小镇做题家 - TypeScript 类型大挑战（中等篇 - 中）

前端开发
2022年09月25日
0

Medium 组（中）

KebabCase

Replace the camelCase or PascalCase string with kebab-case.

typescript
type KebabCase<S> = any

/* _____________ Test Cases _____________ */
import type { Equal, Expect } from '@type-challenges/utils'

type cases = [
  Expect<Equal<KebabCase<'FooBarBaz'>, 'foo-bar-baz'>>,
  Expect<Equal<KebabCase<'fooBarBaz'>, 'foo-bar-baz'>>,
  Expect<Equal<KebabCase<'foo-bar'>, 'foo-bar'>>,
  Expect<Equal<KebabCase<'foo_bar'>, 'foo_bar'>>,
  Expect<Equal<KebabCase<'Foo-Bar'>, 'foo--bar'>>,
  Expect<Equal<KebabCase<'ABC'>, 'a-b-c'>>,
  Expect<Equal<KebabCase<'-'>, '-'>>,
  Expect<Equal<KebabCase<''>, ''>>
]

在 TypeScript 的工具类中，有一个工具是可以把首字母转成小写，它就是 Uncapitalize：

typescript
type A = 'HELLO WORLD'
type E = Expect<Equal<Uncapitalize<A>, 'hELLO WORLD'>>

因为这一题里面首字母是大写时，只需要把它转成小写即可，而不需要再加上 -，所以我们需要对剩余字符进行判断：

typescript
type KebabCase<S> = S extends `${infer F}${infer R}`
  ? R extends Uncapitalize<R>
    ? `${Lowercase<F>}${KebabCase<R>}`
    : `${Lowercase<F>}-${KebabCase<R>}`
  : S

Diff

Get an Object that is the difference between O & O1

typescript
type Diff<O, O1> = any

/* _____________ Test Cases _____________ */
import type { Equal, Expect } from '@type-challenges/utils'

type Foo = {
  name: string
  age: string
}
type Bar = {
  name: string
  age: string
  gender: number
}
type Coo = {
  name: string
  gender: number
}

type cases = [
  Expect<Equal<Diff<Foo, Bar>, { gender: number }>>,
  Expect<Equal<Diff<Bar, Foo>, { gender: number }>>,
  Expect<Equal<Diff<Foo, Coo>, { age: string; gender: number }>>,
  Expect<Equal<Diff<Coo, Foo>, { age: string; gender: number }>>,
]

取差集，基操，只需要去除交集即可：

typescript
type Diff<O, O1> = {
  [P in Exclude<keyof O, keyof O1> | Exclude<keyof O1, keyof O>]: P extends keyof O
    ? O[P]
    : P extends keyof O1
      ? O1[P]
      : never
}

AnyOf

Implement Python liked any function in the type system. A type takes the Array and returns true if any element of the Array is true. If the Array is empty, return false.

typescript
type AnyOf<T extends readonly any[]> = any

/* _____________ Test Cases _____________ */
import type { Equal, Expect } from '@type-challenges/utils'

type cases = [
  Expect<Equal<AnyOf<[1, 'test', true, [1], { name: 'test' }, { 1: 'test' }]>, true>>,
  Expect<Equal<AnyOf<[1, '', false, [], {}]>, true>>,
  Expect<Equal<AnyOf<[0, 'test', false, [], {}]>, true>>,
  Expect<Equal<AnyOf<[0, '', true, [], {}]>, true>>,
  Expect<Equal<AnyOf<[0, '', false, [1], {}]>, true>>,
  Expect<Equal<AnyOf<[0, '', false, [], { name: 'test' }]>, true>>,
  Expect<Equal<AnyOf<[0, '', false, [], { 1: 'test' }]>, true>>,
  Expect<Equal<AnyOf<[0, '', false, [], { name: 'test' }, { 1: 'test' }]>, true>>,
  Expect<Equal<AnyOf<[0, '', false, [], {}]>, false>>,
  Expect<Equal<AnyOf<[]>, false>>,
]

只要传入的数组中有一项为 true，则结果为 true。首先我们需要知道哪些值是 falsy 值，从 cases 中可以得知：

typescript
type Falsy = 0 | '' | [] | false | Record<PropertyKey, never>

需要注意的是，空对象类型得采用 Record<PropertyKey, never> 来表示。如此一来，解题也变得非常简单了：

typescript
type Falsy = 0 | '' | [] | false | Record<PropertyKey, never>

type AnyOf<T extends readonly any[]> = T[number] extends Falsy
  ? false
  : true

IsNever

Implement a type IsNever, which takes input type T.

If the type of resolves to never, return true, otherwise false.

typescript
type IsNever<T> = any

/* _____________ Test Cases _____________ */
import type { Equal, Expect } from '@type-challenges/utils'

type cases = [
  Expect<Equal<IsNever<never>, true>>,
  Expect<Equal<IsNever<never | string>, false>>,
  Expect<Equal<IsNever<''>, false>>,
  Expect<Equal<IsNever<undefined>, false>>,
  Expect<Equal<IsNever<null>, false>>,
  Expect<Equal<IsNever<[]>, false>>,
  Expect<Equal<IsNever<{}>, false>>,
]

我总感觉这题不应该出现在这个栏目：

typescript
type IsNever<T> = [T] extends [never] ? true : false

IsUnion

Implement a type IsUnion, which takes an input type T and returns whether T resolves to a union type.

typescript
type IsUnion<T> = any

/* _____________ Test Cases _____________ */
import type { Equal, Expect } from '@type-challenges/utils'

type cases = [
  Expect<Equal<IsUnion<string>, false>>,
  Expect<Equal<IsUnion<string | number>, true>>,
  Expect<Equal<IsUnion<'a' | 'b' | 'c' | 'd'>, true>>,
  Expect<Equal<IsUnion<undefined | null | void | ''>, true>>,
  Expect<Equal<IsUnion<{ a: string } | { a: number }>, true>>,
  Expect<Equal<IsUnion<{ a: string | number }>, false>>,
  Expect<Equal<IsUnion<[string | number]>, false>>,
  // Cases where T resolves to a non-union type.
  Expect<Equal<IsUnion<string | never>, false>>,
  Expect<Equal<IsUnion<string | unknown>, false>>,
  Expect<Equal<IsUnion<string | any>, false>>,
  Expect<Equal<IsUnion<string | 'a'>, false>>,
  Expect<Equal<IsUnion<never>, false>>,
]

首先，我们先要把 never 排除掉：

typescript
type IsUnion<T, A = T> = [T] extends [never]
  ? false
	: // ...

如果传入的是联合类型，那么就拿第一项和整个联合类型进行对比：

typescript
type IsUnion<T, A = T> = [T] extends [never]
  ? false
  : T extends A
		? // ...
    : // ...

如果 T 是一个非联合类型，那么 T extends A 这条语句永远都会是 true，所以我们还需要进一步判断，通过一个小技巧：

typescript
type IsUnion<T, A = T> = [T] extends [never]
  ? false
  : T extends A
    ? Equal<[T], [A]> extends true
      ? false
      : true
    : false

如果 T 是一个联合类型，假设是：string | number，那么 [T] 就是 [string]，而 [A] 就是 [string | number] ，它们两者必不相等；如果 T 是一个非联合类型，假设是：string，那么 [T] 和 [A] 都是 [string]，两者相等。所以以此来判断它是否为一个联合类型。

ReplaceKeys

Implement a type ReplaceKeys, that replace keys in union types, if some type has not this key, just skip replacing,

A type takes three arguments.

typescript
type ReplaceKeys<U, T, Y> = any

/* _____________ Test Cases _____________ */
import type { Equal, Expect } from '@type-challenges/utils'

type NodeA = {
  type: 'A'
  name: string
  flag: number
}

type NodeB = {
  type: 'B'
  id: number
  flag: number
}

type NodeC = {
  type: 'C'
  name: string
  flag: number
}

type ReplacedNodeA = {
  type: 'A'
  name: number
  flag: string
}

type ReplacedNodeB = {
  type: 'B'
  id: number
  flag: string
}

type ReplacedNodeC = {
  type: 'C'
  name: number
  flag: string
}

type NoNameNodeA = {
  type: 'A'
  flag: number
  name: never
}

type NoNameNodeC = {
  type: 'C'
  flag: number
  name: never
}

type Nodes = NodeA | NodeB | NodeC
type ReplacedNodes = ReplacedNodeA | ReplacedNodeB | ReplacedNodeC
type NodesNoName = NoNameNodeA | NoNameNodeC | NodeB

type cases = [
  Expect<Equal<ReplaceKeys<Nodes, 'name' | 'flag', { name: number; flag: string }>, ReplacedNodes>>,
  Expect<Equal<ReplaceKeys<Nodes, 'name', { aa: number }>, NodesNoName>>,
]

题目中指出需要三个泛型参数，分别是：源类型（U）、需要的键（T）以及替换键组成的接口（Y），而我们要做的是，把 U 中所有符合 T 的键都替换成 Y 里面的类型。

所以解题的方法也很简单了，把需要匹配的条件都列出来即可：

typescript
type ReplaceKeys<U, T, Y> = {
  [P in keyof U]: P extends T
    ? P extends keyof Y
      ? Y[P]
      : never
    : U[P]
}

Remove Index Signature

Implement RemoveIndexSignature<T> , exclude the index signature from object types.

typescript
type RemoveIndexSignature<T> = any

/* _____________ Test Cases _____________ */
import type { Equal, Expect } from '@type-challenges/utils'

type Foo = {
  [key: string]: any
  foo(): void
}

type Bar = {
  [key: number]: any
  bar(): void
  0: string
}

const foobar = Symbol('foobar')
type FooBar = {
  [key: symbol]: any
  [foobar](): void
}

type Baz = {
  bar(): void
  baz: string
}

type cases = [
  Expect<Equal<RemoveIndexSignature<Foo>, { foo(): void }>>,
  Expect<Equal<RemoveIndexSignature<Bar>, { bar(): void; 0: string }>>,
  Expect<Equal<RemoveIndexSignature<FooBar>, { [foobar](): void }>>,
  Expect<Equal<RemoveIndexSignature<Baz>, { bar(): void; baz: string }>>,
]

从对象中移除索引类型，那么我们先需要判断哪些才是索引类型：

typescript
type IsSignature<T> = string extends T
  ? true
  : number extends T
    ? true
    : symbol extends T
      ? true
      : false

从对象中删除一个键，将该键置为 never 即可：

typescript
type Obj = {
  a: string;
  b: number;
}
type DeleteKeys<T> = {
  [K in keyof T as never]: T[K]
}
type EmptyObj = DeleteKeys<Obj> // {}

所以最终答案如下代码所示：

typescript
type IsSignature<T> = string extends T
  ? true
  : number extends T
    ? true
    : symbol extends T
      ? true
      : false

type RemoveIndexSignature<T> = {
  [P in keyof T as IsSignature<P> extends true ? never : P]: T[P]
}

Percentage Parser

Implement PercentageParser.

According to the /^(\+|\-)?(\d*)?(\%)?$/ regularity to match T and get three matches.

typescript
type PercentageParser<A extends string> = any

/* _____________ Test Cases _____________ */
import type { Equal, Expect } from '@type-challenges/utils'

type Case0 = ['', '', '']
type Case1 = ['+', '', '']
type Case2 = ['+', '1', '']
type Case3 = ['+', '100', '']
type Case4 = ['+', '100', '%']
type Case5 = ['', '100', '%']
type Case6 = ['-', '100', '%']
type Case7 = ['-', '100', '']
type Case8 = ['-', '1', '']
type Case9 = ['', '', '%']
type Case10 = ['', '1', '']
type Case11 = ['', '100', '']

type cases = [
  Expect<Equal<PercentageParser<''>, Case0>>,
  Expect<Equal<PercentageParser<'+'>, Case1>>,
  Expect<Equal<PercentageParser<'+1'>, Case2>>,
  Expect<Equal<PercentageParser<'+100'>, Case3>>,
  Expect<Equal<PercentageParser<'+100%'>, Case4>>,
  Expect<Equal<PercentageParser<'100%'>, Case5>>,
  Expect<Equal<PercentageParser<'-100%'>, Case6>>,
  Expect<Equal<PercentageParser<'-100'>, Case7>>,
  Expect<Equal<PercentageParser<'-1'>, Case8>>,
  Expect<Equal<PercentageParser<'%'>, Case9>>,
  Expect<Equal<PercentageParser<'1'>, Case10>>,
  Expect<Equal<PercentageParser<'100'>, Case11>>,
]

这题可以分两步来判断，一个是以 '+' | '-' 开头，另一个是以 '%' 结束：

typescript
type PercentageParser<A extends string> = A extends `${infer S extends '+' | '-'}${infer R}`
  ? R extends `${infer F}%`
    ? [S, F, '%']
    : [S, R, '']
  : A extends `${infer F}%`
    ? ['', F, '%']
    : ['', A, '']

Drop Char

Drop a specified char from a string.

typescript
type DropChar<S, C> = any

/* _____________ Test Cases _____________ */
import type { Equal, Expect } from '@type-challenges/utils'

type cases = [
  // @ts-expect-error
  Expect<Equal<DropChar<'butter fly!', ''>, 'butterfly!'>>,
  Expect<Equal<DropChar<'butter fly!', ' '>, 'butterfly!'>>,
  Expect<Equal<DropChar<'butter fly!', '!'>, 'butter fly'>>,
  Expect<Equal<DropChar<'    butter fly!        ', ' '>, 'butterfly!'>>,
  Expect<Equal<DropChar<' b u t t e r f l y ! ', ' '>, 'butterfly!'>>,
  Expect<Equal<DropChar<' b u t t e r f l y ! ', 'b'>, '  u t t e r f l y ! '>>,
  Expect<Equal<DropChar<' b u t t e r f l y ! ', 't'>, ' b u   e r f l y ! '>>,
]

又是一道字符串操作题，递归处理即可：

typescript
type DropChar<S, C, Result extends string = ''> = S extends `${infer F}${infer R}`
  ? F extends C
    ? DropChar<R, C, Result>
    : DropChar<R, C, `${Result}${F}`>
  : Result

或者：

typescript
type DropChar<S extends string, C extends string> = S extends `${infer F}${infer R}`
  ? `${F extends C ? '' : F}${DropChar<R, C>}`
  : S

这题的 // @ts-expect-error 有点谜，就不处理了。

MinusOne

Given a number (always positive) as a type. Your type should return the number decreased by one.

typescript
type MinusOne<T extends number> = any

/* _____________ Test Cases _____________ */
import type { Equal, Expect } from '@type-challenges/utils'

type cases = [
  Expect<Equal<MinusOne<1>, 0>>,
  Expect<Equal<MinusOne<55>, 54>>,
  Expect<Equal<MinusOne<3>, 2>>,
  Expect<Equal<MinusOne<100>, 99>>,
  Expect<Equal<MinusOne<1101>, 1100>>,
]

数组它承受了它不该承受的东西，只因为它有个 length，所以在做这种加减法的类型体操，我们只需要给个数组，最终拿到它的 Length 即可：

typescript
type Fill<N extends number, R extends number[] = []> = R['length'] extends N
  ? R
  : Fill<N, [...R, 0]>

type MinusOne<T extends number, A extends number[] = []> = Fill<T> extends [infer F, ...infer R]
  ? R['length']
  : never

但这一题，需要注意一下递归溢出问题，所以我们需要用其他办法来创建一个符合题目要求的数组：

typescript
type Dict = {
  '0': [];
  '1': [0];
  '2': [0, 0];
  '3': [0, 0, 0];
  '4': [0, 0, 0, 0];
  '5': [0, 0, 0, 0, 0];
  '6': [0, 0, 0, 0, 0, 0];
  '7': [0, 0, 0, 0, 0, 0, 0];
  '8': [0, 0, 0, 0, 0, 0, 0, 0];
  '9': [0, 0, 0, 0, 0, 0, 0, 0, 0];
}

type FillTenTimes<A extends 0[] = []> = [
  ...A, ...A, ...A, ...A, ...A,
  ...A, ...A, ...A, ...A, ...A
]

type Fill<N extends string, Result extends 0[] = []> = N extends `${infer F extends keyof Dict}${infer R}`
  ? Fill<R, [...FillTenTimes<Result>, ...Dict[F]]>
  : Result

type MinusOne<T extends number> = Fill<`${T}`> extends [infer F, ...infer R]
  ? R['length']
  : T

这里解释一下这个数组是怎么被创建出来的，以最后一个 case 的 1101 为例：

  1. 交给 Fill 来填充的是字符串 1101，此时 F === '1'，R === '101'，初始的 Result 为 []，它经过 FillTenTimes 处理后还是 []，Dict[F] 为 [0]，所以交给第二次递归的是 Fill<'101', [0]>；
  2. 第二次递归时：F === '1'，R === '01'，Result === [0]，然后 Result 经过 FillTenTimes 处理后变成 [10 * 0]，Dict[F] 为 [0]，所以交给第三次递归的是 Fill<'01', [10 * 0, 0]>；
  3. 第三次递归时：F === '0'，R === '1'，Result === [11 * 0]， 然后 Result 经过 FillTenTimes 处理后变成 [110 * 0]，Dict[F] 为 []，所以交给第四次递归的是 Fill<'1', [110 * 0]>；
  4. 第四次递归时：F === '1'，R === ''，Result === [110 * 0]，然后 Result 经过 FillTenTimes 处理后变成 [1100 * 0]，Dict[f] 为 [0]，所以交给第五次递归的是 Fill<'', [1100 * 0, 0]>；
  5. 第五次递归时，由于传入的是一个空字符串，所以条件不满足，此时返回 Result，也就是 [1101 * 1]。

PickByType

From T, pick a set of properties whose type are assignable to U.

typescript
type PickByType<T, U> = any

/* _____________ Test Cases _____________ */
import type { Equal, Expect } from '@type-challenges/utils'

interface Model {
  name: string
  count: number
  isReadonly: boolean
  isEnable: boolean
}

type cases = [
  Expect<Equal<PickByType<Model, boolean>, { isReadonly: boolean; isEnable: boolean }>>,
  Expect<Equal<PickByType<Model, string>, { name: string }>>,
  Expect<Equal<PickByType<Model, number>, { count: number }>>,
]

和 Pick 很类似，只不过是需要判断的是值：

typescript
type PickByType<T, U> = {
  [P in keyof T as T[P] extends U ? P : never]: T[P]
}

StartsWith

Implement StartsWith<T, U> which takes two exact string types and returns whether T starts with U

typescript
type StartsWith<T extends string, U extends string> = any

/* _____________ Test Cases _____________ */
import type { Equal, Expect } from '@type-challenges/utils'

type cases = [
  Expect<Equal<StartsWith<'abc', 'ac'>, false>>,
  Expect<Equal<StartsWith<'abc', 'ab'>, true>>,
  Expect<Equal<StartsWith<'abc', 'abcd'>, false>>,
  Expect<Equal<StartsWith<'abc', ''>, true>>,
  Expect<Equal<StartsWith<'abc', ' '>, false>>,
]

又是一道考验字符串操作的题：

typescript
type StartsWith<T extends string, U extends string> = T extends `${U}${infer R}`
  ? true
  : false

EndsWith

Implement EndsWith<T, U> which takes two exact string types and returns whether T ends with U

typescript
type EndsWith<T extends string, U extends string> = any

/* _____________ Test Cases _____________ */
import type { Equal, Expect } from '@type-challenges/utils'

type cases = [
  Expect<Equal<EndsWith<'abc', 'bc'>, true>>,
  Expect<Equal<EndsWith<'abc', 'abc'>, true>>,
  Expect<Equal<EndsWith<'abc', 'd'>, false>>,
]

和 StartsWith 一样，只需要变换一下位置：

typescript
type EndsWith<T extends string, U extends string> = T extends `${infer R}${U}`
  ? true
  : false

PartialByKeys

Implement a generic PartialByKeys<T, K> which takes two type argument T and K.

typescript
type PartialByKeys<T, K> = any

/* _____________ Test Cases _____________ */
import type { Equal, Expect } from '@type-challenges/utils'

interface User {
  name: string
  age: number
  address: string
}

interface UserPartialName {
  name?: string
  age: number
  address: string
}

interface UserPartialNameAndAge {
  name?: string
  age?: number
  address: string
}

type cases = [
  Expect<Equal<PartialByKeys<User, 'name'>, UserPartialName>>,
  Expect<Equal<PartialByKeys<User, 'name' | 'unknown'>, UserPartialName>>,
  Expect<Equal<PartialByKeys<User, 'name' | 'age'>, UserPartialNameAndAge>>,
  Expect<Equal<PartialByKeys<User>, Partial<User>>>,
]

将符合指定的 Key 的那些项转成可选项，我们可以分成两步来做：一是从源对象中移除指定的 Key（Omit）；二是从源对象中取出指定的 Key （Extract）并将它们转成可选的：

typescript
// 1. 通过 Omit 移除指定的 Key
type RequiredObject<T, K extends PropertyKey = keyof T> = Omit<T, K>
// 2. 通过 Extract 过滤指定的 Key
type PatrialObject<T, K extends PropertyKey = keyof T> = {
  [P in Extract<K, keyof T>]?: T[P]
}

然后再将两者合并即可：

typescript
type Merge<T> = Omit<T, never>

type PartialByKeys<T, K extends PropertyKey = keyof T> = Merge<Omit<T, K> & {
  [P in Extract<K, keyof T>]?: T[P]
}>

RequiredByKeys

Implement a generic RequiredByKeys<T, K> which takes two type argument T and K.

K specify the set of properties of T that should set to be required. When K is not provided, it should make all properties required just like the normal Required<T>.

typescript
type RequiredByKeys<T, K> = any


/* _____________ Test Cases _____________ */
import type { Equal, Expect } from '@type-challenges/utils'

interface User {
  name?: string
  age?: number
  address?: string
}

interface UserRequiredName {
  name: string
  age?: number
  address?: string
}

interface UserRequiredNameAndAge {
  name: string
  age: number
  address?: string
}

type cases = [
  Expect<Equal<RequiredByKeys<User, 'name'>, UserRequiredName>>,
  Expect<Equal<RequiredByKeys<User, 'name' | 'unknown'>, UserRequiredName>>,
  Expect<Equal<RequiredByKeys<User, 'name' | 'age'>, UserRequiredNameAndAge>>,
  Expect<Equal<RequiredByKeys<User>, Required<User>>>,
]

这题的思路和 PartialByKeys 一样，但有点小区别：

typescript
type Merge<T> = Omit<T, never>

type RequiredByKeys<T, K = keyof T> = Merge<{
  [P in keyof T as P extends K ? P : never]-?: T[P]
} & {
  [P in keyof T as P extends K ? never : P]: T[P]
}>

Mutable

Implement the generic Mutable<T> which makes all properties in T mutable (not readonly).

typescript
type Mutable<T> = any

/* _____________ Test Cases _____________ */
import type { Equal, Expect } from '@type-challenges/utils'

interface Todo1 {
  title: string
  description: string
  completed: boolean
  meta: {
    author: string
  }
}

type List = [1, 2, 3]

type cases = [
  Expect<Equal<Mutable<Readonly<Todo1>>, Todo1>>,
  Expect<Equal<Mutable<Readonly<List>>, List>>,
]

type errors = [
  // @ts-expect-error
  Mutable<'string'>,
  // @ts-expect-error
  Mutable<0>,
]

通过 - 号操作可以把 readonly 标识给移除，同时别忘了给泛型加约束处理掉 errors：

typescript
type Mutable<T extends Record<PropertyKey, any>> = {
  -readonly [P in keyof T]: T[P]
}

OmitByType

From T, pick a set of properties whose type are not assignable to U.

typescript
type OmitByType<T, U> = any

/* _____________ Test Cases _____________ */
import type { Equal, Expect } from '@type-challenges/utils'

interface Model {
  name: string
  count: number
  isReadonly: boolean
  isEnable: boolean
}

type cases = [
  Expect<Equal<OmitByType<Model, boolean>, { name: string; count: number }>>,
  Expect<Equal<OmitByType<Model, string>, { count: number; isReadonly: boolean; isEnable: boolean }>>,
  Expect<Equal<OmitByType<Model, number>, { name: string; isReadonly: boolean; isEnable: boolean }>>,
]

只需要把值的类型做一次对比即可：

typescript
type OmitByType<T, U> = {
  [P in keyof T as T[P] extends U ? never : P]: T[P]
}

ObjectEntries

Implement the type version of Object.entries

typescript
type ObjectEntries<T> = any

/* _____________ Test Cases _____________ */
import type { Equal, Expect } from '@type-challenges/utils'

interface Model {
  name: string
  age: number
  locations: string[] | null
}

type ModelEntries = ['name', string] | ['age', number] | ['locations', string[] | null]

type cases = [
  Expect<Equal<ObjectEntries<Model>, ModelEntries>>,
  Expect<Equal<ObjectEntries<Partial<Model>>, ModelEntries>>,
  Expect<Equal<ObjectEntries<{ key?: undefined }>, ['key', undefined]>>,
  Expect<Equal<ObjectEntries<{ key: undefined }>, ['key', undefined]>>,
]

这题需要注意两点，一个是把 T 转成 Required，另一个就是 undefined 的处理：

typescript
type ObjectEntries<T, R = Required<T>, K extends keyof R = keyof R> = K extends keyof R
  ? [K, R[K] extends undefined ? undefined : R[K]]
  : never

Shift

Implement the type version of Array.shift

typescript
type Shift<T> = any

/* _____________ Test Cases _____________ */
import type { Equal, Expect } from '@type-challenges/utils'

type cases = [
  Expect<Equal<Shift<[3, 2, 1]>, [2, 1]>>,
  Expect<Equal<Shift<['a', 'b', 'c', 'd']>, ['b', 'c', 'd']>>,
]

这题是比较简单的：

typescript
type Shift<T> = T extends [infer F, ...infer R]
  ? R
  : never

Tuple to Nested Object

Given a tuple type T that only contains string type, and a type U, build an object recursively.

typescript
type TupleToNestedObject<T, U> = any

/* _____________ Test Cases _____________ */
import type { Equal, Expect } from '@type-challenges/utils'

type cases = [
  Expect<Equal<TupleToNestedObject<['a'], string>, { a: string }>>,
  Expect<Equal<TupleToNestedObject<['a', 'b'], number>, { a: { b: number } }>>,
  Expect<Equal<TupleToNestedObject<['a', 'b', 'c'], boolean>, { a: { b: { c: boolean } } }>>,
  Expect<Equal<TupleToNestedObject<[], boolean>, boolean>>,
]

常规的数组递归操作即可：

typescript
type TupleToNestedObject<T, U> = T extends [infer F extends string, ...infer R]
  ? {
    [P in F]: TupleToNestedObject<R, U>
  }
  : U
TypeScript
小镇做题家 - TypeScript 类型大挑战（中等篇 - 上）
小镇做题家 - TypeScript 类型大挑战（中等篇 - 下）