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‎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)
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[English Version](/solution/2700-2799/2793.Status%20of%20Flight%20Tickets/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>表：&nbsp;<code><font face="monospace">Flights</font></code></p>
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<pre>
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+-------------+------+
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| 列名 | 类型 |
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+-------------+------+
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| flight_id | int |
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| capacity | int |
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+-------------+------+
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flight_id 是该表的主键列。
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每行包含航班 id 和座位容量。
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</pre>
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<p>表：<code>Passengers</code></p>
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<pre>
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+--------------+----------+
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| 列名 | 类型 |
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+--------------+----------+
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| passenger_id | int |
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| flight_id | int |
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| booking_time | datetime |
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+--------------+----------+
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passenger_id 是该表的主键。
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booking_time 包含不同的值。
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每行包含乘客 id、预订时间和所预订的航班 id。
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</pre>
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<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>
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<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>
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<p>按 <code>passenger_id</code> <strong>升序排序&nbsp;</strong>返回结果表。</p>
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<p>查询结果的格式如下所示。</p>
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<p>&nbsp;</p>
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<p><strong class="example">示例 1：</strong></p>
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<pre>
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<b>输入：</b>
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Flights 表:
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+-----------+----------+
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| flight_id | capacity |
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+-----------+----------+
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| 1 | 2 |
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| 2 | 2 |
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| 3 | 1 |
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+-----------+----------+
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Passengers 表:
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+--------------+-----------+---------------------+
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| passenger_id | flight_id | booking_time |
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+--------------+-----------+---------------------+
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| 101 | 1 | 2023-07-10 16:30:00 |
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| 102 | 1 | 2023-07-10 17:45:00 |
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| 103 | 1 | 2023-07-10 12:00:00 |
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| 104 | 2 | 2023-07-05 13:23:00 |
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| 105 | 2 | 2023-07-05 09:00:00 |
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| 106 | 3 | 2023-07-08 11:10:00 |
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| 107 | 3 | 2023-07-08 09:10:00 |
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+--------------+-----------+---------------------+
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<b>输出：</b>
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+--------------+-----------+
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| passenger_id | Status |
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+--------------+-----------+
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| 101 | Confirmed |
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| 102 | Waitlist |
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| 103 | Confirmed |
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| 104 | Confirmed |
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| 105 | Confirmed |
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| 106 | Waitlist |
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| 107 | Confirmed |
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+--------------+-----------+
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<b>解释：</b>
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- 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。
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- 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。
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- 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>
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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/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)
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[中文文档](/solution/2700-2799/2793.Status%20of%20Flight%20Tickets/README.md)
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## Description
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<p>Table: <code><font face="monospace">Flights</font></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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| flight_id | int |
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| capacity | int |
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+-------------+------+
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<code>flight_id</code> is the primary key column for this table.
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Each row of this table contains flight id and capacity.
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</pre>
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<p>Table: <code>Passengers</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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| passenger_id | int |
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| flight_id | int |
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| booking_time | datetime |
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+--------------+----------+
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passenger_id is the primary key column for this table.
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booking_time column contains distinct values.
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Each row of this table contains passenger id, booking time, and their flight id.
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</pre>
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<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>
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<p>Write an SQL query to determine the current status of flight tickets for each passenger.</p>
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<p>Return <em>the result table ordered by</em> <code>passenger_id</code> <em>in <strong>ascending order</strong>.</em></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 class="example">Example 1:</strong></p>
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<pre>
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<strong>Input:</strong>
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Flights table:
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+-----------+----------+
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| flight_id | capacity |
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+-----------+----------+
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| 1 | 2 |
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| 2 | 2 |
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| 3 | 1 |
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+-----------+----------+
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Passengers table:
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+--------------+-----------+---------------------+
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| passenger_id | flight_id | booking_time |
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+--------------+-----------+---------------------+
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| 101 | 1 | 2023-07-10 16:30:00 |
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| 102 | 1 | 2023-07-10 17:45:00 |
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| 103 | 1 | 2023-07-10 12:00:00 |
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| 104 | 2 | 2023-07-05 13:23:00 |
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| 105 | 2 | 2023-07-05 09:00:00 |
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| 106 | 3 | 2023-07-08 11:10:00 |
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| 107 | 3 | 2023-07-08 09:10:00 |
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+--------------+-----------+---------------------+
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<strong>Output:</strong>
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+--------------+-----------+
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| passenger_id | Status |
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+--------------+-----------+
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| 101 | Confirmed |
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| 102 | Waitlist |
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| 103 | Confirmed |
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| 104 | Confirmed |
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| 105 | Confirmed |
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| 106 | Waitlist |
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| 107 | Confirmed |
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+--------------+-----------+
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<strong>Explanation:</strong>
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- 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,
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- 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.
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- 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.
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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/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)
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[English Version](/solution/2700-2799/2794.Create%20Object%20from%20Two%20Arrays/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>
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<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>
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<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>
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<p>&nbsp;</p>
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<p><strong class="example">示例 1：</strong></p>
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<pre>
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<b>输入：</b>arr1 = ["a", "b", "c"], arr2 = [1, 2, 3]
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<b>输出：</b>{"a": 1, "b": 2, "c": 3}
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<b>解释：</b>键 "a"、"b" 和 "c" 分别与值 1、2 和 3 配对。
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</pre>
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<p><strong class="example">示例 2：</strong></p>
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<pre>
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<b>输入：</b>arr1 = ["1", 1, false], arr2 = [4, 5, 6]
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<b>输出：</b>{"1": 4, "false": 6}
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<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。
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</pre>
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<p><strong class="example">示例 3：</strong></p>
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<pre>
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<b>输入：</b>arr1 = [], arr2 = []
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<b>输出：</b>{}
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<b>解释：</b>没有键，因此返回一个空对象。
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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>arr1 和 arr2 都是有效的 JSON 数组</code></li>
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<li><code>2 &lt;= JSON.stringify(arr1).length &lt;= 5 * 10<sup>5</sup></code></li>
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<li><code>2 &lt;= JSON.stringify(arr2).length &lt;= 5 * 10<sup>5</sup></code></li>
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<li><code>arr1.length === arr2.length</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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### **TypeScript**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```ts
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```
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<!-- 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)
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[中文文档](/solution/2700-2799/2794.Create%20Object%20from%20Two%20Arrays/README.md)
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## Description
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<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>
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<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>
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<p>- If the key is not a string, it should be converted into a string by calling `String()` on it.</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> arr1 = [&quot;a&quot;, &quot;b&quot;, &quot;c&quot;], arr2 = [1, 2, 3]
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<strong>Output:</strong> {&quot;a&quot;: 1, &quot;b&quot;: 2, &quot;c&quot;: 3}
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<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.
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</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> arr1 = [&quot;1&quot;, 1, false], arr2 = [4, 5, 6]
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<strong>Output:</strong> {&quot;1&quot;: 4, &quot;false&quot;: 6}
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<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.
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</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> arr1 = [], arr2 = []
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<strong>Output:</strong> {}
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<strong>Explanation:</strong> There are no keys so an empty object is returned.
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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>arr1 and arr2 are valid JSON arrays</code></li>
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<li><code>2 &lt;= JSON.stringify(arr1).length &lt;= 5 * 10<sup>5</sup></code></li>
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<li><code>2 &lt;= JSON.stringify(arr2).length &lt;= 5 * 10<sup>5</sup></code></li>
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<li><code>arr1.length === arr2.length</code></li>
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</ul>
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## Solutions
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<!-- tabs:start -->
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### **TypeScript**
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```ts
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```
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<!-- tabs:end -->

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