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feat: add solutions to lc problem: No.3188 #3124

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199 changes: 199 additions & 0 deletions solution/3100-3199/3188.Find Top Scoring Students II/README.md
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---
comments: true
difficulty: 困难
edit_url: https://github.com/doocs/leetcode/edit/main/solution/3100-3199/3188.Find%20Top%20Scoring%20Students%20II/README.md
---

<!-- problem:start -->

# [3188. Find Top Scoring Students II 🔒](https://leetcode.cn/problems/find-top-scoring-students-ii)

[English Version](/solution/3100-3199/3188.Find%20Top%20Scoring%20Students%20II/README_EN.md)

## 题目描述

<!-- description:start -->

<p>Table: <code>students</code></p>

<pre>
+-------------+----------+
| Column Name | Type |
+-------------+----------+
| student_id | int |
| name | varchar |
| major | varchar |
+-------------+----------+
student_id is the primary key for this table.
Each row contains the student ID, student name, and their major.
</pre>

<p>Table: <code>courses</code></p>

<pre>
+-------------+-------------------+
| Column Name | Type |
+-------------+-------------------+
| course_id | int |
| name | varchar |
| credits | int |
| major | varchar |
| mandatory | enum |
+-------------+-------------------+
course_id is the primary key for this table.
mandatory is an enum type of (&#39;Yes&#39;, &#39;No&#39;).
Each row contains the course ID, course name, credits, major it belongs to, and whether the course is mandatory.
</pre>

<p>Table: <code>enrollments</code></p>

<pre>
+-------------+----------+
| Column Name | Type |
+-------------+----------+
| student_id | int |
| course_id | int |
| semester | varchar |
| grade | varchar |
| GPA | decimal |
+-------------+----------+
(student_id, course_id, semester) is the primary key (combination of columns with unique values) for this table.
Each row contains the student ID, course ID, semester, and grade received.
</pre>

<p>Write a solution to find the students who meet the following criteria:</p>

<ul>
<li>Have<strong> taken all mandatory courses</strong> and <strong>at least two</strong> elective courses offered in <strong>their major.</strong></li>
<li>Achieved a grade of <strong>A</strong>&nbsp;in <strong>all mandatory courses</strong> and at least <strong>B</strong>&nbsp;in<strong> elective courses</strong>.</li>
<li>Maintained an average <code>GPA</code> of at least&nbsp;<code>2.5</code> across all their courses (including those outside their major).</li>
</ul>

<p>Return <em>the result table ordered by</em> <code>student_id</code> <em>in <strong>ascending</strong> order</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example:</strong></p>

<div class="example-block">
<p><strong>Input:</strong></p>

<p>students table:</p>

<pre class="example-io">
+------------+------------------+------------------+
| student_id | name | major |
+------------+------------------+------------------+
| 1 | Alice | Computer Science |
| 2 | Bob | Computer Science |
| 3 | Charlie | Mathematics |
| 4 | David | Mathematics |
+------------+------------------+------------------+
</pre>

<p>courses table:</p>

<pre class="example-io">
+-----------+-------------------+---------+------------------+----------+
| course_id | name | credits | major | mandatory|
+-----------+-------------------+---------+------------------+----------+
| 101 | Algorithms | 3 | Computer Science | yes |
| 102 | Data Structures | 3 | Computer Science | yes |
| 103 | Calculus | 4 | Mathematics | yes |
| 104 | Linear Algebra | 4 | Mathematics | yes |
| 105 | Machine Learning | 3 | Computer Science | no |
| 106 | Probability | 3 | Mathematics | no |
| 107 | Operating Systems | 3 | Computer Science | no |
| 108 | Statistics | 3 | Mathematics | no |
+-----------+-------------------+---------+------------------+----------+
</pre>

<p>enrollments table:</p>

<pre class="example-io">
+------------+-----------+-------------+-------+-----+
| student_id | course_id | semester | grade | GPA |
+------------+-----------+-------------+-------+-----+
| 1 | 101 | Fall 2023 | A | 4.0 |
| 1 | 102 | Spring 2023 | A | 4.0 |
| 1 | 105 | Spring 2023 | A | 4.0 |
| 1 | 107 | Fall 2023 | B | 3.5 |
| 2 | 101 | Fall 2023 | A | 4.0 |
| 2 | 102 | Spring 2023 | B | 3.0 |
| 3 | 103 | Fall 2023 | A | 4.0 |
| 3 | 104 | Spring 2023 | A | 4.0 |
| 3 | 106 | Spring 2023 | A | 4.0 |
| 3 | 108 | Fall 2023 | B | 3.5 |
| 4 | 103 | Fall 2023 | B | 3.0 |
| 4 | 104 | Spring 2023 | B | 3.0 |
+------------+-----------+-------------+-------+-----+
</pre>

<p><strong>Output:</strong></p>

<pre class="example-io">
+------------+
| student_id |
+------------+
| 1 |
| 3 |
+------------+
</pre>

<p><strong>Explanation:</strong></p>

<ul>
<li>Alice (student_id 1) is a Computer Science major and has taken both Algorithms&nbsp;and Data Structures, receiving an A&nbsp;in both. She has also taken Machine Learning&nbsp;and Operating Systems&nbsp;as electives, receiving an A&nbsp;and B&nbsp;respectively.</li>
<li>Bob (student_id 2) is a Computer Science major but did not receive an A&nbsp;in all required courses.</li>
<li>Charlie (student_id 3) is a Mathematics major and has taken both Calculus&nbsp;and Linear Algebra, receiving an A&nbsp;in both. He has also taken Probability&nbsp;and Statistics&nbsp;as electives, receiving an A&nbsp;and B&nbsp;respectively.</li>
<li>David (student_id 4) is a Mathematics major but did not receive an A&nbsp;in all required courses.</li>
</ul>

<p><strong>Note:</strong> Output table is ordered by student_id in ascending order.</p>
</div>

<!-- description:end -->

## 解法

<!-- solution:start -->

### 方法一：连接 + 分组 + 条件过滤

我们首先计算出每个学生的平均 GPA，记录在临时表 `T` 中。

然后，我们将 `students` 表与 `courses` 表按照 `major` 进行连接，然后与 `T` 表按照 `student_id` 进行连接，再与 `enrollments` 表按照 `student_id` 和 `course_id` 进行左连接。

接下来，我们筛选出平均 GPA 大于等于 2.5 的学生，并按照学生 ID 进行分组，然后使用 `HAVING` 子句过滤出符合条件的学生，最后按照学生 ID 进行排序。

<!-- tabs:start -->

#### MySQL

```sql
# Write your MySQL query statement below
WITH
T AS (
SELECT student_id, AVG(GPA) AS avg_gpa
FROM enrollments
GROUP BY 1
)
SELECT student_id
FROM
students
JOIN courses USING (major)
JOIN T USING (student_id)
LEFT JOIN enrollments USING (student_id, course_id)
WHERE avg_gpa >= 2.5
GROUP BY 1
HAVING
SUM(mandatory = 'yes' AND grade = 'A') = SUM(mandatory = 'yes')
AND SUM(mandatory = 'no' AND grade IS NOT NULL) = SUM(mandatory = 'no' AND grade IN ('A', 'B'))
AND SUM(mandatory = 'no' AND grade IS NOT NULL) >= 2
ORDER BY 1;
```

<!-- tabs:end -->

<!-- solution:end -->

<!-- problem:end -->
199 changes: 199 additions & 0 deletions solution/3100-3199/3188.Find Top Scoring Students II/README_EN.md
View file
Open in desktop
Original file line number Diff line number Diff line change
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---
comments: true
difficulty: Hard
edit_url: https://github.com/doocs/leetcode/edit/main/solution/3100-3199/3188.Find%20Top%20Scoring%20Students%20II/README_EN.md
---

<!-- problem:start -->

# [3188. Find Top Scoring Students II 🔒](https://leetcode.com/problems/find-top-scoring-students-ii)

[中文文档](/solution/3100-3199/3188.Find%20Top%20Scoring%20Students%20II/README.md)

## Description

<!-- description:start -->

<p>Table: <code>students</code></p>

<pre>
+-------------+----------+
| Column Name | Type |
+-------------+----------+
| student_id | int |
| name | varchar |
| major | varchar |
+-------------+----------+
student_id is the primary key for this table.
Each row contains the student ID, student name, and their major.
</pre>

<p>Table: <code>courses</code></p>

<pre>
+-------------+-------------------+
| Column Name | Type |
+-------------+-------------------+
| course_id | int |
| name | varchar |
| credits | int |
| major | varchar |
| mandatory | enum |
+-------------+-------------------+
course_id is the primary key for this table.
mandatory is an enum type of (&#39;Yes&#39;, &#39;No&#39;).
Each row contains the course ID, course name, credits, major it belongs to, and whether the course is mandatory.
</pre>

<p>Table: <code>enrollments</code></p>

<pre>
+-------------+----------+
| Column Name | Type |
+-------------+----------+
| student_id | int |
| course_id | int |
| semester | varchar |
| grade | varchar |
| GPA | decimal |
+-------------+----------+
(student_id, course_id, semester) is the primary key (combination of columns with unique values) for this table.
Each row contains the student ID, course ID, semester, and grade received.
</pre>

<p>Write a solution to find the students who meet the following criteria:</p>

<ul>
<li>Have<strong> taken all mandatory courses</strong> and <strong>at least two</strong> elective courses offered in <strong>their major.</strong></li>
<li>Achieved a grade of <strong>A</strong>&nbsp;in <strong>all mandatory courses</strong> and at least <strong>B</strong>&nbsp;in<strong> elective courses</strong>.</li>
<li>Maintained an average <code>GPA</code> of at least&nbsp;<code>2.5</code> across all their courses (including those outside their major).</li>
</ul>

<p>Return <em>the result table ordered by</em> <code>student_id</code> <em>in <strong>ascending</strong> order</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example:</strong></p>

<div class="example-block">
<p><strong>Input:</strong></p>

<p>students table:</p>

<pre class="example-io">
+------------+------------------+------------------+
| student_id | name | major |
+------------+------------------+------------------+
| 1 | Alice | Computer Science |
| 2 | Bob | Computer Science |
| 3 | Charlie | Mathematics |
| 4 | David | Mathematics |
+------------+------------------+------------------+
</pre>

<p>courses table:</p>

<pre class="example-io">
+-----------+-------------------+---------+------------------+----------+
| course_id | name | credits | major | mandatory|
+-----------+-------------------+---------+------------------+----------+
| 101 | Algorithms | 3 | Computer Science | yes |
| 102 | Data Structures | 3 | Computer Science | yes |
| 103 | Calculus | 4 | Mathematics | yes |
| 104 | Linear Algebra | 4 | Mathematics | yes |
| 105 | Machine Learning | 3 | Computer Science | no |
| 106 | Probability | 3 | Mathematics | no |
| 107 | Operating Systems | 3 | Computer Science | no |
| 108 | Statistics | 3 | Mathematics | no |
+-----------+-------------------+---------+------------------+----------+
</pre>

<p>enrollments table:</p>

<pre class="example-io">
+------------+-----------+-------------+-------+-----+
| student_id | course_id | semester | grade | GPA |
+------------+-----------+-------------+-------+-----+
| 1 | 101 | Fall 2023 | A | 4.0 |
| 1 | 102 | Spring 2023 | A | 4.0 |
| 1 | 105 | Spring 2023 | A | 4.0 |
| 1 | 107 | Fall 2023 | B | 3.5 |
| 2 | 101 | Fall 2023 | A | 4.0 |
| 2 | 102 | Spring 2023 | B | 3.0 |
| 3 | 103 | Fall 2023 | A | 4.0 |
| 3 | 104 | Spring 2023 | A | 4.0 |
| 3 | 106 | Spring 2023 | A | 4.0 |
| 3 | 108 | Fall 2023 | B | 3.5 |
| 4 | 103 | Fall 2023 | B | 3.0 |
| 4 | 104 | Spring 2023 | B | 3.0 |
+------------+-----------+-------------+-------+-----+
</pre>

<p><strong>Output:</strong></p>

<pre class="example-io">
+------------+
| student_id |
+------------+
| 1 |
| 3 |
+------------+
</pre>

<p><strong>Explanation:</strong></p>

<ul>
<li>Alice (student_id 1) is a Computer Science major and has taken both Algorithms&nbsp;and Data Structures, receiving an A&nbsp;in both. She has also taken Machine Learning&nbsp;and Operating Systems&nbsp;as electives, receiving an A&nbsp;and B&nbsp;respectively.</li>
<li>Bob (student_id 2) is a Computer Science major but did not receive an A&nbsp;in all required courses.</li>
<li>Charlie (student_id 3) is a Mathematics major and has taken both Calculus&nbsp;and Linear Algebra, receiving an A&nbsp;in both. He has also taken Probability&nbsp;and Statistics&nbsp;as electives, receiving an A&nbsp;and B&nbsp;respectively.</li>
<li>David (student_id 4) is a Mathematics major but did not receive an A&nbsp;in all required courses.</li>
</ul>

<p><strong>Note:</strong> Output table is ordered by student_id in ascending order.</p>
</div>

<!-- description:end -->

## Solutions

<!-- solution:start -->

### Solution 1: Joining + Grouping + Conditional Filtering

First, we calculate the average GPA of each student and store it in a temporary table `T`.

Next, we join the `students` table with the `courses` table based on `major`, and then join with the `T` table based on `student_id`, followed by a left join with the `enrollments` table based on `student_id` and `course_id`.

After that, we filter out students with an average GPA greater than or equal to 2.5, group by student ID, use the `HAVING` clause to filter students who meet the criteria, and finally sort by student ID.

<!-- tabs:start -->

#### MySQL

```sql
# Write your MySQL query statement below
WITH
T AS (
SELECT student_id, AVG(GPA) AS avg_gpa
FROM enrollments
GROUP BY 1
)
SELECT student_id
FROM
students
JOIN courses USING (major)
JOIN T USING (student_id)
LEFT JOIN enrollments USING (student_id, course_id)
WHERE avg_gpa >= 2.5
GROUP BY 1
HAVING
SUM(mandatory = 'yes' AND grade = 'A') = SUM(mandatory = 'yes')
AND SUM(mandatory = 'no' AND grade IS NOT NULL) = SUM(mandatory = 'no' AND grade IN ('A', 'B'))
AND SUM(mandatory = 'no' AND grade IS NOT NULL) >= 2
ORDER BY 1;
```

<!-- tabs:end -->

<!-- solution:end -->

<!-- problem:end -->
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