type Push<T extends any[], U> = [...T, U]

Solution by tomo-local #37582

// your answers

type Push<T extends any[], U> = [...T, U]

Solution by PAVANT009 #37572

type Push<T extends unknown[], U> = [...T, U]

Breaking it down:

• <T extends unknown[], U>

  • Two generic parameters.
  • T must be an array (tuple) of any types.
  • U can be any single type (the element to push).
  • Note: Unlike Concat, this doesn't use readonly, so it's slightly more flexible but works the same way.

• [...T, U]

  • This is a tuple spread type.
  • It creates a new tuple type by:
    • Taking all elements from T using the spread ...T
    • Adding U as the last element
  • The result is a new tuple with U appended to the end.

In plain words:

Push<T, U> constructs a new tuple type by:

1. Taking all elements from tuple T.
2. Adding the single element U at the end.
3. Producing a new tuple type that represents the array after a push operation.

✅ Examples:

type A = [1, 2, 3]
type B = ["hello", "world"]
type Test1 = Push<A, 4>           // [1, 2, 3, 4] ✅
type Test2 = Push<B, "!">         // ["hello", "world", "!"] ✅
type Test3 = Push<[], "first">    // ["first"] ✅
type Test4 = Push<[1], [2, 3]>    // [1, [2, 3]] ✅ (pushes the whole array as one element)

Solution by 3aluw #37492

type Push<T extends any[], U> = [...T, U] 

Solution by Nakamura25257 #37262

type Push<T extends unknown[], U> = [...T, U];

Solution by djdidi #37137

type Push<T extends unknown[], U> = [...T, U]

Solution by Anonymous961 #37046

type Push<T extends any[], U> = [...T, U]

Solution by 359Steve #37020

// that's cool, thanks all Ts developer
type Push<T extends any[] , U> = [...T, U]

Solution by leongaooo #36928

// your answers
type Push<T extends readonly any[], U> = [...T, U];

Solution by AlexBraunMagic #36912

type Push<T extends unknown[], U> = [...T, U];

Solution by shaishab316 #36848

type Push<T extends unknown[], U> = [...T, U]

Solution by Abdullah-Elsayed01 #36791

type Push<T extends any[], U> = [...T, U];

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

type cases = [
  Expect<Equal<Push<[], 1>, [1]>>,
  Expect<Equal<Push<[1, 2], '3'>, [1, 2, '3']>>,
  Expect<Equal<Push<['1', 2, '3'], boolean>, ['1', 2, '3', boolean]>>,
]

type errors = [
  // @ts-expect-error
  Expect<Equal<Push<number[], string>, string[]>>,
  // @ts-expect-error
  Expect<Equal<Push<string[], number>, [string, number]>>,
]

Solution by AnastasiaSv #36768

type Push<T extends any[], U> =  [...T, U]

Solution by tungulin #36719

// 여기 풀이를 입력하세요
type Push<T extends readonly any[], U> = [...T, U];

Solution by seungdeok #36682

// your answers

type Push<T, U> = T extends [...infer TType] ? [...TType,U] : never

Modifications and Suggestions are welcomed 🎉

Solution by SubramaniyanTN #36675

type Push<T extends readonly any[], U> = [...T, U]

This implementation is better for the following reasons: Greater Compatibility: Supports both readonly and regular array types, making it suitable for more scenarios.

Avoids type errors caused by readonly input types.

Aligns with TypeScript Best Practices: Using readonly any[] is more general, reducing potential type conflicts.

It will be better than type Push<T extends any[], U> = [...T, U]

Solution by spike014 #36625

type Push<T extends any[], U> = [...T, U]

Solution by tungulin #36608

type Push<T extends any[], U> = [...T, U]

Solution by ChemieAi #36556

type push<T extends Array<unknow>, U> = [...T, U]

Solution by Divcutu #36455

type Push<T extends readonly any[], U> = [...T, U];

Solution by alirezaprime #36417

// your answers
type Push<T extends readonly any[], U> = [...T, U];

Solution by justBadProgrammer #36392

type Push<T extends readonly any[], U> = [...T , U]

Solution by HelloWook #36384

type Push<T extends any[], U> = [...T,U]

Solution by tac-tac-go #36361

type Push<T extends any[], U> = [...T, U]

Solution by 1Alex4949031 #36316

type Push<T extends unknown[], U> = [...T,U]

Solution by asylbekduldiev #36291

type Push<T extends any[], U> = [...T, U]

Solution by AleksandrShcherbackov #36147

type Push<T extends any[], U> = [...T, U]

Solution by reonce #36050

type Push<T extends any[], U> = [...T, U]

Solution by karazyabko #36032

type Push<T extends unknown[], U> =[...T, U] 

Solution by codingaring #35942

type Push<T extends unknown[], P> = [...T, P]

Solution by Sensuele #35802