feat: add sql solution to lc problem: No.2922

solution/2900-2999/2921.Maximum Profitable Triplets With Increasing Prices II/README.md

@@ -0,0 +1,97 @@
+# [2921. Maximum Profitable Triplets With Increasing Prices II](https://leetcode.cn/problems/maximum-profitable-triplets-with-increasing-prices-ii)
+
+[English Version](/solution/2900-2999/2921.Maximum%20Profitable%20Triplets%20With%20Increasing%20Prices%20II/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>Given the <strong>0-indexed</strong> arrays <code>prices</code> and <code>profits</code> of length <code>n</code>. There are <code>n</code> items in an store where the <code>i<sup>th</sup></code> item has a price of <code>prices[i]</code> and a profit of <code>profits[i]</code>.</p>
+
+<p>We have to pick three items with the following condition:</p>
+
+<ul>
+	<li><code>prices[i] &lt; prices[j] &lt; prices[k]</code> where <code>i &lt; j &lt; k</code>.</li>
+</ul>
+
+<p>If we pick items with indices <code>i</code>, <code>j</code> and <code>k</code> satisfying the above condition, the profit would be <code>profits[i] + profits[j] + profits[k]</code>.</p>
+
+<p>Return<em> the <strong>maximum profit</strong> we can get, and </em><code>-1</code><em> if it&#39;s not possible to pick three items with the given condition.</em></p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> prices = [10,2,3,4], profits = [100,2,7,10]
+<strong>Output:</strong> 19
+<strong>Explanation:</strong> We can&#39;t pick the item with index i=0 since there are no indices j and k such that the condition holds.
+So the only triplet we can pick, are the items with indices 1, 2 and 3 and it&#39;s a valid pick since prices[1] &lt; prices[2] &lt; prices[3].
+The answer would be sum of their profits which is 2 + 7 + 10 = 19.</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> prices = [1,2,3,4,5], profits = [1,5,3,4,6]
+<strong>Output:</strong> 15
+<strong>Explanation:</strong> We can select any triplet of items since for each triplet of indices i, j and k such that i &lt; j &lt; k, the condition holds.
+Therefore the maximum profit we can get would be the 3 most profitable items which are indices 1, 3 and 4.
+The answer would be sum of their profits which is 5 + 4 + 6 = 15.</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> prices = [4,3,2,1], profits = [33,20,19,87]
+<strong>Output:</strong> -1
+<strong>Explanation:</strong> We can&#39;t select any triplet of indices such that the condition holds, so we return -1.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>3 &lt;= prices.length == profits.length &lt;= 50000</code></li>
+	<li><code>1 &lt;= prices[i] &lt;= 5000</code></li>
+	<li><code>1 &lt;= profits[i] &lt;= 10<sup>6</sup></code></li>
+</ul>
+
+## 解法
+
+<!-- 这里可写通用的实现逻辑 -->
+
+<!-- tabs:start -->
+
+### **Python3**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```python
+
+```
+
+### **Java**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```java
+
+```
+
+### **C++**
+
+```cpp
+
+```
+
+### **Go**
+
+```go
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

solution/2900-2999/2921.Maximum Profitable Triplets With Increasing Prices II/README_EN.md

@@ -0,0 +1,89 @@
+# [2921. Maximum Profitable Triplets With Increasing Prices II](https://leetcode.com/problems/maximum-profitable-triplets-with-increasing-prices-ii)
+
+[中文文档](/solution/2900-2999/2921.Maximum%20Profitable%20Triplets%20With%20Increasing%20Prices%20II/README.md)
+
+## Description
+
+<p>Given the <strong>0-indexed</strong> arrays <code>prices</code> and <code>profits</code> of length <code>n</code>. There are <code>n</code> items in an store where the <code>i<sup>th</sup></code> item has a price of <code>prices[i]</code> and a profit of <code>profits[i]</code>.</p>
+
+<p>We have to pick three items with the following condition:</p>
+
+<ul>
+	<li><code>prices[i] &lt; prices[j] &lt; prices[k]</code> where <code>i &lt; j &lt; k</code>.</li>
+</ul>
+
+<p>If we pick items with indices <code>i</code>, <code>j</code> and <code>k</code> satisfying the above condition, the profit would be <code>profits[i] + profits[j] + profits[k]</code>.</p>
+
+<p>Return<em> the <strong>maximum profit</strong> we can get, and </em><code>-1</code><em> if it&#39;s not possible to pick three items with the given condition.</em></p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> prices = [10,2,3,4], profits = [100,2,7,10]
+<strong>Output:</strong> 19
+<strong>Explanation:</strong> We can&#39;t pick the item with index i=0 since there are no indices j and k such that the condition holds.
+So the only triplet we can pick, are the items with indices 1, 2 and 3 and it&#39;s a valid pick since prices[1] &lt; prices[2] &lt; prices[3].
+The answer would be sum of their profits which is 2 + 7 + 10 = 19.</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> prices = [1,2,3,4,5], profits = [1,5,3,4,6]
+<strong>Output:</strong> 15
+<strong>Explanation:</strong> We can select any triplet of items since for each triplet of indices i, j and k such that i &lt; j &lt; k, the condition holds.
+Therefore the maximum profit we can get would be the 3 most profitable items which are indices 1, 3 and 4.
+The answer would be sum of their profits which is 5 + 4 + 6 = 15.</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> prices = [4,3,2,1], profits = [33,20,19,87]
+<strong>Output:</strong> -1
+<strong>Explanation:</strong> We can&#39;t select any triplet of indices such that the condition holds, so we return -1.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>3 &lt;= prices.length == profits.length &lt;= 50000</code></li>
+	<li><code>1 &lt;= prices[i] &lt;= 5000</code></li>
+	<li><code>1 &lt;= profits[i] &lt;= 10<sup>6</sup></code></li>
+</ul>
+
+## Solutions
+
+<!-- tabs:start -->
+
+### **Python3**
+
+```python
+
+```
+
+### **Java**
+
+```java
+
+```
+
+### **C++**
+
+```cpp
+
+```
+
+### **Go**
+
+```go
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

solution/2900-2999/2922.Market Analysis III/README.md

@@ -0,0 +1,133 @@
+# [2922. Market Analysis III](https://leetcode.cn/problems/market-analysis-iii)
+
+[English Version](/solution/2900-2999/2922.Market%20Analysis%20III/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>Table: <code>Users</code></p>
+
+<pre>
++----------------+---------+
+| Column Name    | Type    |
++----------------+---------+
+| seller_id      | int     |
+| join_date      | date    |
+| favorite_brand | varchar |
++----------------+---------+
+seller_id is the primary key for this table.
+This table contains seller id, join date, and favorite brand of sellers.
+</pre>
+
+<p>Table: <code>Items</code></p>
+
+<pre>
++---------------+---------+
+| Column Name   | Type    |
++---------------+---------+
+| item_id       | int     |
+| item_brand    | varchar |
++---------------+---------+
+item_id is the primary key for this table.
+This table contains item id and item brand.</pre>
+
+<p>Table: <code>Orders</code></p>
+
+<pre>
++---------------+---------+
+| Column Name   | Type    |
++---------------+---------+
+| order_id      | int     |
+| order_date    | date    |
+| item_id       | int     |
+| seller_id     | int     |
++---------------+---------+
+order_id is the primary key for this table.
+item_id is a foreign key to the Items table.
+seller_id is a foreign key to the Users table.
+This table contains order id, order date, item id and seller id.</pre>
+
+<p>Write a solution to find the <strong>top seller</strong> who has sold the highest number of<strong> unique</strong> items with a <strong>different</strong> brand than their favorite brand. If there are multiple sellers with the same highest count, return all of them.</p>
+
+<p>Return <em>the result table ordered by</em> <code>seller_id</code> <em>in <strong>ascending</strong> order.</em></p>
+
+<p>The result format is in the following example.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> 
+Users table:
++-----------+------------+----------------+
+| seller_id | join_date  | favorite_brand |
++-----------+------------+----------------+
+| 1         | 2019-01-01 | Lenovo         |
+| 2         | 2019-02-09 | Samsung        |
+| 3         | 2019-01-19 | LG             |
++-----------+------------+----------------+
+Orders table:
++----------+------------+---------+-----------+
+| order_id | order_date | item_id | seller_id |
++----------+------------+---------+-----------+
+| 1        | 2019-08-01 | 4       | 2         |
+| 2        | 2019-08-02 | 2       | 3         |
+| 3        | 2019-08-03 | 3       | 3         |
+| 4        | 2019-08-04 | 1       | 2         |
+| 5        | 2019-08-04 | 4       | 2         |
++----------+------------+---------+-----------+
+Items table:
++---------+------------+
+| item_id | item_brand |
++---------+------------+
+| 1       | Samsung    |
+| 2       | Lenovo     |
+| 3       | LG         |
+| 4       | HP         |
++---------+------------+
+<strong>Output:</strong> 
++-----------+-----------+
+| seller_id | num_items |
++-----------+-----------+
+| 2         | 1         |
+| 3         | 1         |
++-----------+-----------+
+<strong>Explanation:</strong> 
+- The user with seller_id 2 has sold three items, but only two of them are not marked as a favorite. We will include a unique count of 1 because both of these items are identical.
+- The user with seller_id 3 has sold two items, but only one of them is not marked as a favorite. We will include just that non-favorite item in our count.
+Since seller_ids 2 and 3 have the same count of one item each, they both will be displayed in the output.</pre>
+
+## 解法
+
+<!-- 这里可写通用的实现逻辑 -->
+
+**方法一：等值连接 + 分组 + 子查询**
+
+我们可以使用等值连接，将 `Orders` 表和 `Users` 表按照 `seller_id` 进行连接，接着再按照 `item_id` 连接 `Items`，筛选出 `item_brand` 不等于 `favorite_brand` 的记录，然后按照 `seller_id` 进行分组，统计每个 `seller_id` 对应的 `item_id` 的个数，最后再使用子查询，找出 `item_id` 个数最多的 `seller_id`。
+
+<!-- tabs:start -->
+
+### **SQL**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```sql
+# Write your MySQL query statement below
+WITH
+    T AS (
+        SELECT seller_id, COUNT(DISTINCT item_id) AS num_items
+        FROM
+            Orders
+            JOIN Users USING (seller_id)
+            JOIN Items USING (item_id)
+        WHERE item_brand != favorite_brand
+        GROUP BY 1
+    )
+SELECT seller_id, num_items
+FROM T
+WHERE num_items = (SELECT MAX(num_items) FROM T)
+ORDER BY 1;
+```
+
+<!-- tabs:end -->