Intensive reading "Trim Right, Without, Trunc..."

The best way to solve the TS problem is to practice more. This time, the difficulty of interpreting type-challenges Medium is 57~62 questions.

intensive reading

Trim Right

Implementation TrimRight remove the space on the right:

 type Trimed = TrimRight<'   Hello World    '> // expected to be '   Hello World'

Use infer to find the string before the space recursively:

 type TrimRight<S extends string> = S extends `${infer R}${' '}`
  ? TrimRight<R>
  : S

Then add the boundary conditions of the test case, \n and \t is the complete answer:

 // 本题答案
type TrimRight<S extends string> = S extends `${infer R}${' ' | '\n' | '\t'}`
  ? TrimRight<R>
  : S

Without

Implements Without<T, U> , removes elements U 6b1a633641b7a6a43d1f682f48071c9f--- from array T :

 type Res = Without<[1, 2], 1> // expected to be [2]
type Res1 = Without<[1, 2, 4, 1, 5], [1, 2]> // expected to be [4, 5]
type Res2 = Without<[2, 3, 2, 3, 2, 3, 2, 3], [2, 3]> // expected to be []

The most difficult point of this question is that the parameter U may be a string or an array of strings. We can only use extends to judge whether it exists, so there are two problems:

How to judge whether it is both a string and an array, judged together or judged separately?
[1] extends [1, 2] is false, how to judge the array mode?

This problem can be solved by converting the array to Union:

 type ToUnion<T> = T extends any[] ? T[number] : T

In this way, whether it is a number or an array, it will be converted into a union type, and the union type is very convenient to judge extends includes the relationship:

 // 本题答案
type Without<T, U> = T extends [infer H, ...infer R]
  ? H extends ToUnion<U>
    ? Without<R, U>
    : [H, ...Without<R, U>]
  : []

Each time the first item of the array is taken, it is judged whether it is included by U , and if so, it is discarded (the action of discarding is to discard H and continue the recursion), otherwise it is included (the action of inclusion is to form new array [H, ...] and deconstruct the recursive content to the back).

Trunc

Implement Math.trunc a function of the same function Trunc :

 type A = Trunc<12.34> // 12

If the input parameter is a string, it is very simple:

 type Trunc<T> = T extends `${infer H}.${infer R}` ? H : ''

If it's not a string, just convert it to a string:

 // 本题答案
type Trunc<T extends string | number> = `${T}` extends `${infer H}.${infer R}`
  ? H
  : `${T}`

IndexOf

Implement IndexOf find the subscript where the element is located, but can't find it and return -1 :

 type Res = IndexOf<[1, 2, 3], 2>; // expected to be 1
type Res1 = IndexOf<[2,6, 3,8,4,1,7, 3,9], 3>; // expected to be 2
type Res2 = IndexOf<[0, 0, 0], 2>; // expected to be -1

You need to use an auxiliary variable to store the hit index, and determine whether it matches one by one in a recursive way:

 type IndexOf<T, U, Index extends any[] = []> = 
  T extends [infer F, ...infer R]
    ? F extends U
      ? Index['length']
      : IndexOf<R, U, [...Index, 0]>
    : -1

But it did not pass the test case IndexOf<[string, 1, number, 'a'], number> , the reason is that the result of 1 extends number is true, so we have to replace it with Equal function to judge the equality:

 // 本题答案
type IndexOf<T, U, Index extends any[] = []> = 
  T extends [infer F, ...infer R]
    ? Equal<F, U> extends true
      ? Index['length']
      : IndexOf<R, U, [...Index, 0]>
    : -1

Join

Implement TS version Join<T, P> :

 type Res = Join<["a", "p", "p", "l", "e"], "-">; // expected to be 'a-p-p-l-e'
type Res1 = Join<["Hello", "World"], " ">; // expected to be 'Hello World'
type Res2 = Join<["2", "2", "2"], 1>; // expected to be '21212'
type Res3 = Join<["o"], "u">; // expected to be 'o'

Recursion T Take the first element each time, use an auxiliary string to store the answer, and splicing it together:

 // 本题答案
type Join<T, U extends string | number> =
  T extends [infer F extends string, ...infer R extends string[]]
  ? R['length'] extends 0
    ? F
    : `${F}${U}${Join<R, U>}`
  : ''

The only thing to note is that when the last item is processed, do not add U , it can be judged by R['length'] extends 0 .

LastIndexOf

Implementation LastIndexOf find the last matching subscript:

 type Res1 = LastIndexOf<[1, 2, 3, 2, 1], 2> // 3
type Res2 = LastIndexOf<[0, 0, 0], 2> // -1

Similar to IndexOf , you can judge from the last subscript forward. It should be noted that we cannot use the conventional method to reduce the subscript of Index by one, but fortunately R length of the array can replace the current subscript:

 // 本题答案
type LastIndexOf<T, U> = T extends [...infer R, infer L]
  ? Equal<L, U> extends true
    ? R['length']
    : LastIndexOf<R, U>
  : -1

Summarize

There are no new knowledge points in the six questions this week, and there are still 6 medium problems left. If you can feel a dull feeling after learning this, it means that you have learned very solidly.

The discussion address is: Intensive Reading "Trim Right, Without, Trunc..." Issue #433 dt-fe/weekly

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Intensive reading "Trim Right, Without, Trunc..."