Type Challenges: 05310-medium-join Solution

type Join<T, U extends string | number = ',', Res extends string = ''> = T extends [infer F extends string, ...infer R extends string[]]
  ? Join<R, U, Res extends '' ? `${F}` : `${Res}${U}${F}`> : Res

Solution by tani-fumiya-ktc #35202

type Join<T, U extends string | number = ',', S extends string = ''> = T extends [infer F, ...infer Rest] ? Join<Rest, U, `${S}${S extends '' ? '' : U}${F & string}`> : S

type Join<T, U extends string | number = ',', S extends string = ''> = T extends [infer F extends string, ...infer Rest] ? Join<Rest, U, `${S}${S extends '' ? '' : U}${F}`> : S

Solution by wendao-liu #35077

// your answers
type Join<T, U = ','> =  T extends [infer P, ...infer L] ? `${P & string}${L['length'] extends 0 ? '' : U & (string | number)}${Join<L, U> & string}`: ''

Solution by gobielJonathan #34486

type Join<T extends any[], U extends number | string = ','> = T extends [] ? '' : 
T extends [infer R, ...infer rest] ? 
rest['length'] extends 0 ? R & string : 
`${R & string}${U}${Join<rest, U>}` : never

Solution by ouzexi #34093

Constrained type parameters
Tail call

Try it in TS Playground with additional test case

type Stringifiable = string | number | bigint | boolean | null | undefined;

type Join<
  T extends readonly Stringifiable[],
  U extends Stringifiable = ",",
> = T extends []
  ? ""
  : T extends [infer First extends Stringifiable]
  ? `${First}`
  : T extends [
    infer First extends Stringifiable,
    infer Second extends Stringifiable,
    ...infer Rest extends Stringifiable[],
  ]
  ? Join<[`${First}${U}${Second}`, ...Rest], U>
  : string;

Solution by PinkChampagne17 #33930

type Join<T, U extends string | number, V extends string = ""> = T extends []
	? V extends ""
		? ""
		: V
	: T extends [infer F, ...infer R]
	? R extends []
		? Join<R, U, `${V}${F & string}`>
		: Join<R, U, `${V}${F & string}${U}`>
	: never;

Solution by gasmg #32253

type Join<T extends unknown[], U extends string|number, R extends any = ''> = T extends [infer F, ...infer Rest] ? Rest extends [] ? 
Join<Rest, U, `${R}${F}`> : Join<Rest, U, `${R}${F}${U}`> : `${R}`

Solution by Zhen-code #32040

type Join<T extends any[], U extends string | number> = T extends [
  infer F extends string,
  ...infer R
]
  ? R["length"] extends 0
    ? `${F}`
    : `${F}${U}${Join<R, U>}`
  : "";

Solution by sunsunmonkey #31917

// your answers
type SupportedType = string | number | bigint | boolean | null | undefined
type Join<T extends SupportedType[], U extends SupportedType> = T extends [infer F extends SupportedType, ...infer Rest extends SupportedType[]] ? Rest extends [] ? `${F}` : `${F}${U}${Join<Rest, U>}` : ''

Solution by Kakeru-Miyazaki #30893

type Join<T extends any[], U extends string | number> =
  T extends [infer F, ...infer R]
    ? `${F & string}${R extends [] ? '' : U}${Join<R, U>}`
    : ''

Solution by matallui #30812

type Join<T extends unknown[], U extends string | number> = T extends [infer F extends string, ...infer R]
  ? `${F}${R extends [] ? '' : U}${Join<R, U>}`
  : ''

Solution by wangly19 #30756

type Join<T extends any[], U extends string | number> = T extends [infer A, ...infer B] ? `${A & string}${ Join<B, U> extends '' ? '' : U}${Join<B, U>}` : ''

Solution by dreamluo-plus #30663

type Join<T extends Array<string|number>, U extends string|number,F extends string = ''> = T extends [] ? '': T extends [infer S extends string,...infer O extends string[]] ? O extends [] ? `${F}${S}`: Join<O,U,`${F}${S}${U}`>:never;

Solution by jiangxd2016 #30553

type Join<T extends unknown[], U extends string | number> = T extends [infer F extends string, ...infer R]
  ? `${F}${R extends [] ? '' : U}${Join<R, U>}`
  : ''

Solution by iclaxton #30343

type Join<T extends unknown[], U extends string | number, S extends string = ''> = 
  T extends [infer A extends string, ...infer Rest]
    ? 
      Rest extends []
        ?
          `${S}${A}`
        : Join<Rest, U, `${S}${A}${U}`>
    : S

Solution by zhangqiangzgz #30201

type Join<T extends any[], U extends string | number> = T["length"] extends 1
  ? `${T[0]}`
  : T extends [infer S extends string, ...infer Tail]
  ? Tail extends string[]
    ? Join<Tail, U> extends string
      ? `${S}${U}${Join<Tail, U>}`
      : never
    : `${S}`
  : T

Solution by Kying-star #29042

type Join<T, U extends number | string> = T extends [infer F, ...infer R]
  ? R extends []
    ? `${F & string}`
    : `${F & string}${U}${Join<R, U>}`
  : "";

Solution by DoubleWoodLin #28780

// your answers

type Join<T, U extends string | number> = 
  T extends [infer S extends string, ...infer Tail] ? 
    Tail["length"] extends 0 ? `${S}`
      :`${S}${U}${Join<Tail, U>}` 
    :never

Solution by kerolossamir165 #28467

// 定义一个类型"变量"R存储拼接后的字符
type Join<T, U extends string | number, R extends string = ""> = T extends [
  // 取出第一个字符
  infer F,
  ...infer Rest
]
  // 递归拼接
  ? Join<Rest, U, `${R}${R extends "" ? "" : U}${F & string}`>
  : R;

Solution by jiaowoxiaobala #27984

type Join<T, U extends string | number, R extends string = ""> = T extends [ infer F, ...infer Rest ] ? Join<Rest, U, ${R}${R extends "" ? "" : U}${F & string}> : R;

Solution by jiaowoxiaobala #27869

type Join<T extends unknown[], S extends string, Result extends string = ''> = T['length'] extends 1 ? ${Result}${T[0] & string} : T extends [infer Pre, ...infer Next] ? Join<Next, S, ${Result}${Pre & string}${S}> : Result

Solution by jiechliu #27804

// your answers
type Join<T extends unknown[], K extends string | number> = T extends [infer First extends string, ...infer Rest] ? Rest['length'] extends 0 ? `${First}${Join<Rest, K>}` : `${First}${K}${Join<Rest, K>}` : '';

Solution by Hencky #27638


type Join<
  T extends string[],
  U extends string | number,
  Result extends string = ""
> = T extends [infer F extends string, ...infer R extends string[]]
  ? Join<R, U, Result extends "" ? F : `${Result}${U}${F}`>
  : Result;

Solution by alythobani #27610

type Join<T extends string[], S extends string> = T extends [
  infer Start extends string,
  ...infer Rest
]
  ? `${Start}${Rest extends [string] ? `${S}${Join<Rest, S>}` : ""}`
  : "";

Solution by slemchik03 #27574

type Join<T extends (number | string)[], U extends string | number, _Acc extends string = ''> = 
  T extends [infer Char extends string | number, ...infer Rest extends (string | number)[]] ? 
  Join<Rest, U, `${_Acc}${Char}${Rest extends [] ? '' : U}`> : _Acc

Solution by jjswifty #27539

type Join<T extends Array<number | string>, U extends number | string, S extends string = ''> =
    T extends [infer F, ...infer R extends Array<number | string>]
    ? Join<R, U, `${S}${F & string}${R extends [] ? '' : `${U}`}`> : S

Solution by smileboyi #27109

type Join<A extends any[], I extends number | string> = A extends [infer F, ...infer L] ? L["length"] extends 0 ? `${F & string}` : `${F & string}${I}${Join<L, I>}` : never

Solution by 20yuteo #26977

type ToString<T>= T extends string ? T : never;
type ToStringTuple<T>= T extends string[] ? T : never;
type Join<T extends string[], U extends string | number> = 
  T extends [infer F, ...infer Rest] ?
    Rest extends [] ?
      `${ToString<F>}` :
      `${ToString<F>}${U}${Join<ToStringTuple<Rest>, U>}` :
    '';

Solution by kakasoo #26441

type Join<T extends string[], U extends string | number> = T extends [] ? '' : T extends [string] ? `${T[0]}` : T extends [string, ...infer Args] ? Args extends string[] ? `${T[0]}${U}${Join<Args,U>}`: never : never

Solution by ScarboroughCoral #26340

type Join<TArray extends string[], Divider extends string | number> = TArray extends [
  infer Head extends string,
  ...infer Tail extends [] | string[],
]
  ? IsEmptyArray<Tail> extends true
    ? `${Head}`
    : `${Head}${Divider}${Join<Tail, Divider>}`
  : ''

Solution by valentynpodkradylin #26243