MediumFrontend6 min read

Implement Promise.all() from scratch

Q: Implement Promise.all() from scratch

**Target Audience:** Engineers preparing for practical JavaScript coding interviews. **Difficulty:** Medium > **How to read this doc:** the "order is preserved regardless of completion order" claim — the single most commonly-missed detail in a naive first attempt — was verified directly with a slower promise listed BEFORE a faster one. ## 1. The problem, restated Given an array of promises, return one promise that resolves with an array of ALL their results, in the SAME order as the input array — genuinely independent of which one actually finishes first — and rejects immediately the moment any single input rejects. ## 2. Clarifying questions a strong candidate asks | Question | Why it matters | | :--- | :--- | | Does "fail fast" reject while others are still pending? | Confirms the real, documented behavior — yes, immediately, without waiting for the rest. | | Non-promise values in the array? | The real spec wraps every input with `Promise.resolve()`, treating a plain value as al

1.0k views

89 of 154

The problem

Problem, as an interviewer would state it: "Implement Promise.all() yourself. Given an array of promises, return a single promise that resolves with an array of all their results — in the ORIGINAL input order, even if they finish in a different real order — and rejects immediately if any one of them rejects."

Examples:

Code

await myPromiseAll([slowPromise, fastPromise]);
// [slowResult, fastResult] - original order, even though fast finished first

Clarifying questions expected:

Does "fail fast" mean reject the moment the FIRST rejection happens, even while other promises are still pending?
Should a plain, non-promise value in the array be handled, or is every element guaranteed to already be a real promise?
What should happen with an empty input array?

Code / implementation expected: Yes — real, direct proof that results preserve original input order regardless of real completion order, and that a rejection genuinely short-circuits immediately.

promise
promiseAll
concurrency

Try it yourself

Sketch your approach before you read the solution. That is what the interview tests.

Stuck? Get a hint.

A short nudge toward the approach. It will not give the answer away.

Solution

Target Audience: Engineers preparing for practical JavaScript coding interviews. Difficulty: Medium

How to read this doc: the "order is preserved regardless of completion order" claim — the single most commonly-missed detail in a naive first attempt — was verified directly with a slower promise listed BEFORE a faster one.

1. The problem, restated

Given an array of promises, return one promise that resolves with an array of ALL their results, in the SAME order as the input array — genuinely independent of which one actually finishes first — and rejects immediately the moment any single input rejects.

2. Clarifying questions a strong candidate asks

Question	Why it matters
Does "fail fast" reject while others are still pending?	Confirms the real, documented behavior — yes, immediately, without waiting for the rest.
Non-promise values in

The solution is hidden so you can try first

6 min read · 9 sections

Run the code

Edit the code and run it here. Each console line with an expected value becomes a test.

JavaScriptReal, direct proof: Promise.all() preserves original input order regardless of real completion order, and genuinely fails fast on rejection

Practice

Discussion

Other approaches, follow-ups and how it went in real interviews.

No comments yet. Share how you would answer it.

Question 89 of 154 in JavaScript coding

Keep your momentum going.

Mark this question solved to track your progress, then take the next one while it is fresh.

All JavaScript coding questions

Up nextMedium→

Implement throttle(func, wait, {leading, trailing})

Continue