Is Customer Purchasing Behavior Analysis easy or hard?

The problem is rated Medium because it requires combining multiple SQL concepts. Basic aggregation alone is straightforward, but the correct solution requires understanding window functions, partitioning logic, and how to join derived results back to the base table.

Customer Purchasing Behavior Analysis Python/Java solution

This problem is designed as a SQL query challenge rather than a Python or Java algorithm implementation. The solution is typically written in MySQL using GROUP BY, window functions like ROW_NUMBER or RANK, and joins to filter the final result set.

How to solve Customer Purchasing Behavior Analysis in O(n)?

Pure O(n) solutions are only possible for simple aggregation using GROUP BY when no ranking or intra-group comparison is required. Once the query needs ordering within partitions, window functions such as ROW_NUMBER or RANK are necessary, which generally introduce O(n log n) complexity due to sorting.

What is the best approach for Customer Purchasing Behavior Analysis?

The most effective solution uses SQL grouping combined with window functions and joins. First aggregate purchase metrics per customer, then apply window functions such as ROW_NUMBER or RANK with PARTITION BY to compare transactions within each customer group. Finally filter or join the ranked results to produce the required rows. This approach handles complex behavioral comparisons efficiently.

Is Customer Purchasing Behavior Analysis asked at Google/Amazon/Meta?

Database questions involving aggregations, joins, and window functions frequently appear in interviews at companies like Amazon, Google, and Meta. While the exact problem may vary, the pattern of analyzing transactional data using GROUP BY and window functions is a common SQL interview theme.

What data structure is used in Customer Purchasing Behavior Analysis?

The problem is solved using relational database operations rather than traditional in-memory data structures. Key SQL mechanisms include GROUP BY for aggregation, window functions for ranking within partitions, and JOIN operations to combine filtered results with the original dataset.

What is the time complexity of Customer Purchasing Behavior Analysis?

The optimal SQL approach runs in approximately O(n log n) time because window functions require sorting rows within each partition. Aggregation with GROUP BY scans the dataset in O(n) time, while the window operation introduces the log n sorting cost. Space complexity is typically O(n) due to intermediate result sets created during ranking.

3230. Customer Purchasing Behavior Analysis - Practice Online

Table: Transactions

+------------------+---------+
| Column Name      | Type    |
+------------------+---------+
| transaction_id   | int     |
| customer_id      | int     |
| product_id       | int     |
| transaction_date | date    |
| amount           | decimal |
+------------------+---------+
transaction_id is the unique identifier for this table.
Each row of this table contains information about a transaction, including the customer ID, product ID, date, and amount spent.

Table: Products

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| product_id  | int     |
| category    | varchar |
| price       | decimal |
+-------------+---------+
product_id is the unique identifier for this table.
Each row of this table contains information about a product, including its category and price.

Write a solution to analyze customer purchasing behavior. For each customer, calculate:

The total amount spent.
The number of transactions.
The number of unique product categories purchased.
The average amount spent.
The most frequently purchased product category (if there is a tie, choose the one with the most recent transaction).
A loyalty score defined as: (Number of transactions * 10) + (Total amount spent / 100).

Round total_amount, avg_transaction_amount, and loyalty_score to 2 decimal places.

Return the result table ordered by loyalty_score in descending order, then by customer_id in ascending order.

The query result format is in the following example.

Example:

Input:

Transactions table:

+----------------+-------------+------------+------------------+--------+
| transaction_id | customer_id | product_id | transaction_date | amount |
+----------------+-------------+------------+------------------+--------+
| 1              | 101         | 1          | 2023-01-01       | 100.00 |
| 2              | 101         | 2          | 2023-01-15       | 150.00 |
| 3              | 102         | 1          | 2023-01-01       | 100.00 |
| 4              | 102         | 3          | 2023-01-22       | 200.00 |
| 5              | 101         | 3          | 2023-02-10       | 200.00 |
+----------------+-------------+------------+------------------+--------+

Products table:

+------------+----------+--------+
| product_id | category | price  |
+------------+----------+--------+
| 1          | A        | 100.00 |
| 2          | B        | 150.00 |
| 3          | C        | 200.00 |
+------------+----------+--------+

Output:

+-------------+--------------+-------------------+-------------------+------------------------+--------------+---------------+
| customer_id | total_amount | transaction_count | unique_categories | avg_transaction_amount | top_category | loyalty_score |
+-------------+--------------+-------------------+-------------------+------------------------+--------------+---------------+
| 101         | 450.00       | 3                 | 3                 | 150.00                 | C            | 34.50         |
| 102         | 300.00       | 2                 | 2                 | 150.00                 | C            | 23.00         |
+-------------+--------------+-------------------+-------------------+------------------------+--------------+---------------+

Explanation:

For customer 101:
- Total amount spent: 100.00 + 150.00 + 200.00 = 450.00
- Number of transactions: 3
- Unique categories: A, B, C (3 categories)
- Average transaction amount: 450.00 / 3 = 150.00
- Top category: C (Customer 101 made 1 purchase each in categories A, B, and C. Since the count is the same for all categories, we choose the most recent transaction, which is category C on 2023-02-10)
- Loyalty score: (3 * 10) + (450.00 / 100) = 34.50
For customer 102:
- Total amount spent: 100.00 + 200.00 = 300.00
- Number of transactions: 2
- Unique categories: A, C (2 categories)
- Average transaction amount: 300.00 / 2 = 150.00
- Top category: C (Customer 102 made 1 purchase each in categories A and C. Since the count is the same for both categories, we choose the most recent transaction, which is category C on 2023-01-22)
- Loyalty score: (2 * 10) + (300.00 / 100) = 23.00

Note: The output is ordered by loyalty_score in descending order, then by customer_id in ascending order.