feat: add solutions to lc problems: No.2853,2854

solution/2800-2899/2853.Highest Salaries Difference/README.md

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+# [2853. Highest Salaries Difference](https://leetcode.cn/problems/highest-salaries-difference)
+
+[English Version](/solution/2800-2899/2853.Highest%20Salaries%20Difference/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>Table: <code><font face="monospace">Salaries</font></code></p>
+
+<pre>
++-------------+---------+ 
+| Column Name | Type    | 
++-------------+---------+ 
+| emp_name    | varchar | 
+| department  | varchar | 
+| salary      | int     |
++-------------+---------+
+(emp_name, department) is the primary key for this table.
+Each row of this table contains emp_name, department and salary. There will be <strong>at least one</strong> entry for the engineering and marketing departments.
+</pre>
+
+<p>Write an SQL query to calculate the difference between the <strong>highest</strong> salaries in the <strong>marketing</strong> and <strong>engineering</strong> <code>department</code>. Output the absolute difference in salaries.</p>
+
+<p>Return<em> the result table.</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> 
+Salaries table:
++----------+-------------+--------+
+| emp_name | department  | salary |
++----------+-------------+--------+
+| Kathy    | Engineering | 50000  |
+| Roy      | Marketing   | 30000  |
+| Charles  | Engineering | 45000  |
+| Jack     | Engineering | 85000  | 
+| Benjamin | Marketing   | 34000  |
+| Anthony  | Marketing   | 42000  |
+| Edward   | Engineering | 102000 |
+| Terry    | Engineering | 44000  |
+| Evelyn   | Marketing   | 53000  |
+| Arthur   | Engineering | 32000  |
++----------+-------------+--------+
+<strong>Output:</strong> 
++-------------------+
+| salary_difference | 
++-------------------+
+| 49000             | 
++-------------------+
+<strong>Explanation:</strong> 
+- The Engineering and Marketing departments have the highest salaries of 102,000 and 53,000, respectively. Resulting in an absolute difference of 49,000.
+</pre>
+
+## 解法
+
+<!-- 这里可写通用的实现逻辑 -->
+
+**方法一：GROUP BY 分组**
+
+我们可以先分别计算出每个部门的最高工资，然后再计算两个最高工资的差值。
+
+<!-- tabs:start -->
+
+### **SQL**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```sql
+# Write your MySQL query statement below
+SELECT max(s) - min(s) AS salary_difference
+FROM
+    (
+        SELECT max(salary) AS s
+        FROM Salaries
+        GROUP BY department
+    ) AS t;
+```
+
+<!-- tabs:end -->

solution/2800-2899/2853.Highest Salaries Difference/README_EN.md

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+# [2853. Highest Salaries Difference](https://leetcode.com/problems/highest-salaries-difference)
+
+[中文文档](/solution/2800-2899/2853.Highest%20Salaries%20Difference/README.md)
+
+## Description
+
+<p>Table: <code><font face="monospace">Salaries</font></code></p>
+
+<pre>
++-------------+---------+ 
+| Column Name | Type    | 
++-------------+---------+ 
+| emp_name    | varchar | 
+| department  | varchar | 
+| salary      | int     |
++-------------+---------+
+(emp_name, department) is the primary key for this table.
+Each row of this table contains emp_name, department and salary. There will be <strong>at least one</strong> entry for the engineering and marketing departments.
+</pre>
+
+<p>Write an SQL query to calculate the difference between the <strong>highest</strong> salaries in the <strong>marketing</strong> and <strong>engineering</strong> <code>department</code>. Output the absolute difference in salaries.</p>
+
+<p>Return<em> the result table.</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> 
+Salaries table:
++----------+-------------+--------+
+| emp_name | department  | salary |
++----------+-------------+--------+
+| Kathy    | Engineering | 50000  |
+| Roy      | Marketing   | 30000  |
+| Charles  | Engineering | 45000  |
+| Jack     | Engineering | 85000  | 
+| Benjamin | Marketing   | 34000  |
+| Anthony  | Marketing   | 42000  |
+| Edward   | Engineering | 102000 |
+| Terry    | Engineering | 44000  |
+| Evelyn   | Marketing   | 53000  |
+| Arthur   | Engineering | 32000  |
++----------+-------------+--------+
+<strong>Output:</strong> 
++-------------------+
+| salary_difference | 
++-------------------+
+| 49000             | 
++-------------------+
+<strong>Explanation:</strong> 
+- The Engineering and Marketing departments have the highest salaries of 102,000 and 53,000, respectively. Resulting in an absolute difference of 49,000.
+</pre>
+
+## Solutions
+
+<!-- tabs:start -->
+
+### **SQL**
+
+```sql
+# Write your MySQL query statement below
+SELECT max(s) - min(s) AS salary_difference
+FROM
+    (
+        SELECT max(salary) AS s
+        FROM Salaries
+        GROUP BY department
+    ) AS t;
+```
+
+<!-- tabs:end -->

solution/2800-2899/2853.Highest Salaries Difference/Solution.sql

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+# Write your MySQL query statement below
+SELECT max(s) - min(s) AS salary_difference
+FROM
+    (
+        SELECT max(salary) AS s
+        FROM Salaries
+        GROUP BY department
+    ) AS t;

solution/2800-2899/2854.Rolling Average Steps/README.md

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+# [2854. Rolling Average Steps](https://leetcode.cn/problems/rolling-average-steps)
+
+[English Version](/solution/2800-2899/2854.Rolling%20Average%20Steps/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>Table: <code><font face="monospace">Steps</font></code></p>
+
+<pre>
++-------------+------+ 
+| Column Name | Type | 
++-------------+------+ 
+| user_id     | int  | 
+| steps_count | int  |
+| steps_date  | date |
++-------------+------+
+(user_id, steps_date) is the primary key for this table.
+Each row of this table contains user_id, steps_count, and steps_date.
+</pre>
+
+<p>Write a solution to calculate <code>3-day</code> <strong>rolling averages</strong> of steps for each user.</p>
+
+<p>We calculate the <code>n-day</code> <strong>rolling average</strong> this way:</p>
+
+<ul>
+	<li>For each day, we calculate the average of <code>n</code> consecutive days of step counts ending on that day if available, otherwise, <code>n-day</code> rolling average is not defined for it.</li>
+</ul>
+
+<p>Output the <code>user_id</code>, <code>steps_date</code>, and rolling average. Round the rolling average to <strong>two decimal places</strong>.</p>
+
+<p>Return<em> the result table ordered by </em><code>user_id</code><em>, </em><code>steps_date</code><em> in <strong>ascending</strong> order.</em></p>
+
+<p>The result format is in the following example.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> 
+Steps table:
++---------+-------------+------------+
+| user_id | steps_count | steps_date |
++---------+-------------+------------+
+| 1       | 687         | 2021-09-02 |
+| 1       | 395         | 2021-09-04 |
+| 1       | 499         | 2021-09-05 |
+| 1       | 712         | 2021-09-06 |
+| 1       | 576         | 2021-09-07 |
+| 2       | 153         | 2021-09-06 |
+| 2       | 171         | 2021-09-07 |
+| 2       | 530         | 2021-09-08 |
+| 3       | 945         | 2021-09-04 |
+| 3       | 120         | 2021-09-07 |
+| 3       | 557         | 2021-09-08 |
+| 3       | 840         | 2021-09-09 |
+| 3       | 627         | 2021-09-10 |
+| 5       | 382         | 2021-09-05 |
+| 6       | 480         | 2021-09-01 |
+| 6       | 191         | 2021-09-02 |
+| 6       | 303         | 2021-09-05 |
++---------+-------------+------------+
+<strong>Output:</strong> 
++---------+------------+-----------------+
+| user_id | steps_date | rolling_average | 
++---------+------------+-----------------+
+| 1       | 2021-09-06 | 535.33          | 
+| 1       | 2021-09-07 | 595.67          | 
+| 2       | 2021-09-08 | 284.67          |
+| 3       | 2021-09-09 | 505.67          |
+| 3       | 2021-09-10 | 674.67          |    
++---------+------------+-----------------+
+<strong>Explanation:</strong> 
+- For user id 1, the step counts for the three consecutive days up to 2021-09-06 are available. Consequently, the rolling average for this particular date is computed as (395 + 499 + 712) / 3 = 535.33.
+- For user id 1, the step counts for the three consecutive days up to 2021-09-07 are available. Consequently, the rolling average for this particular date is computed as (499 + 712 + 576) / 3 = 595.67.
+- For user id 2, the step counts for the three consecutive days up to 2021-09-08 are available. Consequently, the rolling average for this particular date is computed as (153 + 171 + 530) / 3 = 284.67.
+- For user id 3, the step counts for the three consecutive days up to 2021-09-09 are available. Consequently, the rolling average for this particular date is computed as (120 + 557 + 840) / 3 = 505.67.
+- For user id 3, the step counts for the three consecutive days up to 2021-09-10 are available. Consequently, the rolling average for this particular date is computed as (557 + 840 + 627) / 3 = 674.67.
+- For user id 4 and 5, the calculation of the rolling average is not viable as there is insufficient data for the consecutive three days. Output table ordered by user_id and steps_date in ascending order.</pre>
+
+## 解法
+
+<!-- 这里可写通用的实现逻辑 -->
+
+**方法一：窗口函数**
+
+我们用窗口函数 `LAG() OVER()` 来计算每个用户当前日期与上上个日期之间的天数差，如果为 $2$，说明这两个日期之间有连续 $3$ 天的数据，我们可以利用窗口函数 `AVG() OVER()` 来计算这 $3$ 个数据的平均值。
+
+<!-- tabs:start -->
+
+### **SQL**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```sql
+# Write your MySQL query statement below
+WITH
+    T AS (
+        SELECT
+            user_id,
+            steps_date,
+            round(
+                avg(steps_count) OVER (
+                    PARTITION BY user_id
+                    ORDER BY steps_date
+                    ROWS 2 PRECEDING
+                ),
+                2
+            ) AS rolling_average,
+            datediff(
+                steps_date,
+                lag(steps_date, 2) OVER (
+                    PARTITION BY user_id
+                    ORDER BY steps_date
+                )
+            ) = 2 AS st
+        FROM Steps
+    )
+SELECT
+    user_id,
+    steps_date,
+    rolling_average
+FROM T
+WHERE st = 1
+ORDER BY 1, 2;
+```
+
+<!-- tabs:end -->