L2 · Working engineerDictionaries & hashing~4 min · 4 tests#34

Find Duplicate Values in a Dictionary

List every value that appears under more than one key in a Python dictionary, in first-seen order, with a counting pass in O(n). Live tests included.

The problem

Return the values that appear under more than one key, each once, in the order they are first seen.

Examples

Example 1

Input

duplicate_values({'a': 10, 'b': 20, 'c': 10, 'd': 30, 'e': 20})

Expected output

[10, 20]

Example 2
Input
```
duplicate_values({'a': 1, 'b': 2})
```
Expected output
```
[]
```

+ 2 hidden tests on Submit.

Edge cases to ask about

No duplicates
All the same value
Empty dict

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 · duplicate_values

def duplicate_values(d):
    """Values shared by 2+ keys, 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(values).count per value	O(n²)	O(n)	Recounts each time.
bestCount values once	O(n)	O(n)	Dict insertion order gives first-seen order for free.

Walkthrough of the optimal approach (try it yourself first)

Count the values, keep those with count > 1. Because dicts keep insertion order, the result is already in first-seen order.

Complexity: O(n) time, O(n) space. One pass to count values and one over the distinct values.

Reveal the reference solution

def duplicate_values(d):
    counts = {}
    for v in d.values():
        counts[v] = counts.get(v, 0) + 1
    return [v for v, c in counts.items() if c > 1]

Follow-ups interviewers ask

Return {value: [keys]} for the duplicates.
Values are unhashable lists.

Frequently asked interview questions

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

What is the time complexity of Find Duplicate Values in a Dictionary in Python?

The optimal solution runs in O(n) time and O(n) auxiliary space. One pass to count values and one over the distinct values.

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

list(values).count per value: O(n²) time, O(n) space. Recounts each time. Count values once: O(n) time, O(n) space. Dict insertion order gives first-seen order for free.

What follow-up questions do interviewers ask about Find Duplicate Values in a Dictionary?

Return {value: [keys]} for the duplicates. Values are unhashable lists.