# Market Analysis II Problem

## Description

LeetCode Problem 1159.

Table: Users

``````1
2
3
4
5
6
7
8
9
+----------------+---------+
| Column Name    | Type    |
+----------------+---------+
| user_id        | int     |
| join_date      | date    |
| favorite_brand | varchar |
+----------------+---------+
user_id is the primary key of this table.
This table has the info of the users of an online shopping website where users can sell and buy items.
``````

Table: Orders

``````1
2
3
4
5
6
7
8
9
10
11
12
+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| order_id      | int     |
| order_date    | date    |
| item_id       | int     |
| seller_id     | int     |
+---------------+---------+
order_id is the primary key of this table.
item_id is a foreign key to the Items table.
buyer_id and seller_id are foreign keys to the Users table.
``````

Table: Items

``````1
2
3
4
5
6
7
+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| item_id       | int     |
| item_brand    | varchar |
+---------------+---------+
item_id is the primary key of this table.
``````

Write an SQL query to find for each user, whether the brand of the second item (by date) they sold is their favorite brand. If a user sold less than two items, report the answer for that user as no.

It is guaranteed that no seller sold more than one item on a day.

The query result format is in the following example:

``````1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
Users table:
+---------+------------+----------------+
| user_id | join_date  | favorite_brand |
+---------+------------+----------------+
| 1       | 2019-01-01 | Lenovo         |
| 2       | 2019-02-09 | Samsung        |
| 3       | 2019-01-19 | LG             |
| 4       | 2019-05-21 | HP             |
+---------+------------+----------------+

Orders table:
+----------+------------+---------+----------+-----------+
| order_id | order_date | item_id | buyer_id | seller_id |
+----------+------------+---------+----------+-----------+
| 1        | 2019-08-01 | 4       | 1        | 2         |
| 2        | 2019-08-02 | 2       | 1        | 3         |
| 3        | 2019-08-03 | 3       | 2        | 3         |
| 4        | 2019-08-04 | 1       | 4        | 2         |
| 5        | 2019-08-04 | 1       | 3        | 4         |
| 6        | 2019-08-05 | 2       | 2        | 4         |
+----------+------------+---------+----------+-----------+

Items table:
+---------+------------+
| item_id | item_brand |
+---------+------------+
| 1       | Samsung    |
| 2       | Lenovo     |
| 3       | LG         |
| 4       | HP         |
+---------+------------+

Result table:
+-----------+--------------------+
| seller_id | 2nd_item_fav_brand |
+-----------+--------------------+
| 1         | no                 |
| 2         | yes                |
| 3         | yes                |
| 4         | no                 |
+-----------+--------------------+

The answer for the user with id 1 is no because they sold nothing.
The answer for the users with id 2 and 3 is yes because the brands of their second sold items are their favorite brands.
The answer for the user with id 4 is no because the brand of their second sold item is not their favorite brand.
``````

## MySQL Solution

``````1
2
3
4
5
6
7
8
9
10
11
select u.user_id as seller_id,
case when u.favorite_brand = b.item_brand then 'yes' else 'no' end as 2nd_item_fav_brand
from users u
left join
(select seller_id, a.item_id, r, item_brand
from (select *, row_number() over (partition by seller_id order by order_date) as r
from orders) a
join items i
on a.item_id = i.item_id
where r = 2) b
on u.user_id = b.seller_id;
``````