// your answers
type Combs<T extends any[]> = T extends [infer L extends string, ...infer R extends string[]]
? `${L} ${R[number]}` | Combs<R>
: never
Solution by pea-sys #33543
type Combs<T extends string[]> = T['length'] extends 0 | 1
? never
: T extends [infer First, ...infer Rest extends string[]]
? `${First & string} ${Rest[number]}` | Combs<Rest>
: never;
Solution by sunupupup #33479
type Combs<T extends any[], Ret = never> =
T extends [infer F extends string, ...infer R extends string[]]
? Combs<R, Ret | `${F} ${R[number]}`>
: Ret
Solution by bkmashiro #33283
// your answers
type Combs<T extends string[]> = T extends [infer Item extends string, ...infer Rest extends string[]]
? `${Item} ${Rest[number]}` | Combs<Rest>
: never
Solution by DevilTea #33258
type Combs<T extends string[]> =
T extends [infer First extends string, ...infer Rest extends string[]]
? `${First} ${Rest[number]}` | Combs<Rest>
: never
Solution by teamchong #33063
type GetUnion<T extends string[], Res extends string = T[0]> = T extends [infer H, ...infer R]
? Res extends H
? GetUnion<string[] & R, Res>
: `${string & Res} ${string & H}` | GetUnion<string[] & R, Res>
: never
// 实现 Combs
type Combs<T extends any[]> = T extends [string, ...infer R]
? GetUnion<T> | Combs<R>
: never
Solution by keyurparalkar #32793
// 你的答案
type Combs<T extends string[]> = T extends [infer S extends string, ...infer O extends string[]] ? `${S} ${O[number]}` | Combs<O> : never;
Solution by tarotlwei #31759
// your answers
// 全部列挙するならこれでいいが、重複が入るので今回は NG
// This is fine for listing everything, but it's not suitable for this case because it includes duplicates.
type All<T extends string[], S extends string = T[number]> = S extends S ? `${S} ${Exclude<T[number], S>}` : never
// the answer for this question
type Combs<T extends string[]> = T extends [infer First extends string, ...infer Rest extends string[]] ? `${First} ${Rest[number]}` | Combs<Rest> : never
Solution by Kakeru-Miyazaki #30921
// 你的答案
type Combs<T extends string[]> =
T extends [infer C extends string, ...infer M extends string[]]
? `${C} ${M[number]}` | Combs<M>
: never
Solution by milletlovemouse #30919
type Combs<T extends string[], R extends unknown[] = []> = T extends [
infer F extends string,
...infer Rest extends string[],
]
? Combs<Rest, [...R, `${F} ${Rest[number]}`]>
: R[number];
Solution by leejaehyup #30876
type Combs<T extends any[]> = T extends [infer A extends string, ...infer B extends any[]] ? `${A} ${B[number]}` | Combs<B> : never
Solution by dreamluo-plus #30695
// your answers
type Combs<T extends string[], RES = ''> = T extends [
infer HEAD extends string,
...infer TAIL extends string[],
]
? RES extends ''
? Combs<TAIL, `${HEAD} ${TAIL[number]}`>
: Combs<TAIL, RES | `${HEAD} ${TAIL[number]}`>
: RES
Solution by playitsafe #30344
type Combs<T extends any[]> = T extends [infer F extends string, ...infer R extends string[]] ? `${F} ${R[number]}` | Combs<R> : never
Solution by sv-98-maxin #30191
type Unshift<T extends any[]> = T extends [infer _,...infer R] ? R: []
type Combs<T extends any[], U extends any[]= Unshift<T>> =
T extends [infer A extends string, ...infer E]
? E extends [infer B extends string, ...infer R]
? `${A} ${B}` | Combs<[A, ...R], U> : Combs<U, Unshift<U>>
: never
Solution by sv-98-maxin #30190
type Shift<T extends any[]> = T extends [any, ...infer Rest] ? Rest : never;
type Combs<
T extends any[],
All extends any[] = Shift<Shift<T>>,
R extends any = `${T[0]} ${T[1]}`
> = T[0] extends undefined
? R
: All[0] extends undefined
? Combs<Shift<T>, Shift<Shift<T>>, R>
: Combs<T, Shift<All>, R | `${T[0]} ${All[0]}`>;
Solution by idebbarh #30012
// your answers
type Combs<T extends any[]> = T extends [infer First extends string, ...infer Rest extends any[] ] ? `${First} ${Rest[number]}` | Combs<Rest> : never
Solution by kerolossamir165 #29775
type Combs<T extends any[]> = T extends [
infer Item extends string,
...infer Rest extends string[]
]
? `${Item} ${Rest[number]}` | Combs<Rest>
: never;
Solution by DoubleWoodLin #28896
type Combs<StringArray extends string[]> = StringArray extends [
infer Head extends string,
...infer Tail extends string[]
]
? `${Head} ${Tail[number]}` | Combs<Tail>
: never;
Solution by yevhenpavliuk #28500
这题大概意思就是把数组的每一项和后面的一一组合起来，那么我们可以遍历数组的每一项，利用数组转联合类型，得到结果
type Combs<T extends any[]> = T extends [infer F extends string, ...infer Rest extends string[]]
? `${F} ${Rest[number]}` | Combs<Rest>
: never
Solution by linjunc #28360
type Combs<
T extends string[],
Result extends string = never,
Key1 extends string = never
> = T extends [infer F extends string, ...infer Rest extends string[]]
? [Key1] extends [never]
? Combs<Rest, Result, never> | Combs<Rest, Result, F>
: Combs<Rest, Result | `${Key1} ${F}`, never> | Combs<Rest, Result, Key1>
: Result;
Solution by alythobani #27621
// It’s recommended to use 'tail recursion'
type Combination<T extends string[], R extends string = never> =
T extends [infer A extends string, ...infer Rest extends string[]]
? Combination<Rest, R | `${A} ${Rest[number]}`>
: R
type ModifierKeys = ['cmd', 'ctrl', 'opt', 'fn']
type Combs = Combination<ModifierKeys>
Solution by drylint #27401
type ModifierKeys = ['cmd', 'ctrl', 'opt', 'fn']
type Combs<T extends string[] = ModifierKeys> =
T extends [infer F extends string, ...infer R extends string[]] ? (`${F} ${R[number]}` | Combs<R>) : never
Solution by smileboyi #27275
type Combs<T extends string[], R = never> = T extends [infer F extends string, ...infer Rest extends string[]]
? Combs<Rest, R | `${F} ${Rest[number]}`>
: R
Solution by XkSuperCool #26096
type CombItemAndArray<T extends string, Arr extends string[], Res = never> =
Arr extends [infer X extends string, ...infer Y extends string[]]
? CombItemAndArray<T, Y, Res | `${T} ${X}`>
: Res
type Combs<Arr extends string[] = ModifierKeys, Res = never> =
Arr extends [infer X extends string, ...infer Y extends string[]]
? Combs<Y, Res | CombItemAndArray<X, Y>>
: Res
Solution by kiki-zjq #25848
// your answers
type Combs<T extends string[] = []> = T extends [infer Left extends string, ...infer Rest extends string[]] ? `${Left} ${Rest[number]}` | Combs<Rest> : never
Solution by studymachiney #24946
type Combs<
T extends string[]
> =
T extends [infer F extends string,...infer R extends string[] ]?
`${F} ${R[number]}` | Combs<R>
:never
Solution by jiangshanmeta #24838
type Combs<T extends string[] = ModifierKeys> = T extends [infer F extends string, ...infer R extends string[]] ? `${F} ${R[number]}` | Combs<R> : never;
Solution by E-uler #24564
type ModifierKeys = ['cmd', 'ctrl', 'opt', 'fn']
// 实现 Combs
type Helper<T> = T extends [infer F extends string, ...infer R extends string[]]
? `${F} ${R[number]}` | Helper<R>
: never
type Combs = Helper<ModifierKeys>
Solution by Sun79 #24496