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feat: add solutions to lc problems: No.2363~2370
* No.2363.Merge Similar Items * No.2364.Count Number of Bad Pairs * No.2365.Task Scheduler II * No.2366.Minimum Replacements to Sort the Array * No.2367.Number of Arithmetic Triplets * No.2368.Reachable Nodes With Restrictions * No.2369.Check if There is a Valid Partition For The Array * No.2370.Longest Ideal Subsequence
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‎solution/1400-1499/1401.Circle and Rectangle Overlapping/Solution.py

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class Solution:
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def checkOverlap(self, radius: int, xCenter: int, yCenter: int, x1: int, y1: int, x2: int, y2: int) -> bool:
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def checkOverlap(
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self,
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radius: int,
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xCenter: int,
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yCenter: int,
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x1: int,
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y1: int,
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x2: int,
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y2: int,
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) -> bool:
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dx = dy = 0
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if x1 > xCenter:
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dx = xCenter - x1

‎solution/2300-2399/2362.Generate the Invoice/README.md

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# [2362. Generate the Invoice](https://leetcode.cn/problems/generate-the-invoice)
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[English Version](/solution/2300-2399/2362.Generate%20the%20Invoice/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>Products</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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| product_id | int |
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| price | int |
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+-------------+------+
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product_id is the primary key for this table.
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Each row in this table shows the ID of a product and the price of one unit.
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</pre>
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<p>&nbsp;</p>
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<p>Table: <code>Purchases</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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| invoice_id | int |
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| product_id | int |
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| quantity | int |
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+-------------+------+
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(invoice_id, product_id) is the primary key for this table.
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Each row in this table shows the quantity ordered from one product in an invoice.
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</pre>
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<p>&nbsp;</p>
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<p>Write an SQL query to show the details of the invoice with the highest price. If two or more invoices have the same price, return the details of the one with the smallest <code>invoice_id</code>.</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 shown 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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Products table:
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+------------+-------+
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| product_id | price |
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+------------+-------+
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| 1 | 100 |
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| 2 | 200 |
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+------------+-------+
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Purchases table:
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+------------+------------+----------+
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| invoice_id | product_id | quantity |
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+------------+------------+----------+
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| 1 | 1 | 2 |
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| 3 | 2 | 1 |
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| 2 | 2 | 3 |
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| 2 | 1 | 4 |
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| 4 | 1 | 10 |
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+------------+------------+----------+
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<strong>Output:</strong>
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+------------+----------+-------+
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| product_id | quantity | price |
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+------------+----------+-------+
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| 2 | 3 | 600 |
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| 1 | 4 | 400 |
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+------------+----------+-------+
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<strong>Explanation:</strong>
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Invoice 1: price = (2 * 100) = $200
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Invoice 2: price = (4 * 100) + (3 * 200) = $1000
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Invoice 3: price = (1 * 200) = $200
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Invoice 4: price = (10 * 100) = $1000
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The highest price is $1000, and the invoices with the highest prices are 2 and 4. We return the details of the one with the smallest ID, which is invoice 2.
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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 -->

‎solution/2300-2399/2362.Generate the Invoice/README_EN.md

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# [2362. Generate the Invoice](https://leetcode.com/problems/generate-the-invoice)
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[中文文档](/solution/2300-2399/2362.Generate%20the%20Invoice/README.md)
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## Description
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<p>Table: <code>Products</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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| product_id | int |
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| price | int |
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+-------------+------+
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product_id is the primary key for this table.
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Each row in this table shows the ID of a product and the price of one unit.
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</pre>
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<p>&nbsp;</p>
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<p>Table: <code>Purchases</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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| invoice_id | int |
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| product_id | int |
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| quantity | int |
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+-------------+------+
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(invoice_id, product_id) is the primary key for this table.
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Each row in this table shows the quantity ordered from one product in an invoice.
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</pre>
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<p>&nbsp;</p>
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<p>Write an SQL query to show the details of the invoice with the highest price. If two or more invoices have the same price, return the details of the one with the smallest <code>invoice_id</code>.</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 shown 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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Products table:
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+------------+-------+
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| product_id | price |
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+------------+-------+
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| 1 | 100 |
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| 2 | 200 |
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+------------+-------+
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Purchases table:
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+------------+------------+----------+
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| invoice_id | product_id | quantity |
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+------------+------------+----------+
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| 1 | 1 | 2 |
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| 3 | 2 | 1 |
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| 2 | 2 | 3 |
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| 2 | 1 | 4 |
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| 4 | 1 | 10 |
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+------------+------------+----------+
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<strong>Output:</strong>
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+------------+----------+-------+
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| product_id | quantity | price |
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+------------+----------+-------+
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| 2 | 3 | 600 |
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| 1 | 4 | 400 |
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+------------+----------+-------+
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<strong>Explanation:</strong>
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Invoice 1: price = (2 * 100) = $200
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Invoice 2: price = (4 * 100) + (3 * 200) = $1000
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Invoice 3: price = (1 * 200) = $200
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Invoice 4: price = (10 * 100) = $1000
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The highest price is $1000, and the invoices with the highest prices are 2 and 4. We return the details of the one with the smallest ID, which is invoice 2.
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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 -->

‎solution/2300-2399/2363.Merge Similar Items/README.md

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# [2363. 合并相似的物品](https://leetcode.cn/problems/merge-similar-items)
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[English Version](/solution/2300-2399/2363.Merge%20Similar%20Items/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>给你两个二维整数数组&nbsp;<code>items1</code> 和&nbsp;<code>items2</code>&nbsp;，表示两个物品集合。每个数组&nbsp;<code>items</code>&nbsp;有以下特质：</p>
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<ul>
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<li><code>items[i] = [value<sub>i</sub>, weight<sub>i</sub>]</code> 其中&nbsp;<code>value<sub>i</sub></code>&nbsp;表示第&nbsp;<code>i</code>&nbsp;件物品的&nbsp;<strong>价值</strong>&nbsp;，<code>weight<sub>i</sub></code>&nbsp;表示第 <code>i</code>&nbsp;件物品的 <strong>重量</strong>&nbsp;。</li>
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<li><code>items</code>&nbsp;中每件物品的价值都是 <strong>唯一的</strong>&nbsp;。</li>
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</ul>
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<p>请你返回一个二维数组&nbsp;<code>ret</code>，其中&nbsp;<code>ret[i] = [value<sub>i</sub>, weight<sub>i</sub>]</code>，&nbsp;<code>weight<sub>i</sub></code>&nbsp;是所有价值为&nbsp;<code>value<sub>i</sub></code><sub>&nbsp;</sub>物品的&nbsp;<strong>重量之和</strong>&nbsp;。</p>
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<p><strong>注意：</strong><code>ret</code>&nbsp;应该按价值 <strong>升序</strong>&nbsp;排序后返回。</p>
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<p>&nbsp;</p>
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<p><strong>示例 1：</strong></p>
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<pre>
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<b>输入：</b>items1 = [[1,1],[4,5],[3,8]], items2 = [[3,1],[1,5]]
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<b>输出：</b>[[1,6],[3,9],[4,5]]
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<b>解释：</b>
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value = 1 的物品在 items1 中 weight = 1 ，在 items2 中 weight = 5 ，总重量为 1 + 5 = 6 。
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value = 3 的物品再 items1 中 weight = 8 ，在 items2 中 weight = 1 ，总重量为 8 + 1 = 9 。
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value = 4 的物品在 items1 中 weight = 5 ，总重量为 5 。
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所以，我们返回 [[1,6],[3,9],[4,5]] 。
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</pre>
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<p><strong>示例 2：</strong></p>
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<pre>
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<b>输入：</b>items1 = [[1,1],[3,2],[2,3]], items2 = [[2,1],[3,2],[1,3]]
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<b>输出：</b>[[1,4],[2,4],[3,4]]
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<b>解释：</b>
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value = 1 的物品在 items1 中 weight = 1 ，在 items2 中 weight = 3 ，总重量为 1 + 3 = 4 。
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value = 2 的物品在 items1 中 weight = 3 ，在 items2 中 weight = 1 ，总重量为 3 + 1 = 4 。
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value = 3 的物品在 items1 中 weight = 2 ，在 items2 中 weight = 2 ，总重量为 2 + 2 = 4 。
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所以，我们返回 [[1,4],[2,4],[3,4]] 。</pre>
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<p><strong>示例 3：</strong></p>
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<pre>
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<b>输入：</b>items1 = [[1,3],[2,2]], items2 = [[7,1],[2,2],[1,4]]
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<b>输出：</b>[[1,7],[2,4],[7,1]]
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<strong>解释：
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</strong>value = 1 的物品在 items1 中 weight = 3 ，在 items2 中 weight = 4 ，总重量为 3 + 4 = 7 。
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value = 2 的物品在 items1 中 weight = 2 ，在 items2 中 weight = 2 ，总重量为 2 + 2 = 4 。
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value = 7 的物品在 items2 中 weight = 1 ，总重量为 1 。
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所以，我们返回 [[1,7],[2,4],[7,1]] 。
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</pre>
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<p>&nbsp;</p>
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<p><strong>提示：</strong></p>
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<ul>
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<li><code>1 &lt;= items1.length, items2.length &lt;= 1000</code></li>
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<li><code>items1[i].length == items2[i].length == 2</code></li>
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<li><code>1 &lt;= value<sub>i</sub>, weight<sub>i</sub> &lt;= 1000</code></li>
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<li><code>items1</code>&nbsp;中每个 <code>value<sub>i</sub></code>&nbsp;都是 <b>唯一的</b>&nbsp;。</li>
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<li><code>items2</code>&nbsp;中每个 <code>value<sub>i</sub></code>&nbsp;都是 <b>唯一的</b>&nbsp;。</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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class Solution:
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def mergeSimilarItems(self, items1: List[List[int]], items2: List[List[int]]) -> List[List[int]]:
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cnt = [0] * 1010
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for v, w in chain(items1, items2):
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cnt[v] += w
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return [[i, v] for i, v in enumerate(cnt) if v]
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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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class Solution {
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public List<List<Integer>> mergeSimilarItems(int[][] items1, int[][] items2) {
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int[] cnt = new int[1010];
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for (int[] e : items1) {
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cnt[e[0]] += e[1];
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}
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for (int[] e : items2) {
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cnt[e[0]] += e[1];
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}
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List<List<Integer>> ans = new ArrayList<>();
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for (int i = 0; i < cnt.length; ++i) {
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if (cnt[i] > 0) {
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ans.add(Arrays.asList(i, cnt[i]));
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}
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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vector<vector<int>> mergeSimilarItems(vector<vector<int>>& items1, vector<vector<int>>& items2) {
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vector<int> cnt(1010);
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for (auto& e : items1) cnt[e[0]] += e[1];
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for (auto& e : items2) cnt[e[0]] += e[1];
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vector<vector<int>> ans;
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for (int i = 0; i < cnt.size(); ++i) if (cnt[i]) ans.push_back({i, cnt[i]});
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return ans;
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}
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};
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```
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### **Go**
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```go
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func mergeSimilarItems(items1 [][]int, items2 [][]int) [][]int {
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cnt := make([]int, 1010)
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for _, e := range items1 {
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cnt[e[0]] += e[1]
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}
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for _, e := range items2 {
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cnt[e[0]] += e[1]
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}
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ans := [][]int{}
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for i, v := range cnt {
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if v > 0 {
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ans = append(ans, []int{i, v})
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}
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}
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return ans
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}
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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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