Compute average unique pins per user

Q: Compute average unique pins per user

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.

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Question

Task: Average Unique Pins Per User

You are given a mapping from user IDs to their lists of pin IDs. Implement a function that computes the mean number of unique pin IDs per user.

Requirements

Function signature: avg_unique_pins(user_to_pins: dict[str, list[int]]) -> float
Treat a missing or empty list as 0 unique pins.
Ignore None values within lists.
Input can be large; target:
- Time complexity: O(total_items) where total_items is the sum of list lengths across users.
- Space complexity: O(U) where U is the number of unique pins observed (use only a temporary set per user).
Return a float rounded to two decimals.

Example

Input: {"user1": [1, 2, 2], "user2": [1, 2, 3]} → Output: 2.50

Unit Tests to Include

Empty dict → expect 0.00 .
Users with empty lists (and lists containing only None ).
Users sharing overlapping pins (e.g., the example above).

Compute average unique pins per user

Task: Average Unique Pins Per User

Requirements

Example

Unit Tests to Include

Solution

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Compute average unique pins per user

Overview

Task: Average Unique Pins Per User

Requirements

Example

Unit Tests to Include

Solution

Comments (0)