L2 · Working engineerSets~4 min · 4 tests#36

Elements in One List but Not Another

Return values in list A that are missing from list B, preserving A's order, using a set for O(n+m) lookups. Python set difference explained with tests.

The problem

Return the values of a that are not in b, in the order they appear in a, each once.

Examples

Example 1

Input

difference([1, 2, 3, 4, 5], [3, 4, 5, 6, 7])

Expected output

[1, 2]

Example 2
Input
```
difference([5, 1, 5], [])
```
Expected output
```
[5, 1]
```

+ 2 hidden tests on Submit.

Edge cases to ask about

b empty
Everything excluded
Duplicates in a

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 · difference

def difference(a, b):
    """Values of a not present in b, ordered as in a."""
    # 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
List lookups	O(n × m)	O(1)	Scans b for every element.
bestset(b) once	O(n + m)	O(m)	Order-preserving set difference.

Walkthrough of the optimal approach (try it yourself first)

set(a) - set(b) is the textbook answer but loses order. Build set(b) once and filter a — O(n + m) and order-preserving.

Complexity: O(n + m) time, O(m) space. Building set(b) is O(m); scanning a with O(1) lookups is O(n).

Reveal the reference solution

def difference(a, b):
    exclude = set(b)
    seen = set()
    out = []
    for x in a:
        if x not in exclude and x not in seen:
            seen.add(x)
            out.append(x)
    return out

The brute force, for comparison

def difference(a, b):
    out = []
    for x in a:
        if x not in b and x not in out:
            out.append(x)
    return out

Follow-ups interviewers ask

Symmetric difference (in exactly one list).

Frequently asked interview questions

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

What is the time complexity of Elements in One List but Not Another in Python?

The optimal solution runs in O(n + m) time and O(m) auxiliary space. Building set(b) is O(m); scanning a with O(1) lookups is O(n).

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

List lookups: O(n × m) time, O(1) space. Scans b for every element. set(b) once: O(n + m) time, O(m) space. Order-preserving set difference.

What follow-up questions do interviewers ask about Elements in One List but Not Another?

Symmetric difference (in exactly one list).