Type Challenges: 18220-medium-filter Solution

// your answers
type Filter<T extends any[], P> =   T extends [infer L, ...infer R]
  ? [...L extends P ? [L] : [], ...Filter<R, P>]
  : []

Solution by pea-sys #33495

type Filter<T, Condition, Ret extends any[] = []> = T extends [
  infer First,
  ...infer Rest
]
  ? Filter<Rest, Condition, First extends Condition ? [...Ret, First] : Ret>
  : Ret;

Solution by sunupupup #33477

// your answers
type Filter<T extends any[], P> = T extends [infer Item, ...infer Rest]
  ? [...Item extends P ? [Item] : [], ...Filter<Rest, P>]
  : []

Solution by DevilTea #33259

type Filter<T extends unknown[], P> = T extends [infer F, ...infer R]
  ? F extends P
    ? [F, ...Filter<R, P>]
    : Filter<R, P>
  : T

Solution by keyurparalkar #32781

type Filter<T extends any[], P, Acc extends any[] = []> = T extends [infer F, ...infer Rest]
  ? F extends P
    ? Filter<Rest, P, [...Acc, F]>
    : Filter<Rest, P, Acc>
  : Acc

Solution by ZhulinskiiDanil #32724

type Filter<T extends any[], P, R extends any[] = []> = T["length"] extends 0
  ? R
  : T[0] extends P
  ? Filter<T extends [any, ...infer U] ? U : never, P, [...R, T[0]]>
  : Filter<T extends [any, ...infer U] ? U : never, P, R>;

Solution by vangie #32193

type Filter<T extends any[], P, Result extends T[number][] = []> = T extends [infer First, ...infer Rest]
  ? First extends P ? Filter<Rest, P, [...Result, First]> : Filter<Rest, P, Result>
  : Result

Solution by hwasurr #31430

type Filter<T extends any[], P, U extends any[] = []> = T extends [infer A, ...infer B] ? A extends P ? Filter<B, P, [...U, A]> : Filter<B, P, U> : U

Solution by dreamluo-plus #30691

// your answers
// 一个个筛，想了两个方式
// 除了加一个U做记录，也可以直接返回数组
// type Filter<T extends unknown[], P, U extends unknown[] = []> = T extends [
// 	infer F,
// 	...infer R
// ]
// 	? F extends P
// 		? Filter<R, P, [...U, F]>
// 		: Filter<R, P, U>
// 	: U

type Filter<T extends unknown[], P> = T extends [infer F, ...infer R]
	? F extends P
		? [F, ...Filter<R, P>]
		: [...Filter<R, P>]
	: []

Solution by bebusy007 #30031

type Filter<T extends unknown[], K> = T extends [infer F, ...infer R]
  ? F extends K
    ? [F, ...Filter<R, K>]
    : Filter<R, K>
  : [];

Solution by kanishev #29881

// your answers

type Filter<T extends any[], P> = 
            T extends [infer F , ...infer Rest] ? 
                F extends P ? [F, ...Filter<Rest, P>]
                   : Filter<Rest, P> 
                : []

Solution by kerolossamir165 #29626

type Filter<T extends any[], P> = T extends [infer Item, ...infer Rest]
  ? Item extends P
    ? [Item, ...Filter<Rest, P>]
    : Filter<Rest, P>
  : [];

Solution by DoubleWoodLin #28895

type Filter<ArrayType extends unknown[], Type> = ArrayType extends [
  infer Head,
  ...infer Tail
]
  ? [...(Head extends Type ? [Head] : []), ...Filter<Tail, Type>]
  : [];

Solution by yevhenpavliuk #28499

这题和上一题很类型，基本一样，通过添加多一个参数 Res 来存储返回结果，遍历生成 Res 数组

type Filter<T extends any[], P, Res extends any[] = []> = T extends [infer F, ...infer Rest] 
  ? F extends P
    ? Filter<Rest, P, [...Res, F]>
    : Filter<Rest, P, Res>
  : Res

在 github 上还看到了这个回答，可以直接在数组中递归，这样可以减少掉 Res 参数

type Filter<T extends unknown[], P> = T extends [infer F, ...infer R]
  ? F extends P
    ? [F, ...Filter<R, P>]
    : Filter<R, P>
  : [];

Solution by linjunc #28342

type Filter<T extends any[], P> = T extends [infer F, ...infer R]
  ? F extends P
    ? [F, ...Filter<R, P>]
    : Filter<R, P>
  : [];

Solution by slemchik03 #27575

type Filter<T extends any[], P> = T extends [infer F, ...infer R] ? F extends P ? [F, ...Filter<R, P>] : Filter<R, P> : [];

Solution by RainbowIsPerfect #27427

type Filter<T extends any[], P> =
  T extends [infer F, ...infer R] ? [...(F extends P ? [F] : []), ...Filter<R, P>] : []

Solution by smileboyi #27273

type Filter<T extends unknown[], P, Result extends unknown[] = []> = T extends [infer First, ...infer Rest]
  ? [First] extends [P]
    ? Filter<Rest, P, [...Result, First]>
    : Filter<Rest, P, Result>
  : Result;

Solution by rldnd #27015

This works with duplicates as well.

type Filter<T extends any[], P, Acc extends any[] = []> = T extends [
  infer F,
  ...infer L
]
  ? F extends P
    ? Filter<L, Exclude<P, F>, [...Acc, F]>
    : Filter<L, P, Acc>
  : Acc;

Solution by tany1 #26714

type Filter<T extends unknown[], P extends unknown> = T extends [infer X, ...infer Tail] ? X extends P ? [X, ...Filter<Tail,P>] : Filter<Tail,P>  : []

Solution by ScarboroughCoral #26572

type Filter<TArray extends any[], TTarget> = TArray extends [
  infer Head,
  ...infer Tail,
]
  ? Head extends TTarget
    ? [Head, ...Filter<Tail, TTarget>]
    : [...Filter<Tail, TTarget>]
  : []

Solution by valentynpodkradylin #26265

// your answers
type Filter<T extends any[], P, Res extends any[] = []> = T extends [
  infer F,
  ...infer Rest
]
  ? F extends P
    ? Filter<Rest, P, [...Res, F]>
    : Filter<Rest, P, Res>
  : Res;

test playground

Solution by DvYinPo #26173

// your answers
type Filter<T extends any[], P, Res extends unknown[] = []> = 
  T extends [infer X, ...infer Y]
  ? X extends P 
    ? Filter<Y, P, [...Res, X]>
    : Filter<Y, P, Res>
  : Res

Solution by kiki-zjq #25847

// your answers
type Filter<T extends any[], P, Res extends unknown[] = []> = T extends [infer Left, ...infer Rest]
                                                              ? Left extends P
                                                              ? Filter<Rest, P, [...Res, Left]>
                                                              : Filter<Rest, P, Res>
                                                              : Res

Solution by studymachiney #24945

type Filter<T extends any[], P> = T extends [infer F, ...infer R] ? [...(F extends P ? [F] : []), ...Filter<R, P>] : [];

// old way
// type Filter<T extends any[], P> = T extends [infer F, ...infer R] ? (F extends P ? [F, ...Filter<R, P>] : Filter<R, P>) : [];

Solution by E-uler #24561

type Filter<T extends any[], P,R extends any[] = []> = T extends [infer F,...infer L]? Filter<L,P,F extends P?[...R,F]:R>:R

Solution by jiangshanmeta #24478

type Filter<T extends unknown[], P> = T extends [infer A, ...infer rest] ? [...(A extends P ? [A] : []), ...Filter<rest, P>] : [];

Solution by sabercc #24439

type Filter<T extends unknown[], P> = T extends [infer First, ...infer Rest]
? First extends P 
  ? [First , ...Filter<Rest, P>]
  : Filter<Rest,P>
: T

Solution by NeylonR #24413

// your answers
type Filter<T extends any[], P, Acc extends unknown[] = []> = T extends [infer F, ...infer R] ? 
F extends P ? Filter<R, P, [...Acc, F]> : Filter<R, P, [...Acc]> : Acc

Solution by snakeUni #23346

type Filter<
  Arr extends any[],
  T,
  Res extends any[] = []
> = Arr extends []
  ? Res
  : Arr extends [infer F, ...infer Rest]
    ? Bool<F> extends T
      ? Filter<Rest, T, [...Res, F]>
      : Filter<Rest, T, [...Res]>
    : never

type Bool<T> = T extends boolean ? Extract<boolean, T> : T

Solution by DedaVnuk #23151