L2 · Working engineerLists & arrays~5 min · 5 tests#9

Union of Two Lists Without Duplicates

Merge two Python lists into their union with no duplicates and first-seen order preserved. Uses a seen-set for O(n+m) time.

The problem

Return every value that appears in either list, each once, in the order it is first seen (all of list1, then list2).

Examples

Example 1

Input

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

Expected output

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

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

+ 3 hidden tests on Submit.

Edge cases to ask about

Both empty
Fully overlapping lists
Duplicates inside one 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 · union

def union(list1, list2):
    """All values from both lists, once each, first-seen order."""
    # Write your solution here
    pass

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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 membership	O((n+m)²)	O(n+m)	Each `not in out` scans the result.
bestSeen set	O(n + m)	O(n + m)	Set lookups are O(1); the list keeps order.

Walkthrough of the optimal approach (try it yourself first)

Union is just dedupe applied to list1 + list2: keep a seen set, append each value the first time it shows up. set(a) | set(b) is the one-liner, but it throws away order — say so in the interview and offer both.

Complexity: O(n + m) time, O(n + m) space. Each of the n + m values is visited once with O(1) set operations; the set and output can hold all of them.

Reveal the reference solution

def union(list1, list2):
    seen = set()
    out = []
    for x in list1 + list2:
        if x not in seen:
            seen.add(x)
            out.append(x)
    return out

The brute force, for comparison

def union(list1, list2):
    out = []
    for x in list1 + list2:
        if x not in out:
            out.append(x)
    return out

Follow-ups interviewers ask

Return it sorted.
Union of k lists.

Frequently asked interview questions

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

What is the time complexity of Union of Two Lists Without Duplicates in Python?

The optimal solution runs in O(n + m) time and O(n + m) auxiliary space. Each of the n + m values is visited once with O(1) set operations; the set and output can hold all of them.

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

List membership: O((n+m)²) time, O(n+m) space. Each `not in out` scans the result. Seen set: O(n + m) time, O(n + m) space. Set lookups are O(1); the list keeps order.

What follow-up questions do interviewers ask about Union of Two Lists Without Duplicates?

Return it sorted. Union of k lists.