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Partition Array

Q: Partition Array

**Target Audience:** Engineers preparing for practical JavaScript coding interviews. **Difficulty:** Medium > **How to read this doc:** two facts were measured rather than assumed. First, calling `filter` twice invokes the predicate 2n times against n for a real partition, which matters for costly or side-effecting predicates. Second, the async trap: `[1,2,3].filter(async n => n > 5)` keeps every element, because each callback returns a truthy promise. ## 1. The problem, restated Walk the array once, sending each element to a "pass" list or a "fail" list based on a predicate, preserving order within each list. The async version must wait for the predicate results before deciding, without losing the order of the original array. ## 2. Clarifying questions a strong candidate asks | Question | Why it matters | | :--- | :--- | | Order preserved? | Yes, within each group; both versions must keep the original relative order. | | Predicate call count? | Two `filter` calls run it 2n times,

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The problem

Problem, as an interviewer would state it: "Split an array into two lists: one with the elements that satisfy a predicate and one with those that do not. Support asynchronous predicates."

Examples:

Code

partition([1, 2, 3, 4, 5, 6], n => n % 2 === 0); // [[2, 4, 6], [1, 3, 5]]
await partitionAsync(users, u => isActive(u.id)); // [[active...], [inactive...]]

Clarifying questions expected:

Must the original order be preserved inside each group?
For an async predicate, should the checks run in parallel or one at a time, and what happens if one rejects?
Is it acceptable to call the predicate twice per element (for example by filtering twice)?

Code / implementation expected: Yes — real, direct proof for both versions, including the out-of-order-completion and rejection cases.

array

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Solution

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

How to read this doc: two facts were measured rather than assumed. First, calling filter twice invokes the predicate 2n times against n for a real partition, which matters for costly or side-effecting predicates. Second, the async trap: [1,2,3].filter(async n => n > 5) keeps every element, because each callback returns a truthy promise.

1. The problem, restated

Walk the array once, sending each element to a "pass" list or a "fail" list based on a predicate, preserving order within each list. The async version must wait for the predicate results before deciding, without losing the order of the original array.

2. Clarifying questions a strong candidate asks

Question	Why it matters
Order preserved?	Yes, within each group; both versions mu

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: sync and async partition preserve order, call the predicate once, and avoid the filter(async) trap that keeps every element

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