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Rotate array N times

Q: Rotate array N times

**Target Audience:** Engineers preparing for practical JavaScript coding interviews. **Difficulty:** Hard > **How to read this doc:** this question looks like a one-liner and hides two traps: JavaScript's remainder operator returns negative values and `NaN`, and the in-place solution is a non-obvious three-reversal trick. Both implementations were checked exhaustively (n = 0..9, k = −25..25, 0 mismatches), and the cost of the naive one-step approach was measured growing linearly with k while the reversal approach stayed flat. ## 1. The problem, restated Shift every element `k` positions to the right, wrapping the last `k` elements around to the front. Rotating by the length (or a multiple of it) returns the original array, so only `k mod n` matters, and a negative `k` means rotating left. ## 2. Clarifying questions a strong candidate asks | Question | Why it matters | | :--- | :--- | | Direction and negative `k`? | Define right as positive. A negative `k` then rotates left, which

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The problem

Problem, as an interviewer would state it: "Rotate an array to the right by k positions. Discuss edge cases (negative k, k larger than the length), runtime, and how to do it in place."

Examples:

Code

rotate([1, 2, 3, 4, 5, 6, 7], 3); // [5, 6, 7, 1, 2, 3, 4]
rotate([1, 2, 3, 4, 5], -2);      // [3, 4, 5, 1, 2]  (a negative k rotates left)

Clarifying questions expected:

Rotate right or left, and how should a negative k or a k larger than the length behave?
May I return a new array, or must the rotation happen in place with O(1) extra space?
What about an empty array?

Code / implementation expected: Yes — real, direct proof against an independent reference, and a measurement of why "rotate one step, k times" is the wrong approach.

js-coding
utility
algorithm

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Solution

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

How to read this doc: this question looks like a one-liner and hides two traps: JavaScript's remainder operator returns negative values and NaN, and the in-place solution is a non-obvious three-reversal trick. Both implementations were checked exhaustively (n = 0..9, k = −25..25, 0 mismatches), and the cost of the naive one-step approach was measured growing linearly with k while the reversal approach stayed flat.

1. The problem, restated

Shift every element k positions to the right, wrapping the last k elements around to the front. Rotating by the length (or a multiple of it) returns the original array, so only k mod n matters, and a negative k means rotating left.

2. Clarifying questions a strong candidate asks

Question	Why it matters

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: slice and in-place three-reversal rotations agree with an index-mapping reference for negative and oversized k, and one-step rotation grows with k

Practice

Discussion

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