type StringToUnion<T extends string> = T extends `${infer First}${infer Rest}` ? First | StringToUnion<Rest> : never

Solution by DevShinnThant #34609

type StringToUnion<T extends string> = T extends `${infer R}${infer rest}` ? R | StringToUnion<rest> : never

Solution by ouzexi #34018

531 - String to Union

by Andrey Krasovsky (@bre30kra69cs) #medium #union #string

Question

Implement the String to Union type. Type take string argument. The output should be a union of input letters

For example

type Test = "123"
type Result = StringToUnion<Test> // expected to be "1" | "2" | "3"

View on GitHub: https://tsch.js.org/531

...

type StringToUnion<T extends string> = T extends `${infer L}${infer R}`
  ? L | StringToUnion<R>
  : never 

Solution by veralex #33787

type StringToUnion<T extends string,R extends any=never> = T extends `${infer F}${infer B}` ? StringToUnion<B,F | R> : R;

Solution by knife710 #33762

type StringToUnion<T extends string, A extends string[] = []> = T extends ${infer S}${infer E} ? StringToUnion<E, [...A, S]> : A[number]

Solution by rookiewxy #33748

// 你的答案

type Tuple2Union<T> = T extends any[] ? T[number] : never;
type StringToUnion<T extends string, U extends any[]= []> = T extends `${infer H}${infer R}` ? StringToUnion<R, [...U, H]> : Tuple2Union<U>;

Solution by HelloGGG #33415

type StringToUnion<T extends string> = T extends `${infer F}${infer L}` ? F | StringToUnion<L> : never;

Solution by ZhipengYang0605 #32831

type StringToUnion<T extends string> = T extends `${infer First}${infer Rest}`
    ? First | StringToUnion<Rest>
    : never

Solution by ZhulinskiiDanil #32701

type StringToUnion<T extends string, R extends string[] = []> = 
  T extends `${infer Shift}${infer Rest}` 
    ? StringToUnion<Rest, [Shift, ...R]> 
    : R[number]

Solution by dev-hobin #32414

// 解答をここに記入
type StringToUnion<T extends string, K = never> = T extends `${infer F}${infer R}`
  ? StringToUnion<R, K | F>
  : K;

Solution by pea-sys #32195

type StringToUnion<T extends string, U extends string[] = []> = T extends ""
	? never
	: T extends `${infer F}${infer L}`
	? L extends ""
		? [...U, F][number]
		: StringToUnion<L, [...U, F]>
	: never;

Solution by gasmg #32146

type StrintToUnion<T extends string> = T extends `${infer Letter}${infer Rest}`
  ? Letter | StrintToUnion<Rest>
  : never;

Solution by rkamely #32061

type StringToUnion<T extends string, K = never> = T extends `${infer F}${infer R}`
  ? StringToUnion<R, K | F>
  : K;

Solution by jinyoung234 #31813

type StringToUnion<T extends string, R extends any[] = []> = T extends `${infer F}${infer Rest}` ? StringToUnion<Rest, [...R, F]> : R[number]

Solution by kai-phan #31647

type StringToUnion<
  S extends string,
  W extends any[] = []
> = S extends `${infer H}${infer T}` ? StringToUnion<T, [...W, H]> : W[number];

Solution by vipulpathak113 #31501

type StringToUnion<S extends string> = 
    S extends ""
    ? never
    : S extends `${infer F}${infer Rest}` 
        ? F | StringToUnion<Rest> 
        : S;

Solution by ricky-fn #31298

type StringToUnion<T extends string> = T extends `${infer F}${infer L}` ? F | StringToUnion<L> : never;

Solution by RainbowIsPerfect #31281

type StringToUnion<T extends string, R = {}> = T extends `${infer A}${infer B}` ? StringToUnion<B, {[k in keyof R | A]: true}> : keyof R;

Solution by eward957 #31249

// to array then to union

type StringToUnion<T extends string, K extends any[] = []> = T extends ``
  ? K[number]
  : T extends `${infer F}${infer R}`
  ? StringToUnion<R, [...K, F]>
  : never

Solution by gamejoye #30886

type StringToUnion<T extends string, U extends string[]=[]> =
	T extends `${infer F}${infer E}`
		? StringToUnion<E, [...U, F]>
		: U[number]

Solution by neMew5 #30767

type StringToUnion<T extends string> = T extends `${infer First}${infer Other}` ? First | StringToUnion<Other> : never

Solution by MrRabbit1993 #30464

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

Solution by kai-phan #30365

type StringToUnion<T extends string, K extends Array<unknown> = []> = T extends `${infer First}${infer Last}`
  ? First extends ''
    ? K
    : StringToUnion<Last, [...K, First]>
  : K[number]

Solution by agus-wesly #30283

type StringToUnion<T extends string> = T extends `${infer L}${infer R}` ? L | StringToUnion<R> : never;

Solution by kanishev #30131

type Test = '123';

type StringToUnion<T extends string> = T extends `${infer U}${infer R}` ? U | StringToUnion<R> : never
type Result = StringToUnion<Test>;

Solution by rxMATTEO #30098

type StringToUnion<T extends string, P = never> = T extends '' ? P : T extends `${infer X}${infer Y}` ?
  StringToUnion<Y, P | X> : T

Solution by gaoerjing #30000

// your answers

type StringToUnion<T extends string, Acc extends string[] = []> = T extends `${infer F}${infer Rest}` ?  StringToUnion<Rest, [ ...Acc, F]>: Acc[number]

Solution by kerolossamir165 #29973

type StringToUnion<T extends string, TRes extends string = never> = 
  T extends `${infer Letter}${infer Rest}` 
    ? StringToUnion<Rest, Letter | TRes> 
    : TRes

Solution by gstcarv #29912

type StringToUnion<T extends string> = T extends `${infer S}${infer Left}` ? S | StringToUnion<Left> : never;

Solution by Whiskeyi #29850

type StringToUnion<T extends string> =
  T extends `${infer First}${infer Rest}` ?
  First extends '' ?
    never :
    First | StringToUnion<Rest>
  : never;

Solution by maximallain #29599