L1 · FoundationsList basics~2 min · 5 tests

Check If a List Is Sorted

Q: What is the time complexity of Check If a List Is Sorted in Python?

The optimal solution runs in O(n) time and O(1) auxiliary space. At most n − 1 neighbour comparisons, no copy.

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

Compare with sorted(nums): O(n log n) time, O(n) space. Sorts a full copy to answer a yes/no question. Compare neighbours: O(n) time, O(1) space. Stops at the first out-of-order pair.

Q: What follow-up questions do interviewers ask about Check If a List Is Sorted?

Strictly increasing instead. Sorted in either direction?

Check whether a list is in non-decreasing order by comparing neighbours in O(n), stopping at the first violation. Faster than sorting a copy. Run it live.

The problem

Return True if nums is sorted in non-decreasing order (equal neighbours are fine).

Examples

Example 1
Input
```
is_sorted([1, 2, 2, 5])
```
Expected output
```
True
```
Example 2
Input
```
is_sorted([3, 1, 2])
```
Expected output
```
False
```

+ 3 hidden tests on Submit — violation at the end.

Edge cases to ask about

Empty or single element
Equal neighbours

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 · is_sorted

def is_sorted(nums):
    """True if nums is in non-decreasing order."""
    # 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
Compare with sorted(nums)	O(n log n)	O(n)	Sorts a full copy to answer a yes/no question.
bestCompare neighbours	O(n)	O(1)	Stops at the first out-of-order pair.

Walkthrough of the optimal approach (try it yourself first)

Compare each element with the one before it and return False at the first drop. nums == sorted(nums) works but sorts a whole copy — O(n log n) time and O(n) space for a yes/no answer.

One-liner: all(a <= b for a, b in zip(nums, nums[1:])).

Complexity: O(n) time, O(1) space. At most n − 1 neighbour comparisons, no copy.

Reveal the reference solution

def is_sorted(nums):
    for i in range(1, len(nums)):
        if nums[i] < nums[i - 1]:
            return False
    return True

The brute force, for comparison

def is_sorted(nums):
    return nums == sorted(nums)

Follow-ups interviewers ask

Strictly increasing instead.
Sorted in either direction?

Frequently asked interview questions

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

What is the time complexity of Check If a List Is Sorted in Python?