Compute average unique pins per user
Company: Pinterest
Role: Data Scientist
Category: Coding & Algorithms
Difficulty: easy
Interview Round: Technical Screen
Implement a function avg_unique_pins(user_to_pins: dict[str, list[int]]) -> float that returns the mean number of unique pin IDs per user. Treat a missing or empty list as 0 unique pins; ignore None values within lists; input may be large, so target O(total_items) time and O(U) space where U is total unique pins observed. Return a float rounded to two decimals.
Example: for {"user1": [1,2,2], "user2": [1,2,3]} the result is 2.50. Include unit tests covering: empty dict (expect 0.00), users with empty lists, and users sharing overlapping pins.
Quick Answer: This question evaluates a candidate's ability to process collections and compute aggregated metrics, including per-user deduplication of items, handling of missing/null entries, numeric averaging and rounding.
You are given a dictionary that maps each user ID to a list of pin IDs. Compute the average number of unique, non-None pin IDs per user.
Rules:
- Count unique pins separately for each user.
- Duplicate pin IDs in the same user's list should only be counted once.
- Ignore None values inside a user's list.
- If a user's value is None or an empty list, that user contributes 0 unique pins.
- If the input dictionary is empty, return 0.0.
- Return the final average as a float rounded to two decimal places.
Constraints
- 0 <= number of users
- Each dictionary value is either None or a list that may contain integers and None values
- Count uniqueness per user, not across all users
- Target time complexity: O(total_items), where total_items is the total number of entries across all user lists
- Use only a temporary set per user to track uniqueness
Examples
Input: ({},)
Expected Output: 0.0
Explanation: There are no users, so the average unique pin count is 0.0.
Input: ({'u1': [], 'u2': [None, None], 'u3': None},)
Expected Output: 0.0
Explanation: Each user has 0 valid unique pins, so the average is (0 + 0 + 0) / 3 = 0.0.
Hints
- A set is useful for counting distinct pin IDs for one user while ignoring duplicates.
- Make sure you divide by the total number of users in the dictionary, even if some users have empty lists or only None values.