>, [...A, key]> }[T];

Solution by pfan8 #33831 

type Permutation<T,K = T> = T[] extends never[]?[] : K extends K ? [K, ...Permutation<Exclude<T,K>>] : never

Solution by bananana0118 #32784 

// It's copy+paste, but I understand the logic after wasting 30 minutes of my life.
type Permutation<T, K = T> =
    [T] extends [never]
      ? []
      : K extends K
        ? [K, ...Permutation<Exclude<T, K>>]
        : never

type perm = Permutation<'A' | 'B' | 'C'>;
// ['A', 'B', 'C'] | ['A', 'C', 'B'] | ['B', 'A', 'C']
// | ['B', 'C', 'A'] | ['C', 'A', 'B'] | ['C', 'B', 'A']

Solution by ZhulinskiiDanil #32683 

type Permutation<T, K = T> = 
    [T] extends [never]
      ? [] 
      : K extends unknown
        ? [K, ...Permutation<Exclude<T, K>>] 
        : never
  1. 유니언 타입이 배열의 원소로 들어가면 각각의 유니언 타입에 맞게 여러 유니언 타입이 만들어지는 것을 배웠다.
  2. Exclude<T, K> 부분에서 K가 유니언에서 한가지로 정해진 구체적인 타입으로서 활용 가능하다는 것을 배웠다.

Solution by dev-hobin #32401 

type Permutation<T, U = T> = [T] extends [never] ? [] : T extends any ? [T, ...Permutation<Exclude<U, T>>] : [];

Solution by kai-phan #31639 

// your answers
type Permutation<T, K = T> = [T] extends [never]
  ? []
  : K extends K
  ? [K, ...Permutation<Exclude<T, K>>]
  : never;

Solution by AhmedRagabKamal #31511 

type Permutation<T, U = T> = [ T ] extends [ never ]
  ? []
  : (
    U extends T
      ? [ U, ...Permutation<Exclude<T, U>> ]
      : []
  )

Solution by jbalancer #31173 

// 你的答案
type Permutation<T, K = T> =
  [T] extends [never]
    ? []
    : T extends T
      ? [T, ...Permutation<Exclude<K, T>>]
      : never;

Solution by d1zzzzy #31145 

type Permutation<T, K = T> = [T] extends [never] ? [] : T extends any ? [T, ...Permutation<Exclude<K, T>>] : []

Solution by kai-phan #30348

递归➕剔除

type Permutation<T, U = T> = [T] extends [never]
  ? []
  : (T extends U
    ? [T, ...Permutation<Exclude<U, T>>]
    : [])

Solution by 1587315093 #29930 

type Possible = string | number | bigint | boolean | null | undefined

type Permutation<T, K extends Array<unknown> = []> = {
  [P in T as P extends Possible ? `${P}` : never]: Exclude<T, P> extends never ? [...K, P] : Permutation<Exclude<T, P>, [...K, P]>
} extends Record<any, infer U>
  ? U extends {}
    ? U
    : []
  : never

Solution by agus-wesly #29789 

type Permutation<T, K=T> = 
    [T] extends [never] 
      ? []
      : K extends K
        ? [K, ...Permutation<Exclude<T, K>>] 
        : []

Solution by MohammadArasteh #29409 

type Permutation<T, K = T> = [T] extends [never] ? [] : T extends K ?  [T, ...Permutation<Exclude<K, T>>] : never

Notes:

  1. When comparing a union type to never in generic types, the result of T extends never (when T is never) would always be never because there is no member in the union can be distrubted over. Hence, use straight brackets on both sides of extends to escape the distributive behavior.
  2. The second param K in Permutation is to reserve the original union type so we can use Exclude on it against the distrubuted member T in Exclude<K, T>
  3. An example of how Permutation works:
// example
Permutation<'A' | 'B' | 'C'>

// step 1
['A', ...Permutation<'B' | 'C'>] | 
['B', ...Permutation<'A' | 'C'>] | 
['C', ...Permutation<'A' | 'B'>]

// step 2, in a recursive call, computing the second item in the above arraies
['B', ...Permutation<'C'>] | 
['C', ...Permutation<'B'>] |
// ...omiting the rest for ['B', ...Permutation<'A' | 'C'>] | ['C', ...Permutation<'A' | 'B'>]

// step 3
['C' | ...Permutation<never>] | 
['B' | ...Permutation<never>] |
// ...omiting the rest 

// step 4
['A', 'B', 'C'] |
['A', 'C', 'B'] |
// ...omiting the rest

Solution by qianzhong516 #28941 

type Permutation<T, U = T> = [T] extends [never] ? [] : T extends U ? [T, ...Permutation<Exclude<U, T>>] : [];

Solution by kai-phan #28838 

type Permutation<T, C = T> = [T] extends [never] ? [] : T extends unknown ? [T, ...Permutation<Exclude<C, T>>] : []

Solution by hajeonghun #28825

type Permutation<T, Copy = T> = [T] extends [never] ? [] : T extends T ? [T, ...Permutation<Exclude<Copy, T>>] : [];

Solution by DoubleWoodLin #28614 

    type Permutation<T, K = T> = [T] extends [never]
        ? []
        : K extends K
        ? [K, ...Permutation<Exclude<T, K>>]
        : never;

Solution by de-novo #28556 

type Permutation<T, U = T> = [U] extends [never]
  ? []
  : T extends U
  ? [T, ...Permutation<Exclude<U, T>>]
  : [];

Solution by flt3150sk #28540 

type Permutation<Union, Item=Union> =
  [Union] extends [never] ? [] :
    Item extends Item ?
    [Item, ...Permutation<Exclude<Union, Item>>] :
  []

thanks to help 🙏

Solution by maximallain #28343 

type IsBooleanExact<T> = boolean extends T ? true : false;
type IsNever<T> = [T] extends [never] ? true : false;

type Permutation<T, K = T> = 
  IsBooleanExact<T> extends true 
    ? [false, true] | [true, false] 
    : IsNever<T> extends true 
      ? [] 
      : T extends unknown
        ? [T, ...Permutation<Exclude<K, T>>]
        : never;

Solution by viktor-sakhno-saritasa #27671 

type Permutation<T, K = T> = [T] extends [never] ? [] : K extends K ? [K, ...Permutation<Exclude<T, K>>] : never;

Solution by jjswifty #27473

This one was a really difficult one!

The thing is, it is conceptually quite easy. But the syntax of the conditional types in Typescript is what made it quite challenging for me. Let me explain:

type DistributiveConditionalType<UnionType> = UnionType extends any ? UnionType : never

What you are looking at above is a generic type that takes a union type (e.g. '1' | '2' | '3') and returns the same union type. But how does it do it?

In the answer of the conditional type (following the question mark), UnionType DOES NO LONGER REFER to the union type as a whole but rather only one of its constituent literal types. Namely, '1', '2' or '3'. So, what happens if we want, as part of our conditional typing, to use the union type itself?

This is exactly the challenge here in my opinion. And as much as I did try to use a pseudo generic type for it equating it to the union type. I did not manage to do it in the elegant way shown in other solutions.

My solution was more lengthy which I would like to share below.

type Exclude<U, L> = U extends L ? never : U // Exclude a literal type from a union type

// This is an auxiliary generic type I introduced to solve the problem
type PermutatorCore<U, U2, T extends any[]> = // U2 here is passed the same type as U
  U extends any ? // Now U becomes distributive, U2 still refers to the union type
    Permutation<Exclude<U2, U>, [...T, U]> : never

type Permutation<U, T extends any[] = []> = 
  [U] extends [never] ? [...T] : 
    PermutatorCore<U, U, [...T]> // Utilise the generic type above to pass U twice as U and U2

Solution by diraneyya #26995 

/* _____________ Your Code Here _____________ */

type Permutation<U, C = U> = [U] extends [never] ? [] : U extends C ? [U, ...Permutation<Exclude<C, U>>] : never;

Solution by phinixsdsd #26894 


type GA<S, T = S> = S extends any ? ([T] extends [infer R] ? [S, ...(GA<Exclude<R, S>> extends never ? [] : GA<Exclude<R, S>>)] : []) : []

type Permutation<T> = GA<T> extends never ? [] : GA<T>

Solution by Acoooooooer #26783 

type Permutation<T, K = T> = [T] extends [never] ? [] : K extends K ? [K, ...Permutation<Exclude<T, K>>] : never

Solution by smileboyi #26746