LeetCode 2228. Users With Two Purchases Within Seven Days SQL Solution | Explanation + Code

Description

Table: Purchases

+---------------+------+
| Column Name   | Type |
+---------------+------+
| purchase_id   | int  |
| user_id       | int  |
| purchase_date | date |
+---------------+------+
purchase_id contains unique values.
This table contains logs of the dates that users purchased from a certain retailer.

 

Write a solution to report the IDs of the users that made any two purchases at most 7 days apart.

Return the result table ordered by user_id.

The result format is in the following example.

 

Example 1:

Input: 
Purchases table:
+-------------+---------+---------------+
| purchase_id | user_id | purchase_date |
+-------------+---------+---------------+
| 4           | 2       | 2022-03-13    |
| 1           | 5       | 2022-02-11    |
| 3           | 7       | 2022-06-19    |
| 6           | 2       | 2022-03-20    |
| 5           | 7       | 2022-06-19    |
| 2           | 2       | 2022-06-08    |
+-------------+---------+---------------+
Output: 
+---------+
| user_id |
+---------+
| 2       |
| 7       |
+---------+
Explanation: 
User 2 had two purchases on 2022-03-13 and 2022-03-20. Since the second purchase is within 7 days of the first purchase, we add their ID.
User 5 had only 1 purchase.
User 7 had two purchases on the same day so we add their ID.

Solutions

Solution 1

sql

# Write your MySQL query statement below WITH t AS ( SELECT user_id, DATEDIFF( purchase_date, LAG(purchase_date, 1) OVER ( PARTITION BY user_id ORDER BY purchase_date ) ) AS d FROM Purchases ) SELECT DISTINCT user_id FROM t WHERE d <= 7 ORDER BY user_id;(code-box)