L2 · Working engineerLists & arrays~8 min · 6 tests#7

Rotate a List Right by K Steps

Rotate a Python list to the right by k positions, including k larger than the list. Slicing vs the reversal trick, with O(n) analysis.

The problem

Rotate nums to the right by k steps and return the rotated list.

k can be larger than len(nums), and the list can be empty.

Examples

Example 1

Input

rotate_right([1, 2, 3, 4, 5, 6], 3)

Expected output

[4, 5, 6, 1, 2, 3]

Example 2
Input
```
rotate_right([1, 2, 3], 1)
```
Expected output
```
[3, 1, 2]
```

+ 4 hidden tests on Submit — k == len, k > len, empty.

Edge cases to ask about

k = 0
k equal to length
k larger than length
Empty list

Hints

0/3

How an interviewer scores this

0/9

1. Understand the problemRestate it, ask about edge cases and input size.
2. Write a brute-force solutionCorrect first, clever later.
3. Get the correct outputautoAll visible examples pass.
4. Analyse time complexityautoCommit to a Big-O in the Complexity Lab.
5. Analyse space complexityautoExtra memory, not the input.
6. Identify the bottleneckautoMeasure your code — where does it bend?
7. OptimiseUsually: a set or dict instead of a scan.
8. Explain why it worksSay it out loud to Solvi.
9. Test the edge casesautoSubmit passes every hidden test.

Python 3.13 · rotate_right

def rotate_right(nums, k):
    """Return nums rotated right by k positions."""
    # Write your solution here
    pass

⌘/Ctrl + Enter runs the examples

Your code runs in real CPython inside your browser — nothing is sent anywhere. The first run downloads the interpreter (about 6 MB, once). Your code is saved on this device as you type.

Complexity Lab

What does this cost as n grows?

Interviewers score the analysis as much as the code. Commit to an answer first — then check it, and measure your code against the optimal one at growing input sizes.

Time complexity of the optimal solution

Space complexity (extra memory)

Pick both to reveal the answer.

Measure it

Runs the function on inputs of size 250 up to 16,000 and records the time and peak memory. Slow solutions stop early — a short curve is itself the answer.

From brute force to optimal

The progression an interviewer wants to hear, one step at a time.

Approach	Time	Space	Idea
Rotate one step k times	O(n·k)	O(n)	insert(0, …) shifts every element, k times over.
Slice and join	O(n)	O(n)	After k %= n, the last k elements move to the front.
bestThree reversals (in place)	O(n)	O(1)	Reverse all, reverse the first k, reverse the rest.

Walkthrough of the optimal approach (try it yourself first)

First reduce k with k %= len(nums) — rotating a full lap changes nothing. Then the answer is the last k elements followed by the rest: nums[-k:] + nums[:-k].

Watch the k == 0 trap: nums[-0:] is the whole list. For an in-place O(1)-space answer, reverse the whole list, then reverse the first k, then the remaining n − k.

Complexity: O(n) time, O(n) space. Slicing copies each element once. The in-place three-reversal version gets space down to O(1).

Reveal the reference solution

def rotate_right(nums, k):
    if not nums:
        return []
    k %= len(nums)
    return nums[-k:] + nums[:-k] if k else nums[:]

The brute force, for comparison

def rotate_right(nums, k):
    out = nums[:]
    for _ in range(k):
        if out:
            out.insert(0, out.pop())    # one step at a time
    return out

Follow-ups interviewers ask

Do it in place with O(1) extra memory.
Rotate left instead.

Frequently asked interview questions

Core interview concepts, complexities, and follow-ups scored by hiring teams.

What is the time complexity of Rotate a List Right by K Steps in Python?

The optimal solution runs in O(n) time and O(n) auxiliary space. Slicing copies each element once. The in-place three-reversal version gets space down to O(1).

What is the brute-force approach, and how do you optimise it?

Rotate one step k times: O(n·k) time, O(n) space. insert(0, …) shifts every element, k times over. Slice and join: O(n) time, O(n) space. After k %= n, the last k elements move to the front. Three reversals (in place): O(n) time, O(1) space. Reverse all, reverse the first k, reverse the rest.

What follow-up questions do interviewers ask about Rotate a List Right by K Steps?

Do it in place with O(1) extra memory. Rotate left instead.