Leetcode #2051: The Category of Each Member in the Store
In this guide, we solve Leetcode #2051 The Category of Each Member in the Store in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Table: Members +-------------+---------+ | Column Name | Type | +-------------+---------+ | member_id | int | | name | varchar | +-------------+---------+ member_id is the column with unique values for this table. Each row of this table indicates the name and the ID of a member.
Quick Facts
- Difficulty: Medium
- Premium: Yes
- Tags: Database
Intuition
The task is relational in nature, which maps cleanly to DataFrame operations in Python.
By treating tables as DataFrames, joins and group-bys become concise and readable.
Approach
Load the inputs as DataFrames and apply the appropriate merge, filter, or group-by.
Select or rename the columns to match the required output.
Steps:
- Load inputs as DataFrames.
- Apply merge/groupby/filter operations.
- Select the output columns.
Example
+-------------+---------+
| Column Name | Type |
+-------------+---------+
| member_id | int |
| name | varchar |
+-------------+---------+
member_id is the column with unique values for this table.
Each row of this table indicates the name and the ID of a member.
Python Solution
import duckdb
import pandas as pd
def solution(members: pd.DataFrame, visits: pd.DataFrame, purchases: pd.DataFrame) -> pd.DataFrame:
con = duckdb.connect()
con.register("Members", members)
con.register("Visits", visits)
con.register("Purchases", purchases)
return con.execute("""SELECT
m.member_id,
name,
CASE
WHEN COUNT(v.visit_id) = 0 THEN 'Bronze'
WHEN 100 * COUNT(charged_amount) / COUNT(v.visit_id) >= 80 THEN 'Diamond'
WHEN 100 * COUNT(charged_amount) / COUNT(v.visit_id) >= 50 THEN 'Gold'
ELSE 'Silver'
END AS category
FROM
Members AS m
LEFT JOIN Visits AS v ON m.member_id = v.member_id
LEFT JOIN Purchases AS p ON v.visit_id = p.visit_id
GROUP BY member_id;""").df()
Complexity
The time complexity is O(n log n) (typical). The space complexity is O(n).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.