L2 · Working engineerDictionaries & hashing~5 min · 4 tests#32

Invert a Dictionary (Swap Keys and Values)

Swap keys and values of a Python dictionary with a dict comprehension, and handle duplicate values by grouping keys. O(n) time with live tests.

The problem

Return a new dict whose keys are the values of d and whose values are the lists of keys that had them, in insertion order.

{'a': 1, 'b': 2, 'c': 1} → {1: ['a', 'c'], 2: ['b']}. Grouping keeps duplicate values from silently overwriting each other.

Examples

Example 1

Input

invert({'a': 1, 'b': 2, 'c': 3})

Expected output

{1: ['a'], 2: ['b'], 3: ['c']}

Example 2

Input

invert({'a': 1, 'b': 2, 'c': 1})

Expected output

{1: ['a', 'c'], 2: ['b']}

+ 2 hidden tests on Submit.

Edge cases to ask about

Duplicate values
Unhashable values (lists)
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 · invert

def invert(d):
    """Return {value: [keys with that value]}."""
    # 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 read why.

Time complexity of the optimal solution

Space complexity (extra memory)

Pick both to reveal the answer.

From brute force to optimal

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

Approach	Time	Space	Idea
{v: k for k, v in d.items()}	O(n)	O(n)	Fine only when values are unique — duplicates overwrite.
bestGroup with setdefault	O(n)	O(n)	Every key survives.

Walkthrough of the optimal approach (try it yourself first)

The one-line comprehension {v: k for k, v in d.items()} is the answer most people give — and it silently loses data when two keys share a value. Grouping keys into lists with setdefault keeps everything. Mention that values must be hashable to become keys.

Complexity: O(n) time, O(n) space. One pass over the items; the new dict stores every key once.

Reveal the reference solution

def invert(d):
    out = {}
    for k, v in d.items():
        out.setdefault(v, []).append(k)
    return out

Follow-ups interviewers ask

Assume unique values — write the one-liner.
Raise an error on duplicates instead.

Frequently asked interview questions

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

What is the time complexity of Invert a Dictionary (Swap Keys and Values) in Python?

The optimal solution runs in O(n) time and O(n) auxiliary space. One pass over the items; the new dict stores every key once.

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

{v: k for k, v in d.items()}: O(n) time, O(n) space. Fine only when values are unique — duplicates overwrite. Group with setdefault: O(n) time, O(n) space. Every key survives.

What follow-up questions do interviewers ask about Invert a Dictionary (Swap Keys and Values)?

Assume unique values — write the one-liner. Raise an error on duplicates instead.