chore: add new lc problems

solution/2700-2799/2793.Status of Flight Tickets/README.md

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+# [2793. 航班机票状态](https://leetcode.cn/problems/status-of-flight-tickets)
+
+[English Version](/solution/2700-2799/2793.Status%20of%20Flight%20Tickets/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>表：&nbsp;<code><font face="monospace">Flights</font></code></p>
+
+<pre>
++-------------+------+
+| 列名        | 类型 |
++-------------+------+
+| flight_id   | int  |
+| capacity    | int  |
++-------------+------+
+flight_id 是该表的主键列。 
+每行包含航班 id 和座位容量。
+</pre>
+
+<p>表：<code>Passengers</code></p>
+
+<pre>
++--------------+----------+
+| 列名         | 类型 |
++--------------+----------+
+| passenger_id | int      |
+| flight_id    | int      |
+| booking_time | datetime |
++--------------+----------+
+passenger_id 是该表的主键。 
+booking_time 包含不同的值。
+每行包含乘客 id、预订时间和所预订的航班 id。
+</pre>
+
+<p>乘客提前预订航班机票。如果乘客预订了一张航班机票，并且航班上还有空座位，则乘客的机票将 <strong>得到确认</strong> 。然而，如果航班已经满员，乘客将被列入 <strong>等候名单</strong> 。</p>
+
+<p>编写一个 SQL 查询来确定每个乘客航班机票的当前状态。</p>
+
+<p>按 <code>passenger_id</code> <strong>升序排序&nbsp;</strong>返回结果表。</p>
+
+<p>查询结果的格式如下所示。</p>
+
+<p>&nbsp;</p>
+
+<p><strong class="example">示例 1：</strong></p>
+
+<pre>
+<b>输入：</b>
+Flights 表:
++-----------+----------+
+| flight_id | capacity |
++-----------+----------+
+| 1         | 2        |
+| 2         | 2        |
+| 3         | 1        |
++-----------+----------+
+Passengers 表:
++--------------+-----------+---------------------+
+| passenger_id | flight_id | booking_time        |
++--------------+-----------+---------------------+
+| 101          | 1         | 2023-07-10 16:30:00 |
+| 102          | 1         | 2023-07-10 17:45:00 |
+| 103          | 1         | 2023-07-10 12:00:00 |
+| 104          | 2         | 2023-07-05 13:23:00 |
+| 105          | 2         | 2023-07-05 09:00:00 |
+| 106          | 3         | 2023-07-08 11:10:00 |
+| 107          | 3         | 2023-07-08 09:10:00 |
++--------------+-----------+---------------------+
+<b>输出：</b>
++--------------+-----------+
+| passenger_id | Status    |
++--------------+-----------+
+| 101          | Confirmed | 
+| 102          | Waitlist  | 
+| 103          | Confirmed | 
+| 104          | Confirmed | 
+| 105          | Confirmed | 
+| 106          | Waitlist  | 
+| 107          | Confirmed | 
++--------------+-----------+
+<b>解释：</b>
+- 航班 1 的容量为 2 位乘客。乘客 101 和乘客 103 是最先预订机票的，已经确认他们的预订。然而，乘客 102 是第三位预订该航班的乘客，这意味着没有更多的可用座位。乘客 102 现在被列入等候名单。
+- 航班 2 的容量为 2 位乘客，已经有两位乘客预订了机票，乘客 104 和乘客 105。由于预订机票的乘客数与可用座位数相符，这两个预订都得到了确认。
+- 航班 3 的容量为 1 位乘客，乘客 107 先预订并获得了唯一的可用座位，确认了他们的预订。预订时间在乘客 107 之后的乘客 106 被列入等候名单。</pre>
+
+## 解法
+
+<!-- 这里可写通用的实现逻辑 -->
+
+<!-- tabs:start -->
+
+### **SQL**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```sql
+
+```
+
+<!-- tabs:end -->

solution/2700-2799/2793.Status of Flight Tickets/README_EN.md

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+# [2793. Status of Flight Tickets](https://leetcode.com/problems/status-of-flight-tickets)
+
+[中文文档](/solution/2700-2799/2793.Status%20of%20Flight%20Tickets/README.md)
+
+## Description
+
+<p>Table: <code><font face="monospace">Flights</font></code></p>
+
+<pre>
++-------------+------+
+| Column Name | Type |
++-------------+------+
+| flight_id   | int  |
+| capacity    | int  |
++-------------+------+
+<code>flight_id</code> is the primary key column for this table.
+Each row of this table contains flight id and capacity.
+</pre>
+
+<p>Table: <code>Passengers</code></p>
+
+<pre>
++--------------+----------+
+| Column Name  | Type     |
++--------------+----------+
+| passenger_id | int      |
+| flight_id    | int      |
+| booking_time | datetime |
++--------------+----------+
+passenger_id is the primary key column for this table.
+booking_time column contains distinct values.
+Each row of this table contains passenger id, booking time, and their flight id.
+</pre>
+
+<p>Passengers book tickets for flights in advance. If a passenger books a ticket for a flight and there are still empty seats available on the flight, the passenger&#39;s ticket will be <strong>confirmed</strong>. However, the passenger will be on a <strong>waitlist</strong> if the flight is already at full capacity.</p>
+
+<p>Write an SQL query to determine the current status of flight tickets for each passenger.</p>
+
+<p>Return <em>the result table ordered by</em> <code>passenger_id</code> <em>in <strong>ascending order</strong>.</em></p>
+
+<p>The query result format is in the following example.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> 
+Flights table:
++-----------+----------+
+| flight_id | capacity |
++-----------+----------+
+| 1         | 2        |
+| 2         | 2        |
+| 3         | 1        |
++-----------+----------+
+Passengers table:
++--------------+-----------+---------------------+
+| passenger_id | flight_id | booking_time        |
++--------------+-----------+---------------------+
+| 101          | 1         | 2023-07-10 16:30:00 |
+| 102          | 1         | 2023-07-10 17:45:00 |
+| 103          | 1         | 2023-07-10 12:00:00 |
+| 104          | 2         | 2023-07-05 13:23:00 |
+| 105          | 2         | 2023-07-05 09:00:00 |
+| 106          | 3         | 2023-07-08 11:10:00 |
+| 107          | 3         | 2023-07-08 09:10:00 |
++--------------+-----------+---------------------+
+<strong>Output:</strong> 
++--------------+-----------+
+| passenger_id | Status    |
++--------------+-----------+
+| 101          | Confirmed | 
+| 102          | Waitlist  | 
+| 103          | Confirmed | 
+| 104          | Confirmed | 
+| 105          | Confirmed | 
+| 106          | Waitlist  | 
+| 107          | Confirmed | 
++--------------+-----------+
+<strong>Explanation:</strong> 
+- Flight 1 has a capacity of 2 passengers. Passenger 101 and Passenger 103 were the first to book tickets, securing the available seats. Therefore, their bookings are confirmed. However, Passenger 102 was the third person to book a ticket for this flight, which means there are no more available seats. Passenger 102 is now placed on the waitlist, 
+- Flight 2 has a capacity of 2 passengers, Flight 2 has exactly two passengers who booked tickets,  Passenger 104 and Passenger 105. Since the number of passengers who booked tickets matches the available seats, both bookings are confirmed.
+- Flight 3 has a capacity of 1 passenger. Passenger 107 booked earlier and secured the only available seat, confirming their booking. Passenger 106, who booked after Passenger 107, is on the waitlist.
+</pre>
+
+## Solutions
+
+<!-- tabs:start -->
+
+### **SQL**
+
+```sql
+
+```
+
+<!-- tabs:end -->

solution/2700-2799/2794.Create Object from Two Arrays/README.md

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+# [2794. 从两个数组中创建对象](https://leetcode.cn/problems/create-object-from-two-arrays)
+
+[English Version](/solution/2700-2799/2794.Create%20Object%20from%20Two%20Arrays/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>给定两个数组 <code>keysArr </code>和 <code>valuesArr</code>，返回一个新的对象 <code>obj</code>。<code>obj</code> 中的每个键值对都来自 <code>keysArr[i]</code> 和 <code>valuesArr[i]</code>。</p>
+
+<p>- 如果前面的索引中存在重复的键，则应该跳过该键值对。换句话说，只有第一次出现的键会被添加到对象中。</p>
+
+<p>- 如果键不是字符串，则应通过调用 <code>String()</code> 方法将其转换为字符串。</p>
+
+<p>&nbsp;</p>
+
+<p><strong class="example">示例 1：</strong></p>
+
+<pre>
+<b>输入：</b>arr1 = ["a", "b", "c"], arr2 = [1, 2, 3]
+<b>输出：</b>{"a": 1, "b": 2, "c": 3}
+<b>解释：</b>键 "a"、"b" 和 "c" 分别与值 1、2 和 3 配对。
+</pre>
+
+<p><strong class="example">示例 2：</strong></p>
+
+<pre>
+<b>输入：</b>arr1 = ["1", 1, false], arr2 = [4, 5, 6]
+<b>输出：</b>{"1": 4, "false": 6}
+<b>解释：</b>首先，将 arr1 中的所有元素转换为字符串。我们可以看到有两个 "1" 的出现。使用第一次出现 "1" 的关联值：4。
+</pre>
+
+<p><strong class="example">示例 3：</strong></p>
+
+<pre>
+<b>输入：</b>arr1 = [], arr2 = []
+<b>输出：</b>{}
+<b>解释：</b>没有键，因此返回一个空对象。
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>arr1 和 arr2 都是有效的 JSON 数组</code></li>
+	<li><code>2 &lt;= JSON.stringify(arr1).length &lt;= 5 * 10<sup>5</sup></code></li>
+	<li><code>2 &lt;= JSON.stringify(arr2).length &lt;= 5 * 10<sup>5</sup></code></li>
+	<li><code>arr1.length === arr2.length</code></li>
+</ul>
+
+## 解法
+
+<!-- 这里可写通用的实现逻辑 -->
+
+<!-- tabs:start -->
+
+### **TypeScript**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```ts
+
+```
+
+<!-- tabs:end -->

solution/2700-2799/2794.Create Object from Two Arrays/README_EN.md

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+# [2794. Create Object from Two Arrays](https://leetcode.com/problems/create-object-from-two-arrays)
+
+[中文文档](/solution/2700-2799/2794.Create%20Object%20from%20Two%20Arrays/README.md)
+
+## Description
+
+<p>Given two arrays <code>keysArr</code> and <code>valuesArr</code>, return a new object <code>obj</code>. Each key-value pair in&nbsp;<code>obj</code>&nbsp;should come from&nbsp;<code>keysArr[i]</code>&nbsp;and&nbsp;<code>valuesArr[i]</code>.</p>
+
+<p>- If a duplicate key exists at a previous index, that key-value should be excluded. In other words, only the first key should be added to the object.</p>
+
+<p>- If the key is not a string, it should be converted into a string by calling `String()` on it.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> arr1 = [&quot;a&quot;, &quot;b&quot;, &quot;c&quot;], arr2 = [1, 2, 3]
+<strong>Output:</strong> {&quot;a&quot;: 1, &quot;b&quot;: 2, &quot;c&quot;: 3}
+<strong>Explanation:</strong> The keys &quot;a&quot;, &quot;b&quot;, and &quot;c&quot; are paired with the values 1, 2, and 3 respectively.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> arr1 = [&quot;1&quot;, 1, false], arr2 = [4, 5, 6]
+<strong>Output:</strong> {&quot;1&quot;: 4, &quot;false&quot;: 6}
+<strong>Explanation:</strong> First, all the elements in arr1 are converted into strings. We can see there are two occurrences of &quot;1&quot;. The value associated with the first occurrence of &quot;1&quot; is used: 4.
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> arr1 = [], arr2 = []
+<strong>Output:</strong> {}
+<strong>Explanation:</strong> There are no keys so an empty object is returned.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>arr1 and arr2 are valid JSON arrays</code></li>
+	<li><code>2 &lt;= JSON.stringify(arr1).length &lt;= 5 * 10<sup>5</sup></code></li>
+	<li><code>2 &lt;= JSON.stringify(arr2).length &lt;= 5 * 10<sup>5</sup></code></li>
+	<li><code>arr1.length === arr2.length</code></li>
+</ul>
+
+## Solutions
+
+<!-- tabs:start -->
+
+### **TypeScript**
+
+```ts
+
+```
+
+<!-- tabs:end -->