The Silver Screen Summit

A medium Python interview practice problem on DataDriven. Write the Python and run it against test cases, with instant feedback.

Domain
Python
Difficulty
medium
Seniority
L3
Asked in
Visa

Problem

We rank actors on our box-office leaderboard by the sum of earnings across every movie they appear in. Return the names of the top n actors, biggest totals first, with tied totals in alphabetical order.

Box office totals decide who makes the top of the marquee.

Example

Input:

movies = [{"actor":"Alice","earnings":5000000},{"actor":"Bob","earnings":3000000},{"actor":"Alice","earnings":2000000},{"actor":"Charlie","earnings":8000000}], n = 2

Output:

["Charlie","Alice"]

Worked solution and explanation

Why this problem exists in real interviews

This tests dict-based aggregation and sorting by computed values, a pattern that mirrors GROUP BY + ORDER BY in SQL. It probes whether a candidate can accumulate totals and then extract the top N entries efficiently.

Break down the requirements

Step 1: Aggregate earnings by actor name

Iterate through movie records, accumulating each actor's total box office earnings in a dict.

Step 2: Sort by total descending, then by name

Order the per-actor totals by total descending, and give the sort a second key on the actor name so equal totals come out alphabetically. Leaning on stable sort and dict insertion order only looks right when the input happens to list tied actors alphabetically; feed it Evan before Diane and the tie flips.

Step 3: Return the top N actor names

Slice the sorted list to get the first N entries and extract just the names.

The solution

Dict aggregation with an explicit tie-break
def top_box_office(movies: list[dict], n: int) -> list[str]:
    totals = {}
    for movie in movies:
        actor = movie["actor"]
        totals[actor] = totals.get(actor, 0) + movie["earnings"]

    # Negate the total so one ascending sort gives biggest first, names A to Z on ties.
    ranked = sorted(totals.items(), key=lambda item: (-item[1], item[0]))

    top = ranked[:n]
    return [actor for actor, _ in top]

Common follow-up questions