The Deep Dictionary

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

Domain
Python
Difficulty
easy
Seniority
L6

Problem

Find the restaurant with the most reviews in data, which maps each restaurant name to the list of individual scores it has received. Every score counts as one review whatever its value, and data always holds at least one restaurant.

One key goes further than the rest.

Example 1

Input:

data = {"Diner":[2,3,4],"Bistro":[5,5]}

Output:

"Diner"

Example 2

Input:

data = {"east":[3,4],"west":[5,5],"north":[1,2]}

Output:

"east"

Worked solution and explanation

What this problem really is

Beneath the restaurant framing, this asks which key has the largest value of a metric you have to compute. The metric is not stored in the dict: each value is a list of scores, and you rank by how many scores there are, len(reviews), not by what they say. Anyone can loop over the items. Candidates separate on two things: seeing that a restaurant with two glowing reviews loses to a mediocre one with three, and settling ties by name instead of by whichever entry the dict happened to list first. Rank by the sum and you return the wrong name whenever the best-scored list is not the longest. Skip the tie-break and you return the wrong name whenever the alphabetically first name is not listed first.

Break down the requirements

Step 1: Count reviews per restaurant

Each key is a name and each value is a list of review scores. The review count is the list's length; the values only tell you how good the reviews were, and that is not the question.

Step 2: Track the winner with a name tie-break

Compare each count against the running best and take the new name when its count is larger. On an equal count, take the new name only if it sorts before the current winner. The second example shows why: all three restaurants have two reviews, and 'east' wins even though 'west' is listed before it.

Step 3: Seed the best count below zero

Start the running best count at -1, not 0. Then a restaurant with an empty list still beats the starting value, and a single restaurant with no reviews is returned instead of None.

The solution

Linear scan for the longest-list key
def the_deep_dictionary(data: dict[str, list[int]]) -> str:
    best_name = None
    best_count = -1
    for name, reviews in data.items():
        count = len(reviews)
        if count > best_count or (count == best_count and name < best_name):
            best_count = count
            best_name = name
    return best_name

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