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feat: add sql solution to lc problem: No.2922 (#1918)
No.2922.Market Analysis III
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‎solution/2900-2999/2921.Maximum Profitable Triplets With Increasing Prices II/README.md

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# [2921. Maximum Profitable Triplets With Increasing Prices II](https://leetcode.cn/problems/maximum-profitable-triplets-with-increasing-prices-ii)
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[English Version](/solution/2900-2999/2921.Maximum%20Profitable%20Triplets%20With%20Increasing%20Prices%20II/README_EN.md)
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## 题目描述
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<!-- 这里写题目描述 -->
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<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>
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<p>We have to pick three items with the following condition:</p>
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<ul>
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<li><code>prices[i] &lt; prices[j] &lt; prices[k]</code> where <code>i &lt; j &lt; k</code>.</li>
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</ul>
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<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>
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<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>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> prices = [10,2,3,4], profits = [100,2,7,10]
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<strong>Output:</strong> 19
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<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.
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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].
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The answer would be sum of their profits which is 2 + 7 + 10 = 19.</pre>
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<p><strong class="example">Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> prices = [1,2,3,4,5], profits = [1,5,3,4,6]
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<strong>Output:</strong> 15
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<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.
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Therefore the maximum profit we can get would be the 3 most profitable items which are indices 1, 3 and 4.
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The answer would be sum of their profits which is 5 + 4 + 6 = 15.</pre>
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<p><strong class="example">Example 3:</strong></p>
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<pre>
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<strong>Input:</strong> prices = [4,3,2,1], profits = [33,20,19,87]
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<strong>Output:</strong> -1
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<strong>Explanation:</strong> We can&#39;t select any triplet of indices such that the condition holds, so we return -1.
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</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>3 &lt;= prices.length == profits.length &lt;= 50000</code></li>
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<li><code>1 &lt;= prices[i] &lt;= 5000</code></li>
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<li><code>1 &lt;= profits[i] &lt;= 10<sup>6</sup></code></li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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```
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### **C++**
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```cpp
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```
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### **Go**
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```go
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```
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### **...**
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```
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```
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<!-- tabs:end -->

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

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# [2921. Maximum Profitable Triplets With Increasing Prices II](https://leetcode.com/problems/maximum-profitable-triplets-with-increasing-prices-ii)
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[中文文档](/solution/2900-2999/2921.Maximum%20Profitable%20Triplets%20With%20Increasing%20Prices%20II/README.md)
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## Description
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<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>
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<p>We have to pick three items with the following condition:</p>
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<ul>
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<li><code>prices[i] &lt; prices[j] &lt; prices[k]</code> where <code>i &lt; j &lt; k</code>.</li>
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</ul>
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<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>
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<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>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> prices = [10,2,3,4], profits = [100,2,7,10]
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<strong>Output:</strong> 19
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<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.
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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].
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The answer would be sum of their profits which is 2 + 7 + 10 = 19.</pre>
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<p><strong class="example">Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> prices = [1,2,3,4,5], profits = [1,5,3,4,6]
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<strong>Output:</strong> 15
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<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.
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Therefore the maximum profit we can get would be the 3 most profitable items which are indices 1, 3 and 4.
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The answer would be sum of their profits which is 5 + 4 + 6 = 15.</pre>
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<p><strong class="example">Example 3:</strong></p>
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<pre>
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<strong>Input:</strong> prices = [4,3,2,1], profits = [33,20,19,87]
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<strong>Output:</strong> -1
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<strong>Explanation:</strong> We can&#39;t select any triplet of indices such that the condition holds, so we return -1.
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</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>3 &lt;= prices.length == profits.length &lt;= 50000</code></li>
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<li><code>1 &lt;= prices[i] &lt;= 5000</code></li>
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<li><code>1 &lt;= profits[i] &lt;= 10<sup>6</sup></code></li>
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</ul>
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## Solutions
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<!-- tabs:start -->
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### **Python3**
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```python
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```
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### **Java**
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```java
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```
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### **C++**
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```cpp
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```
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### **Go**
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```go
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```
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### **...**
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```
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```
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<!-- tabs:end -->

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

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# [2922. Market Analysis III](https://leetcode.cn/problems/market-analysis-iii)
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[English Version](/solution/2900-2999/2922.Market%20Analysis%20III/README_EN.md)
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## 题目描述
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<!-- 这里写题目描述 -->
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<p>Table: <code>Users</code></p>
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<pre>
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+----------------+---------+
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| Column Name | Type |
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+----------------+---------+
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| seller_id | int |
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| join_date | date |
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| favorite_brand | varchar |
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+----------------+---------+
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seller_id is the primary key for this table.
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This table contains seller id, join date, and favorite brand of sellers.
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</pre>
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<p>Table: <code>Items</code></p>
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<pre>
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+---------------+---------+
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| Column Name | Type |
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+---------------+---------+
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| item_id | int |
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| item_brand | varchar |
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+---------------+---------+
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item_id is the primary key for this table.
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This table contains item id and item brand.</pre>
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<p>Table: <code>Orders</code></p>
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<pre>
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+---------------+---------+
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| Column Name | Type |
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+---------------+---------+
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| order_id | int |
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| order_date | date |
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| item_id | int |
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| seller_id | int |
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+---------------+---------+
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order_id is the primary key for this table.
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item_id is a foreign key to the Items table.
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seller_id is a foreign key to the Users table.
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This table contains order id, order date, item id and seller id.</pre>
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<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>
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<p>Return <em>the result table ordered by</em> <code>seller_id</code> <em>in <strong>ascending</strong> order.</em></p>
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<p>The result format is in the following example.</p>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<pre>
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<strong>Input:</strong>
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Users table:
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+-----------+------------+----------------+
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| seller_id | join_date | favorite_brand |
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+-----------+------------+----------------+
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| 1 | 2019-01-01 | Lenovo |
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| 2 | 2019-02-09 | Samsung |
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| 3 | 2019-01-19 | LG |
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+-----------+------------+----------------+
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Orders table:
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+----------+------------+---------+-----------+
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| order_id | order_date | item_id | seller_id |
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+----------+------------+---------+-----------+
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| 1 | 2019-08-01 | 4 | 2 |
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| 2 | 2019-08-02 | 2 | 3 |
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| 3 | 2019-08-03 | 3 | 3 |
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| 4 | 2019-08-04 | 1 | 2 |
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| 5 | 2019-08-04 | 4 | 2 |
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+----------+------------+---------+-----------+
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Items table:
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+---------+------------+
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| item_id | item_brand |
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+---------+------------+
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| 1 | Samsung |
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| 2 | Lenovo |
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| 3 | LG |
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| 4 | HP |
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+---------+------------+
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<strong>Output:</strong>
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+-----------+-----------+
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| seller_id | num_items |
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+-----------+-----------+
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| 2 | 1 |
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| 3 | 1 |
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+-----------+-----------+
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<strong>Explanation:</strong>
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- 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.
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- 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.
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Since seller_ids 2 and 3 have the same count of one item each, they both will be displayed in the output.</pre>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：等值连接 + 分组 + 子查询**
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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`。
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<!-- tabs:start -->
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### **SQL**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```sql
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# Write your MySQL query statement below
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WITH
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T AS (
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SELECT seller_id, COUNT(DISTINCT item_id) AS num_items
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FROM
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Orders
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JOIN Users USING (seller_id)
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JOIN Items USING (item_id)
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WHERE item_brand != favorite_brand
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GROUP BY 1
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)
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SELECT seller_id, num_items
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FROM T
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WHERE num_items = (SELECT MAX(num_items) FROM T)
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ORDER BY 1;
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```
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<!-- tabs:end -->

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