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feat: add solutions to lc problems: No.2293~2296
* No.2293.Min Max Game * No.2294.Partition Array Such That Maximum Difference Is K * No.2295.Replace Elements in an Array * No.2296.Design a Text Editor
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# [2291. Maximum Profit From Trading Stocks](https://leetcode.cn/problems/maximum-profit-from-trading-stocks)
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[English Version](/solution/2200-2299/2291.Maximum%20Profit%20From%20Trading%20Stocks/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>You are given two <strong>0-indexed</strong> integer arrays of the same length <code>present</code> and <code>future</code> where <code>present[i]</code> is the current price of the <code>i<sup>th</sup></code> stock and <code>future[i]</code> is the price of the <code>i<sup>th</sup></code> stock a year in the future. You may buy each stock at most <strong>once</strong>. You are also given an integer <code>budget</code> representing the amount of money you currently have.</p>
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<p>Return <em>the maximum amount of profit you can make.</em></p>
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<p>&nbsp;</p>
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<p><strong>Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> present = [5,4,6,2,3], future = [8,5,4,3,5], budget = 10
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<strong>Output:</strong> 6
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<strong>Explanation:</strong> One possible way to maximize your profit is to:
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Buy the 0<sup>th</sup>, 3<sup>rd</sup>, and 4<sup>th</sup> stocks for a total of 5 + 2 + 3 = 10.
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Next year, sell all three stocks for a total of 8 + 3 + 5 = 16.
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The profit you made is 16 - 10 = 6.
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It can be shown that the maximum profit you can make is 6.
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</pre>
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<p><strong>Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> present = [2,2,5], future = [3,4,10], budget = 6
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<strong>Output:</strong> 5
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<strong>Explanation:</strong> The only possible way to maximize your profit is to:
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Buy the 2<sup>nd</sup> stock, and make a profit of 10 - 5 = 5.
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It can be shown that the maximum profit you can make is 5.
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</pre>
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<p><strong>Example 3:</strong></p>
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<pre>
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<strong>Input:</strong> present = [3,3,12], future = [0,3,15], budget = 10
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<strong>Output:</strong> 0
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<strong>Explanation:</strong> One possible way to maximize your profit is to:
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Buy the 1<sup>st</sup> stock, and make a profit of 3 - 3 = 0.
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It can be shown that the maximum profit you can make is 0.
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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>n == present.length == future.length</code></li>
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<li><code>1 &lt;= n &lt;= 1000</code></li>
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<li><code>0 &lt;= present[i], future[i] &lt;= 100</code></li>
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<li><code>0 &lt;= budget &lt;= 1000</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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### **TypeScript**
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```ts
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```
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### **...**
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```
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```
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<!-- tabs:end -->
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# [2291. Maximum Profit From Trading Stocks](https://leetcode.com/problems/maximum-profit-from-trading-stocks)
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[中文文档](/solution/2200-2299/2291.Maximum%20Profit%20From%20Trading%20Stocks/README.md)
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## Description
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<p>You are given two <strong>0-indexed</strong> integer arrays of the same length <code>present</code> and <code>future</code> where <code>present[i]</code> is the current price of the <code>i<sup>th</sup></code> stock and <code>future[i]</code> is the price of the <code>i<sup>th</sup></code> stock a year in the future. You may buy each stock at most <strong>once</strong>. You are also given an integer <code>budget</code> representing the amount of money you currently have.</p>
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<p>Return <em>the maximum amount of profit you can make.</em></p>
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<p>&nbsp;</p>
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<p><strong>Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> present = [5,4,6,2,3], future = [8,5,4,3,5], budget = 10
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<strong>Output:</strong> 6
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<strong>Explanation:</strong> One possible way to maximize your profit is to:
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Buy the 0<sup>th</sup>, 3<sup>rd</sup>, and 4<sup>th</sup> stocks for a total of 5 + 2 + 3 = 10.
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Next year, sell all three stocks for a total of 8 + 3 + 5 = 16.
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The profit you made is 16 - 10 = 6.
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It can be shown that the maximum profit you can make is 6.
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</pre>
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<p><strong>Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> present = [2,2,5], future = [3,4,10], budget = 6
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<strong>Output:</strong> 5
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<strong>Explanation:</strong> The only possible way to maximize your profit is to:
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Buy the 2<sup>nd</sup> stock, and make a profit of 10 - 5 = 5.
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It can be shown that the maximum profit you can make is 5.
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</pre>
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<p><strong>Example 3:</strong></p>
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<pre>
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<strong>Input:</strong> present = [3,3,12], future = [0,3,15], budget = 10
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<strong>Output:</strong> 0
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<strong>Explanation:</strong> One possible way to maximize your profit is to:
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Buy the 1<sup>st</sup> stock, and make a profit of 3 - 3 = 0.
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It can be shown that the maximum profit you can make is 0.
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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>n == present.length == future.length</code></li>
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<li><code>1 &lt;= n &lt;= 1000</code></li>
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<li><code>0 &lt;= present[i], future[i] &lt;= 100</code></li>
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<li><code>0 &lt;= budget &lt;= 1000</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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### **TypeScript**
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```ts
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```
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### **...**
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```
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```
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<!-- tabs:end -->
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# [2292. Products With Three or More Orders in Two Consecutive Years](https://leetcode.cn/problems/products-with-three-or-more-orders-in-two-consecutive-years)
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[English Version](/solution/2200-2299/2292.Products%20With%20Three%20or%20More%20Orders%20in%20Two%20Consecutive%20Years/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>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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| product_id | int |
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| quantity | int |
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| purchase_date | date |
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+---------------+------+
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order_id is the primary key for this table.
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Each row in this table contains the ID of an order, the id of the product purchased, the quantity, and the purchase date.
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</pre>
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<p>&nbsp;</p>
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<p>Write an SQL query to report the IDs of all the products that were purchased three or more times in two consecutive years.</p>
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<p>Return the result table in <strong>any order</strong>.</p>
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<p>The query result format is in the following example.</p>
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<p>&nbsp;</p>
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<p><strong>Example 1:</strong></p>
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<pre>
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<strong>Input:</strong>
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Orders table:
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+----------+------------+----------+---------------+
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| order_id | product_id | quantity | purchase_date |
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+----------+------------+----------+---------------+
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| 1 | 1 | 7 | 2020年03月16日 |
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| 2 | 1 | 4 | 2020年12月02日 |
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| 3 | 1 | 7 | 2020年05月10日 |
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| 4 | 1 | 6 | 2021年12月23日 |
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| 5 | 1 | 5 | 2021年05月21日 |
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| 6 | 1 | 6 | 2021年10月11日 |
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| 7 | 2 | 6 | 2022年10月11日 |
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+----------+------------+----------+---------------+
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<strong>Output:</strong>
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+------------+
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| product_id |
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+------------+
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| 1 |
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+------------+
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<strong>Explanation:</strong>
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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.
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Product 2 was ordered one time in 2022. We do not include it in the answer.
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</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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<!-- 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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```
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<!-- tabs:end -->
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# [2292. Products With Three or More Orders in Two Consecutive Years](https://leetcode.com/problems/products-with-three-or-more-orders-in-two-consecutive-years)
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[中文文档](/solution/2200-2299/2292.Products%20With%20Three%20or%20More%20Orders%20in%20Two%20Consecutive%20Years/README.md)
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## Description
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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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| product_id | int |
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| quantity | int |
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| purchase_date | date |
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+---------------+------+
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order_id is the primary key for this table.
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Each row in this table contains the ID of an order, the id of the product purchased, the quantity, and the purchase date.
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</pre>
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<p>&nbsp;</p>
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<p>Write an SQL query to report the IDs of all the products that were purchased three or more times in two consecutive years.</p>
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<p>Return the result table in <strong>any order</strong>.</p>
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<p>The query result format is in the following example.</p>
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<p>&nbsp;</p>
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<p><strong>Example 1:</strong></p>
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<pre>
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<strong>Input:</strong>
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Orders table:
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+----------+------------+----------+---------------+
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| order_id | product_id | quantity | purchase_date |
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+----------+------------+----------+---------------+
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| 1 | 1 | 7 | 2020年03月16日 |
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| 2 | 1 | 4 | 2020年12月02日 |
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| 3 | 1 | 7 | 2020年05月10日 |
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| 4 | 1 | 6 | 2021年12月23日 |
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| 5 | 1 | 5 | 2021年05月21日 |
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| 6 | 1 | 6 | 2021年10月11日 |
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| 7 | 2 | 6 | 2022年10月11日 |
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+----------+------------+----------+---------------+
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<strong>Output:</strong>
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+------------+
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| product_id |
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+------------+
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| 1 |
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+------------+
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<strong>Explanation:</strong>
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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.
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Product 2 was ordered one time in 2022. We do not include it in the answer.
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</pre>
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## Solutions
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<!-- tabs:start -->
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### **SQL**
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```sql
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```
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<!-- tabs:end -->

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