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‎solution/2600-2699/2686.Immediate Food Delivery III/README.md

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# [2686. Immediate Food Delivery III](https://leetcode.cn/problems/immediate-food-delivery-iii)
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[English Version](/solution/2600-2699/2686.Immediate%20Food%20Delivery%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>Delivery</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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| delivery_id | int |
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| customer_id | int |
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| order_date | date |
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| customer_pref_delivery_date | date |
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+-----------------------------+---------+
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delivery_id is the primary key of this table.
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Each row contains information about food delivery to a customer that makes an order at some date and specifies a preferred delivery date (on the order date or after it).
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</pre>
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<p>If the customer&#39;s preferred delivery date is the same as the order date, then the order is called <strong>immediate,</strong>&nbsp;otherwise, it is <strong>scheduled</strong>.</p>
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<p>Write an SQL query to find the percentage of immediate orders on each unique&nbsp;<code>order_date</code>, <strong>rounded to 2 decimal places</strong>.&nbsp;</p>
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<p>Return <em>the result table ordered by</em> <code>order_date</code> <em>in <strong>ascending</strong> order.</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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Delivery table:
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+-------------+-------------+------------+-----------------------------+
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| delivery_id | customer_id | order_date | customer_pref_delivery_date |
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+-------------+-------------+------------+-----------------------------+
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| 1 | 1 | 2019-08-01 | 2019-08-02 |
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| 2 | 2 | 2019-08-01 | 2019-08-01 |
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| 3 | 1 | 2019-08-01 | 2019-08-01 |
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| 4 | 3 | 2019-08-02 | 2019-08-13 |
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| 5 | 3 | 2019-08-02 | 2019-08-02 |
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| 6 | 2 | 2019-08-02 | 2019-08-02 |
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| 7 | 4 | 2019-08-03 | 2019-08-03 |
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| 8 | 1 | 2019-08-03 | 2019-08-03 |
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| 9 | 5 | 2019-08-04 | 2019-08-08 |
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| 10 | 2 | 2019-08-04 | 2019-08-18 |
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+-------------+-------------+------------+-----------------------------+
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<strong>Output:</strong>
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+------------+----------------------+
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| order_date | immediate_percentage |
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+------------+----------------------+
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| 2019-08-01 | 66.67 |
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| 2019-08-02 | 66.67 |
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| 2019-08-03 | 100.00 |
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| 2019-08-04 | 0.00 |
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+------------+----------------------+
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<strong>Explanation:</strong>
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- On 2019-08-01 there were three orders, out of those, two were immediate and one was scheduled. So, immediate percentage for that date was 66.67.
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- On 2019-08-02 there were three orders, out of those, two were immediate and one was scheduled. So, immediate percentage for that date was 66.67.
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- On 2019-08-03 there were two orders, both were immediate. So, the immediate percentage for that date was 100.00.
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- On 2019-08-04 there were two orders, both were scheduled. So, the immediate percentage for that date was 0.00.
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order_date is sorted in ascending order.
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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/2600-2699/2686.Immediate Food Delivery III/README_EN.md

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# [2686. Immediate Food Delivery III](https://leetcode.com/problems/immediate-food-delivery-iii)
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[中文文档](/solution/2600-2699/2686.Immediate%20Food%20Delivery%20III/README.md)
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## Description
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<p>Table: <code>Delivery</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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| delivery_id | int |
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| customer_id | int |
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| order_date | date |
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| customer_pref_delivery_date | date |
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+-----------------------------+---------+
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delivery_id is the primary key of this table.
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Each row contains information about food delivery to a customer that makes an order at some date and specifies a preferred delivery date (on the order date or after it).
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</pre>
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<p>If the customer&#39;s preferred delivery date is the same as the order date, then the order is called <strong>immediate,</strong>&nbsp;otherwise, it is <strong>scheduled</strong>.</p>
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<p>Write an SQL query to find the percentage of immediate orders on each unique&nbsp;<code>order_date</code>, <strong>rounded to 2 decimal places</strong>.&nbsp;</p>
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<p>Return <em>the result table ordered by</em> <code>order_date</code> <em>in <strong>ascending</strong> order.</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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Delivery table:
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+-------------+-------------+------------+-----------------------------+
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| delivery_id | customer_id | order_date | customer_pref_delivery_date |
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+-------------+-------------+------------+-----------------------------+
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| 1 | 1 | 2019-08-01 | 2019-08-02 |
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| 2 | 2 | 2019-08-01 | 2019-08-01 |
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| 3 | 1 | 2019-08-01 | 2019-08-01 |
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| 4 | 3 | 2019-08-02 | 2019-08-13 |
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| 5 | 3 | 2019-08-02 | 2019-08-02 |
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| 6 | 2 | 2019-08-02 | 2019-08-02 |
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| 7 | 4 | 2019-08-03 | 2019-08-03 |
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| 8 | 1 | 2019-08-03 | 2019-08-03 |
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| 9 | 5 | 2019-08-04 | 2019-08-08 |
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| 10 | 2 | 2019-08-04 | 2019-08-18 |
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+-------------+-------------+------------+-----------------------------+
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<strong>Output:</strong>
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+------------+----------------------+
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| order_date | immediate_percentage |
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+------------+----------------------+
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| 2019-08-01 | 66.67 |
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| 2019-08-02 | 66.67 |
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| 2019-08-03 | 100.00 |
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| 2019-08-04 | 0.00 |
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+------------+----------------------+
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<strong>Explanation:</strong>
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- On 2019-08-01 there were three orders, out of those, two were immediate and one was scheduled. So, immediate percentage for that date was 66.67.
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- On 2019-08-02 there were three orders, out of those, two were immediate and one was scheduled. So, immediate percentage for that date was 66.67.
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- On 2019-08-03 there were two orders, both were immediate. So, the immediate percentage for that date was 100.00.
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- On 2019-08-04 there were two orders, both were scheduled. So, the immediate percentage for that date was 0.00.
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order_date is sorted in ascending order.
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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/2600-2699/2687.Bikes Last Time Used/README.md

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# [2687. Bikes Last Time Used](https://leetcode.cn/problems/bikes-last-time-used)
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[English Version](/solution/2600-2699/2687.Bikes%20Last%20Time%20Used/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><font face="monospace">Bikes</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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| ride_id | int |
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| bike_number | int |
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| start_time | datetime |
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| end_time | datetime |
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+-------------+----------+
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ride_id is the primary key for this table.
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Each row contains a ride information that includes ride_id, bike number, start and end time of the ride.
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</pre>
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<p>Write an SQL query to find the <strong>last</strong> <strong>time</strong> when each bike was used.</p>
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<p>Return<em> the result table ordered by the bikes that were <strong>most recently used</strong>.&nbsp;</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:
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</strong><code>Bikes</code> table:
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+---------+-------------+---------------------+---------------------+
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| ride_id | bike_number | start_time | end_time |
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+---------+-------------+---------------------+---------------------+
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| 1 | W00576 | 2012-03-25 11:30:00 | 2012-03-25 12:40:00 |
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| 2 | W00300 | 2012-03-25 10:30:00 | 2012-03-25 10:50:00 |
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| 3 | W00455 | 2012-03-26 14:30:00 | 2012-03-26 17:40:00 |
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| 4 | W00455 | 2012-03-25 12:30:00 | 2012-03-25 13:40:00 |
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| 5 | W00576 | 2012-03-25 08:10:00 | 2012-03-25 09:10:00 |
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| 6 | W00576 | 2012-03-28 02:30:00 | 2012-03-28 02:50:00 |
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+---------+-------------+---------------------+---------------------+
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<strong>Output:</strong>
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+-------------+---------------------+
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| bike_number | end_time |
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+-------------+---------------------+
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| W00576 | 2012-03-28 02:50:00 |
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| W00455 | 2012-03-26 17:40:00 |
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| W00300 | 2012-03-25 10:50:00 |
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+-------------+---------------------+
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<strong>Explanation:</strong>
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bike with number W00576 has three rides, out of that, most recent ride is with ride_id 6 which ended on 2012-03-28 02:50:00.
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bike with number W00300 has only 1 ride so we will include end_time in output directly.
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bike with number W00455 has two rides, out of that, most recent ride is with ride_id 3 which ended on 2012-03-26 17:40:00.
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Returning output in order by the bike that were most recently used.
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</pre>
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<p>&nbsp;</p>
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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/2600-2699/2687.Bikes Last Time Used/README_EN.md

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# [2687. Bikes Last Time Used](https://leetcode.com/problems/bikes-last-time-used)
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[中文文档](/solution/2600-2699/2687.Bikes%20Last%20Time%20Used/README.md)
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## Description
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<p>Table: <code><font face="monospace">Bikes</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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| ride_id | int |
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| bike_number | int |
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| start_time | datetime |
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| end_time | datetime |
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+-------------+----------+
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ride_id is the primary key for this table.
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Each row contains a ride information that includes ride_id, bike number, start and end time of the ride.
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</pre>
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<p>Write an SQL query to find the <strong>last</strong> <strong>time</strong> when each bike was used.</p>
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<p>Return<em> the result table ordered by the bikes that were <strong>most recently used</strong>.&nbsp;</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:
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</strong><code>Bikes</code> table:
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+---------+-------------+---------------------+---------------------+
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| ride_id | bike_number | start_time | end_time |
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+---------+-------------+---------------------+---------------------+
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| 1 | W00576 | 2012-03-25 11:30:00 | 2012-03-25 12:40:00 |
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| 2 | W00300 | 2012-03-25 10:30:00 | 2012-03-25 10:50:00 |
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| 3 | W00455 | 2012-03-26 14:30:00 | 2012-03-26 17:40:00 |
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| 4 | W00455 | 2012-03-25 12:30:00 | 2012-03-25 13:40:00 |
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| 5 | W00576 | 2012-03-25 08:10:00 | 2012-03-25 09:10:00 |
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| 6 | W00576 | 2012-03-28 02:30:00 | 2012-03-28 02:50:00 |
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+---------+-------------+---------------------+---------------------+
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<strong>Output:</strong>
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+-------------+---------------------+
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| bike_number | end_time |
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+-------------+---------------------+
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| W00576 | 2012-03-28 02:50:00 |
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| W00455 | 2012-03-26 17:40:00 |
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| W00300 | 2012-03-25 10:50:00 |
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+-------------+---------------------+
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<strong>Explanation:</strong>
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bike with number W00576 has three rides, out of that, most recent ride is with ride_id 6 which ended on 2012-03-28 02:50:00.
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bike with number W00300 has only 1 ride so we will include end_time in output directly.
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bike with number W00455 has two rides, out of that, most recent ride is with ride_id 3 which ended on 2012-03-26 17:40:00.
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Returning output in order by the bike that were most recently used.
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</pre>
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<p>&nbsp;</p>
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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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