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feat: add solutions to lc problem: No.3246 (doocs#3365)
No.3246.Premier League Table Ranking
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---
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comments: true
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difficulty: 简单
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edit_url: https://github.com/doocs/leetcode/edit/main/solution/3200-3299/3246.Premier%20League%20Table%20Ranking/README.md
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---
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<!-- problem:start -->
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# [3246. 英超积分榜排名 🔒](https://leetcode.cn/problems/premier-league-table-ranking)
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[English Version](/solution/3200-3299/3246.Premier%20League%20Table%20Ranking/README_EN.md)
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## 题目描述
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<!-- description:start -->
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<p>表:<code>TeamStats</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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| team_id | int |
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| team_name | varchar |
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| matches_played | int |
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| wins | int |
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| draws | int |
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| losses | int |
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+------------------+---------+
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team_id 是这张表的唯一主键。
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这张表包含队伍 id,队伍名,场次,赢局,平局和输局。
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</pre>
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<p>编写一个解决方啊来计算联盟中每支球队的 <strong>得分</strong> 和 <strong>排名</strong>。积分计算方式如下:</p>
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<ul>
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<li><strong>赢局</strong> 有&nbsp;<code>3</code>&nbsp;点得分</li>
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<li><strong>平局</strong> 有&nbsp;<code>1</code>&nbsp;点得分</li>
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<li><strong>输局</strong> 有&nbsp;<code>0</code>&nbsp;点得分</li>
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</ul>
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<p><b>注意:</b>积分相同的球队必须分配相同的排名。</p>
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<p>返回结果表以&nbsp;<code>points</code>&nbsp;<strong>降序</strong>&nbsp;排序,然后以&nbsp;<code>team_name</code> <strong>升序</strong>&nbsp;排序。</p>
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<p>结果格式如下所示。</p>
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<p>&nbsp;</p>
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<p><strong class="example">示例:</strong></p>
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<div class="example-block">
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<p><strong>输入:</strong></p>
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<p><code>TeamStats</code> 表:</p>
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<pre class="example-io">
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+---------+-----------------+----------------+------+-------+--------+
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| team_id | team_name | matches_played | wins | draws | losses |
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+---------+-----------------+----------------+------+-------+--------+
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| 1 | Manchester City | 10 | 6 | 2 | 2 |
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| 2 | Liverpool | 10 | 6 | 2 | 2 |
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| 3 | Chelsea | 10 | 5 | 3 | 2 |
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| 4 | Arsenal | 10 | 4 | 4 | 2 |
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| 5 | Tottenham | 10 | 3 | 5 | 2 |
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+---------+-----------------+----------------+------+-------+--------+
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</pre>
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<p><strong>输出:</strong></p>
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<pre class="example-io">
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+---------+-----------------+--------+----------+
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| team_id | team_name | points | position |
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+---------+-----------------+--------+----------+
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| 2 | Liverpool | 20 | 1 |
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| 1 | Manchester City | 20 | 1 |
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| 3 | Chelsea | 18 | 3 |
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| 4 | Arsenal | 16 | 4 |
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| 5 | Tottenham | 14 | 5 |
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+---------+-----------------+--------+----------+
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</pre>
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<p><strong>解释:</strong></p>
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<ul>
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<li>曼城和利物浦均拿下 20 分(6 赢 * 3 分 + 2 平 * 1 分),所以他们并列第一。</li>
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<li>切尔西拿下&nbsp;18 分(5 赢 * 3 分 + 3 平 * 1 分)所以位列第三。</li>
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<li>阿森纳拿下 16 分(4 赢 * 3 分 + 4 平 * 1 分)位列第四。</li>
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<li>托特纳姆热刺队拿下 14 分(3 赢 * 3 分 + 5 平 * 1 分)位列第五。</li>
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</ul>
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<p>输出表以得分降序排序,然后以&nbsp;team_name 升序排序。</p>
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</div>
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<!-- description:end -->
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## 解法
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<!-- solution:start -->
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### 方法一:窗口函数
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我们可以使用 `RANK()` 窗口函数来计算球队的排名,然后按照得分和球队名进行排序。
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<!-- tabs:start -->
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#### MySQL
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```sql
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# Write your MySQL query statement below
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SELECT
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team_id,
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team_name,
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wins * 3 + draws points,
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RANK() OVER (ORDER BY (wins * 3 + draws) DESC) position
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FROM TeamStats
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ORDER BY 3 DESC, 2;
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```
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#### Pandas
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```python
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import pandas as pd
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def calculate_team_standings(team_stats: pd.DataFrame) -> pd.DataFrame:
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team_stats["points"] = team_stats["wins"] * 3 + team_stats["draws"]
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team_stats["position"] = team_stats["points"].rank(method="min", ascending=False)
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team_stats = team_stats.sort_values(
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by=["points", "team_name"], ascending=[False, True]
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)
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return team_stats[["team_id", "team_name", "points", "position"]]
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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<!-- problem:end -->
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---
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comments: true
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difficulty: Easy
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edit_url: https://github.com/doocs/leetcode/edit/main/solution/3200-3299/3246.Premier%20League%20Table%20Ranking/README_EN.md
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---
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<!-- problem:start -->
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# [3246. Premier League Table Ranking 🔒](https://leetcode.com/problems/premier-league-table-ranking)
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[中文文档](/solution/3200-3299/3246.Premier%20League%20Table%20Ranking/README.md)
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## Description
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<!-- description:start -->
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<p>Table: <code>TeamStats</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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| team_id | int |
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| team_name | varchar |
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| matches_played | int |
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| wins | int |
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| draws | int |
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| losses | int |
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+------------------+---------+
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team_id is the unique key for this table.
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This table contains team id, team name, matches_played, wins, draws, and losses.
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</pre>
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<p>Write a solution to calculate the <strong>points</strong> and <strong>rank</strong> for each team in the league. Points are calculated as follows:</p>
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<ul>
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<li><code>3</code> points for a <strong>win</strong></li>
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<li><code>1</code> point for a <strong>draw</strong></li>
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<li><code>0</code> points for a <strong>loss</strong></li>
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</ul>
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<p><strong>Note:</strong>&nbsp;Teams with the same points must be assigned the same rank.</p>
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<p>Return <em>the result table ordered by</em> <code>points</code>&nbsp;<em>in&nbsp;<strong>descending</strong>,<strong>&nbsp;</strong>and then by</em> <code>team_name</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:</strong></p>
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<div class="example-block">
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<p><strong>Input:</strong></p>
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<p><code>TeamStats</code> table:</p>
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<pre class="example-io">
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+---------+-----------------+----------------+------+-------+--------+
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| team_id | team_name | matches_played | wins | draws | losses |
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+---------+-----------------+----------------+------+-------+--------+
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| 1 | Manchester City | 10 | 6 | 2 | 2 |
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| 2 | Liverpool | 10 | 6 | 2 | 2 |
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| 3 | Chelsea | 10 | 5 | 3 | 2 |
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| 4 | Arsenal | 10 | 4 | 4 | 2 |
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| 5 | Tottenham | 10 | 3 | 5 | 2 |
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+---------+-----------------+----------------+------+-------+--------+
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</pre>
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<p><strong>Output:</strong></p>
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<pre class="example-io">
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+---------+-----------------+--------+----------+
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| team_id | team_name | points | position |
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+---------+-----------------+--------+----------+
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| 2 | Liverpool | 20 | 1 |
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| 1 | Manchester City | 20 | 1 |
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| 3 | Chelsea | 18 | 3 |
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| 4 | Arsenal | 16 | 4 |
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| 5 | Tottenham | 14 | 5 |
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+---------+-----------------+--------+----------+
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</pre>
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<p><strong>Explanation:</strong></p>
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<ul>
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<li>Manchester City and Liverpool both have 20 points (6 wins * 3 points + 2 draws * 1 point), so they share position 1.</li>
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<li>Chelsea has 18 points (5 wins * 3 points + 3 draws * 1 point) and is position 3rd.</li>
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<li>Arsenal has 16 points (4 wins * 3 points + 4 draws * 1 point) and is position 4th.</li>
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<li>Tottenham has 14 points (3 wins * 3 points + 5 draws * 1 point) and is position 5th.</li>
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</ul>
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<p>The output table is ordered by points in descending order, then by team_name in ascending order.</p>
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</div>
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<!-- description:end -->
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## Solutions
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<!-- solution:start -->
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### Solution 1: Window Function
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We can use the `RANK()` window function to calculate the ranking of the teams, and then sort by score and team name.
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<!-- tabs:start -->
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#### MySQL
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```sql
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# Write your MySQL query statement below
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SELECT
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team_id,
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team_name,
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wins * 3 + draws points,
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RANK() OVER (ORDER BY (wins * 3 + draws) DESC) position
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FROM TeamStats
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ORDER BY 3 DESC, 2;
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```
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#### Pandas
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```python
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import pandas as pd
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def calculate_team_standings(team_stats: pd.DataFrame) -> pd.DataFrame:
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team_stats["points"] = team_stats["wins"] * 3 + team_stats["draws"]
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team_stats["position"] = team_stats["points"].rank(method="min", ascending=False)
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team_stats = team_stats.sort_values(
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by=["points", "team_name"], ascending=[False, True]
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)
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return team_stats[["team_id", "team_name", "points", "position"]]
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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<!-- problem:end -->
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import pandas as pd
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def calculate_team_standings(team_stats: pd.DataFrame) -> pd.DataFrame:
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team_stats["points"] = team_stats["wins"] * 3 + team_stats["draws"]
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team_stats["position"] = team_stats["points"].rank(method="min", ascending=False)
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team_stats = team_stats.sort_values(
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by=["points", "team_name"], ascending=[False, True]
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)
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return team_stats[["team_id", "team_name", "points", "position"]]
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# Write your MySQL query statement below
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SELECT
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team_id,
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team_name,
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wins * 3 + draws points,
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RANK() OVER (ORDER BY (wins * 3 + draws) DESC) position
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FROM TeamStats
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ORDER BY 3 DESC, 2;

‎solution/DATABASE_README.md‎

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| 3220 | [奇数和偶数交易](/solution/3200-3299/3220.Odd%20and%20Even%20Transactions/README.md) | `数据库` | 中等 | |
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| 3230 | [客户购买行为分析](/solution/3200-3299/3230.Customer%20Purchasing%20Behavior%20Analysis/README.md) | `数据库` | 中等 | 🔒 |
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| 3236 | [首席执行官下属层级](/solution/3200-3299/3236.CEO%20Subordinate%20Hierarchy/README.md) | `数据库` | 困难 | 🔒 |
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| 3246 | [英超积分榜排名](/solution/3200-3299/3246.Premier%20League%20Table%20Ranking/README.md) | | 简单 | 🔒 |
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## 版权
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‎solution/DATABASE_README_EN.md‎

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| 3220 | [Odd and Even Transactions](/solution/3200-3299/3220.Odd%20and%20Even%20Transactions/README_EN.md) | `Database` | Medium | |
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| 3230 | [Customer Purchasing Behavior Analysis](/solution/3200-3299/3230.Customer%20Purchasing%20Behavior%20Analysis/README_EN.md) | `Database` | Medium | 🔒 |
290290
| 3236 | [CEO Subordinate Hierarchy](/solution/3200-3299/3236.CEO%20Subordinate%20Hierarchy/README_EN.md) | `Database` | Hard | 🔒 |
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| 3246 | [Premier League Table Ranking](/solution/3200-3299/3246.Premier%20League%20Table%20Ranking/README_EN.md) | | Easy | 🔒 |
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## Copyright
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‎solution/README.md‎

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| 3243 | [新增道路查询后的最短距离 I](/solution/3200-3299/3243.Shortest%20Distance%20After%20Road%20Addition%20Queries%20I/README.md) | | 中等 | 第 409 场周赛 |
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| 3244 | [新增道路查询后的最短距离 II](/solution/3200-3299/3244.Shortest%20Distance%20After%20Road%20Addition%20Queries%20II/README.md) | | 困难 | 第 409 场周赛 |
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| 3245 | [交替组 III](/solution/3200-3299/3245.Alternating%20Groups%20III/README.md) | | 困难 | 第 409 场周赛 |
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| 3246 | [英超积分榜排名](/solution/3200-3299/3246.Premier%20League%20Table%20Ranking/README.md) | | 简单 | 🔒 |
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## 版权
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‎solution/README_EN.md‎

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| 3243 | [Shortest Distance After Road Addition Queries I](/solution/3200-3299/3243.Shortest%20Distance%20After%20Road%20Addition%20Queries%20I/README_EN.md) | | Medium | Weekly Contest 409 |
32553255
| 3244 | [Shortest Distance After Road Addition Queries II](/solution/3200-3299/3244.Shortest%20Distance%20After%20Road%20Addition%20Queries%20II/README_EN.md) | | Hard | Weekly Contest 409 |
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| 3245 | [Alternating Groups III](/solution/3200-3299/3245.Alternating%20Groups%20III/README_EN.md) | | Hard | Weekly Contest 409 |
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| 3246 | [Premier League Table Ranking](/solution/3200-3299/3246.Premier%20League%20Table%20Ranking/README_EN.md) | | Easy | 🔒 |
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## Copyright
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