Top N per group

Given a list of (group, score) pairs, return the top N scores for each group, sorted descending.

sorting grouping dictionaries lists

Write a function `top_n_per_group(records, n)` that takes a list of tuples, each containing a group name (string) and a score (integer), and a positive integer `n`. The function should return a dictionary where each key is a group name and the value is a list of the top `n` scores for that group, sorted in descending order. If a group has fewer than `n` scores, include all its scores. If `n` is 0, return an empty dictionary. Groups should appear in the order they first appear in the input. Scores are unique within each group.

Constraints

1 <= len(records) <= 1000; scores are integers; group names are non-empty strings; n >= 0.

Example

>>> top_n_per_group([("A", 10), ("B", 5), ("A", 7)], 1)
{"A": [10], "B": [5]}
>>> top_n_per_group([("x", 1), ("x", 3), ("y", 2)], 2)
{"x": [3, 1], "y": [2]}
>>> top_n_per_group([("g", 4)], 0)
{}

25 points ~25 min

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Hints

Group the scores by group name, preserving the order of first appearance.

Sort each group's scores in descending order.

Slice the sorted list to get the top n elements.

Test Results

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