Table: Orders
+---------------+---------+ | Column Name | Type | +---------------+---------+ | order_id | int | | customer_id | int | | order_date | date | | item_id | varchar | | quantity | int | +---------------+---------+ (ordered_id, item_id) is the primary key (combination of columns with unique values) for this table. This table contains information on the orders placed. order_date is the date item_id was ordered by the customer with id customer_id.
Table: Items
+---------------------+---------+ | Column Name | Type | +---------------------+---------+ | item_id | varchar | | item_name | varchar | | item_category | varchar | +---------------------+---------+ item_id is the primary key (column with unique values) for this table. item_name is the name of the item. item_category is the category of the item.
You are the business owner and would like to obtain a sales report for category items and the day of the week.
Write a solution to report how many units in each category have been ordered on each day of the week.
Return the result table ordered by category.
The result format is in the following example.
Example 1:
Input: Orders table: +------------+--------------+-------------+--------------+-------------+ | order_id | customer_id | order_date | item_id | quantity | +------------+--------------+-------------+--------------+-------------+ | 1 | 1 | 2020-06-01 | 1 | 10 | | 2 | 1 | 2020-06-08 | 2 | 10 | | 3 | 2 | 2020-06-02 | 1 | 5 | | 4 | 3 | 2020-06-03 | 3 | 5 | | 5 | 4 | 2020-06-04 | 4 | 1 | | 6 | 4 | 2020-06-05 | 5 | 5 | | 7 | 5 | 2020-06-05 | 1 | 10 | | 8 | 5 | 2020-06-14 | 4 | 5 | | 9 | 5 | 2020-06-21 | 3 | 5 | +------------+--------------+-------------+--------------+-------------+ Items table: +------------+----------------+---------------+ | item_id | item_name | item_category | +------------+----------------+---------------+ | 1 | LC Alg. Book | Book | | 2 | LC DB. Book | Book | | 3 | LC SmarthPhone | Phone | | 4 | LC Phone 2020 | Phone | | 5 | LC SmartGlass | Glasses | | 6 | LC T-Shirt XL | T-Shirt | +------------+----------------+---------------+ Output: +------------+-----------+-----------+-----------+-----------+-----------+-----------+-----------+ | Category | Monday | Tuesday | Wednesday | Thursday | Friday | Saturday | Sunday | +------------+-----------+-----------+-----------+-----------+-----------+-----------+-----------+ | Book | 20 | 5 | 0 | 0 | 10 | 0 | 0 | | Glasses | 0 | 0 | 0 | 0 | 5 | 0 | 0 | | Phone | 0 | 0 | 5 | 1 | 0 | 0 | 10 | | T-Shirt | 0 | 0 | 0 | 0 | 0 | 0 | 0 | +------------+-----------+-----------+-----------+-----------+-----------+-----------+-----------+ Explanation: On Monday (2020-06-01, 2020-06-08) were sold a total of 20 units (10 + 10) in the category Book (ids: 1, 2). On Tuesday (2020-06-02) were sold a total of 5 units in the category Book (ids: 1, 2). On Wednesday (2020-06-03) were sold a total of 5 units in the category Phone (ids: 3, 4). On Thursday (2020-06-04) were sold a total of 1 unit in the category Phone (ids: 3, 4). On Friday (2020-06-05) were sold 10 units in the category Book (ids: 1, 2) and 5 units in Glasses (ids: 5). On Saturday there are no items sold. On Sunday (2020-06-14, 2020-06-21) were sold a total of 10 units (5 +5) in the category Phone (ids: 3, 4). There are no sales of T-shirts.
Problem Overview: You receive two tables: Orders and Items. Each order records an item_id, order_date, and quantity. The task is to report total quantity sold for each item_category across all seven days of the week (Monday through Sunday) as separate columns.
Approach 1: Join + Conditional Aggregation (O(n) time, O(1) extra space)
Start by joining Orders with Items on item_id so each order is associated with its item_category. Once joined, compute totals for each weekday using conditional aggregation. In MySQL, this is typically written as SUM(CASE WHEN DAYNAME(order_date) = 'Monday' THEN quantity ELSE 0 END). Repeat the same pattern for all seven weekdays. Group the result by item_category. The query scans the dataset once and aggregates results during grouping, which gives O(n) time complexity where n is the number of orders. No additional data structures are required beyond aggregation buffers, so extra space stays O(1).
This pattern is common in SQL analytics queries where you need to convert row-based data into column-based summaries. Conditional aggregation effectively acts as a manual pivot.
Approach 2: Numeric Weekday Mapping with CASE (O(n) time, O(1) space)
Another variation uses DAYOFWEEK(order_date) instead of DAYNAME. MySQL returns weekday numbers (1–7), which you map to Monday–Sunday using CASE expressions inside aggregation. For example: SUM(CASE WHEN DAYOFWEEK(order_date) = 2 THEN quantity ELSE 0 END) for Monday. The rest of the query stays identical: join tables, compute weekday sums, and group by item_category.
This approach avoids string comparison and relies on numeric checks instead. It performs the same logical work and keeps complexity at O(n) time and O(1) extra space. It is often preferred in database workloads because numeric comparisons can be slightly cheaper than string matching.
The core idea across both approaches is transforming row-level sales data into a pivot-style summary using aggregation. This is a fundamental technique in SQL aggregation problems.
Recommended for interviews: Use the join + conditional aggregation approach. It clearly demonstrates understanding of SQL joins, grouping, and pivot-style reporting. Interviewers expect you to recognize that the weekday columns can be computed with SUM(CASE WHEN ...) expressions during grouping rather than running multiple queries.
MySQL
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Join + Conditional Aggregation | O(n) | O(1) | Standard SQL solution for pivot-style reports grouped by category |
| DAYOFWEEK Numeric Mapping | O(n) | O(1) | When preferring numeric weekday checks instead of string comparisons |
LeetCode Hard 1479 "Sales by Day of the Week" Amazon Interview SQL Question with Explanation • Everyday Data Science • 4,411 views views
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