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2292. Products With Three or More Orders in Two Consecutive Years πŸ”’

Description

Table: Orders

+---------------+------+
| Column Name   | Type |
+---------------+------+
| order_id      | int  |
| product_id    | int  |
| quantity      | int  |
| purchase_date | date |
+---------------+------+
order_id contains unique values.
Each row in this table contains the ID of an order, the id of the product purchased, the quantity, and the purchase date.

 

Write a solution to report the IDs of all the products that were ordered three or more times in two consecutive years.

Return the result table in any order.

The result format is shown in the following example.

 

Example 1:

Input: 
Orders table:
+----------+------------+----------+---------------+
| order_id | product_id | quantity | purchase_date |
+----------+------------+----------+---------------+
| 1        | 1          | 7        | 2020-03-16    |
| 2        | 1          | 4        | 2020-12-02    |
| 3        | 1          | 7        | 2020-05-10    |
| 4        | 1          | 6        | 2021-12-23    |
| 5        | 1          | 5        | 2021-05-21    |
| 6        | 1          | 6        | 2021-10-11    |
| 7        | 2          | 6        | 2022-10-11    |
+----------+------------+----------+---------------+
Output: 
+------------+
| product_id |
+------------+
| 1          |
+------------+
Explanation: 
Product 1 was ordered in 2020 three times and in 2021 three times. Since it was ordered three times in two consecutive years, we include it in the answer.
Product 2 was ordered one time in 2022. We do not include it in the answer.

Solutions

Solution 1

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# Write your MySQL query statement below
WITH
    P AS (
        SELECT product_id, YEAR(purchase_date) AS y, COUNT(1) >= 3 AS mark
        FROM Orders
        GROUP BY 1, 2
    )
SELECT DISTINCT p1.product_id
FROM
    P AS p1
    JOIN P AS p2 ON p1.y = p2.y - 1 AND p1.product_id = p2.product_id
WHERE p1.mark AND p2.mark;

Solution 2

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# Write your MySQL query statement below
WITH
    P AS (
        SELECT product_id, YEAR(purchase_date) AS y
        FROM Orders
        GROUP BY 1, 2
        HAVING COUNT(1) >= 3
    )
SELECT DISTINCT p1.product_id
FROM
    P AS p1
    JOIN P AS p2 ON p1.y = p2.y - 1 AND p1.product_id = p2.product_id;

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