{x}
blog image

List the Products Ordered in a Period

Table: Products

+------------------+---------+
| Column Name      | Type    |
+------------------+---------+
| product_id       | int     |
| product_name     | varchar |
| product_category | varchar |
+------------------+---------+
product_id is the primary key (column with unique values) for this table.
This table contains data about the company's products.

 

Table: Orders

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| product_id    | int     |
| order_date    | date    |
| unit          | int     |
+---------------+---------+
This table may have duplicate rows.
product_id is a foreign key (reference column) to the Products table.
unit is the number of products ordered in order_date.

 

Write a solution to get the names of products that have at least 100 units ordered in February 2020 and their amount.

Return the result table in any order.

The result format is in the following example.

 

Example 1:

Input: 
Products table:
+-------------+-----------------------+------------------+
| product_id  | product_name          | product_category |
+-------------+-----------------------+------------------+
| 1           | Leetcode Solutions    | Book             |
| 2           | Jewels of Stringology | Book             |
| 3           | HP                    | Laptop           |
| 4           | Lenovo                | Laptop           |
| 5           | Leetcode Kit          | T-shirt          |
+-------------+-----------------------+------------------+
Orders table:
+--------------+--------------+----------+
| product_id   | order_date   | unit     |
+--------------+--------------+----------+
| 1            | 2020-02-05   | 60       |
| 1            | 2020-02-10   | 70       |
| 2            | 2020-01-18   | 30       |
| 2            | 2020-02-11   | 80       |
| 3            | 2020-02-17   | 2        |
| 3            | 2020-02-24   | 3        |
| 4            | 2020-03-01   | 20       |
| 4            | 2020-03-04   | 30       |
| 4            | 2020-03-04   | 60       |
| 5            | 2020-02-25   | 50       |
| 5            | 2020-02-27   | 50       |
| 5            | 2020-03-01   | 50       |
+--------------+--------------+----------+
Output: 
+--------------------+---------+
| product_name       | unit    |
+--------------------+---------+
| Leetcode Solutions | 130     |
| Leetcode Kit       | 100     |
+--------------------+---------+
Explanation: 
Products with product_id = 1 is ordered in February a total of (60 + 70) = 130.
Products with product_id = 2 is ordered in February a total of 80.
Products with product_id = 3 is ordered in February a total of (2 + 3) = 5.
Products with product_id = 4 was not ordered in February 2020.
Products with product_id = 5 is ordered in February a total of (50 + 50) = 100.

Solution Explanation

This problem requires querying two tables, Products and Orders, to find products with at least 100 units ordered in February 2020. The solution involves joining the tables, filtering by date, grouping by product, aggregating order units, and finally filtering the results based on the total unit count.

Approach

  1. JOIN: We begin by joining the Orders and Products tables using an INNER JOIN on product_id. This combines the order information with the product names.

  2. FILTER (WHERE): A WHERE clause filters the results to include only orders from February 2020. The specific method for date filtering depends on the database system (e.g., DATE_FORMAT in MySQL).

  3. GROUP BY: The GROUP BY clause groups the results by product_id (or equivalently, product_name since product_id is unique in Products). This allows us to aggregate order units for each product.

  4. AGGREGATE (SUM): The SUM(unit) function calculates the total number of units ordered for each product. We alias this as unit for clarity.

  5. FILTER (HAVING): Finally, a HAVING clause filters the grouped results to include only those products with a total unit count greater than or equal to 100. This condition ensures we only select products meeting the requirement.

Time and Space Complexity Analysis

The time complexity of this SQL query depends largely on the database system's query optimizer and the size of the tables. However, we can estimate it as follows:

  • Time Complexity: O(N log N + M log M), where N is the number of rows in the Orders table and M is the number of rows in the Products table. The join operation itself can be considered O(N log N + M log M) in the best case scenario (using efficient indexing). The filtering, grouping, aggregation, and final filtering steps have complexities that depend on the implementation and data size, but generally scale linearly or nearly linearly with the input size.

  • Space Complexity: O(K), where K is the number of unique products that meet the criteria (i.e., having at least 100 units ordered in February 2020). This represents the space needed to store the intermediate results during the grouping and aggregation phases. In the worst case, this could approach the size of the Products table.

Code (MySQL)

SELECT product_name, SUM(unit) AS unit
FROM
    Orders AS o
    JOIN Products AS p ON o.product_id = p.product_id
WHERE DATE_FORMAT(order_date, '%Y-%m') = '2020-02'
GROUP BY o.product_id
HAVING unit >= 100;

This MySQL query directly implements the approach described above. The DATE_FORMAT function is MySQL-specific; other database systems may use different date formatting functions. For example, in PostgreSQL, you might use to_char(order_date, 'YYYY-MM'). The core logic, however, remains the same across different SQL dialects.