Solution Explanation: Finding the Most Frequently Ordered Products for Each Customer

This problem requires analyzing three tables: Customers, Orders, and Products to determine the most frequently ordered product(s) for each customer. The solution leverages SQL's capabilities for efficient data manipulation and aggregation.

Approach:

The core idea is to perform the following steps:

1. Count product orders per customer: Aggregate the Orders table to count how many times each customer ordered each product.
2. Rank products by frequency: For each customer, rank the products based on their order counts. Products with the highest count receive rank 1.
3. Join with product details: Join the ranked results with the Products table to retrieve the product names.
4. Filter for top ranks: Select only the products with rank 1, representing the most frequently ordered items for each customer.

SQL Solution (MySQL):

The provided MySQL solution uses Common Table Expressions (CTEs) to break down the process into logical steps, making it easier to understand.

WITH
    T AS (
        SELECT
            customer_id,
            product_id,
            RANK() OVER (
                PARTITION BY customer_id
                ORDER BY COUNT(1) DESC
            ) AS rk
        FROM Orders
        GROUP BY 1, 2
    )
SELECT customer_id, product_id, product_name
FROM
    T
    JOIN Products USING (product_id)
WHERE rk = 1;

Explanation:

• WITH T AS (...): This defines a CTE named T. CTEs improve readability and allow for modularizing complex queries.
• SELECT customer_id, product_id, COUNT(1) ... FROM Orders GROUP BY 1, 2: This inner query groups the Orders data by customer_id and product_id, counting the occurrences of each product for each customer. COUNT(1) efficiently counts rows in each group.
• RANK() OVER (PARTITION BY customer_id ORDER BY COUNT(1) DESC) AS rk: This is the window function.
  • PARTITION BY customer_id: This divides the data into partitions based on customer_id, applying the ranking independently for each customer.
  • ORDER BY COUNT(1) DESC: This orders the products within each partition by their count in descending order (most frequent first).
  • RANK(): Assigns a rank to each product within its partition. Products with the same count receive the same rank.
• SELECT customer_id, product_id, product_name FROM T JOIN Products USING (product_id): This joins the CTE T (containing the ranked product frequencies) with the Products table using the product_id as the join key. This adds the product_name to the result.
• WHERE rk = 1: This filters the results, selecting only the rows with rank 1 (the most frequently ordered products for each customer).

Time Complexity Analysis:

The time complexity of this SQL query is dominated by the GROUP BY and RANK() operations. The GROUP BY operation's complexity is generally O(N log N) or O(N) depending on the database system's optimization, where N is the number of rows in the Orders table. The RANK() window function also has a complexity related to sorting or indexing, which is typically O(N log N) in the worst case. Therefore, the overall time complexity is approximately O(N log N). The exact complexity may vary depending on the database system's implementation and indexing.

Space Complexity Analysis:

The space complexity depends on the size of intermediate results. The CTE T stores the grouped and ranked data, which could be in the order of the number of unique customer-product combinations. The space complexity is therefore approximately O(M), where M is the number of unique (customer_id, product_id) pairs. This is generally less than or equal to the number of rows in the Orders table.