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feat: add solutions to lc problems: No.2347~2350
* No.2347.Best Poker Hand * No.2348.Number of Zero-Filled Subarrays * No.2349.Design a Number Container System * No.2350.Shortest Impossible Sequence of Rolls
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‎solution/2300-2399/2346.Compute the Rank as a Percentage/README.md

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# [2346. Compute the Rank as a Percentage](https://leetcode.cn/problems/compute-the-rank-as-a-percentage)
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[English Version](/solution/2300-2399/2346.Compute%20the%20Rank%20as%20a%20Percentage/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>Students</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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| student_id | int |
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| department_id | int |
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| mark | int |
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+---------------+------+
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student_id is the primary key of this table.
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Each row of this table indicates a student&#39;s ID, the ID of the department in which the student enrolled, and their mark in the exam.
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</pre>
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<p>&nbsp;</p>
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<p>Write an SQL query that reports the rank of each student in their department as a percentage, where the rank as a percentage is computed using the following formula: <code>(student_rank_in_the_department - 1) * 100 / (the_number_of_students_in_the_department - 1)</code>. The <code>percentage</code> should be <strong>rounded to 2 decimal places</strong>. <code>student_rank_in_the_department</code> is determined by <strong>descending</strong><b> </b><code>mark</code>, such that the student with the highest <code>mark</code> is <code>rank 1</code>. If two students get the same mark, they also get the same rank.</p>
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<p>Return the result table in <strong>any order</strong>.</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>Example 1:</strong></p>
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<pre>
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<strong>Input:</strong>
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Students table:
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+------------+---------------+------+
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| student_id | department_id | mark |
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+------------+---------------+------+
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| 2 | 2 | 650 |
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| 8 | 2 | 650 |
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| 7 | 1 | 920 |
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| 1 | 1 | 610 |
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| 3 | 1 | 530 |
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+------------+---------------+------+
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<strong>Output:</strong>
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+------------+---------------+------------+
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| student_id | department_id | percentage |
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+------------+---------------+------------+
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| 7 | 1 | 0.0 |
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| 1 | 1 | 50.0 |
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| 3 | 1 | 100.0 |
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| 2 | 2 | 0.0 |
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| 8 | 2 | 0.0 |
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+------------+---------------+------------+
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<strong>Explanation:</strong>
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For Department 1:
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- Student 7: percentage = (1 - 1) * 100 / (3 - 1) = 0.0
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- Student 1: percentage = (2 - 1) * 100 / (3 - 1) = 50.0
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- Student 3: percentage = (3 - 1) * 100 / (3 - 1) = 100.0
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For Department 2:
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- Student 2: percentage = (1 - 1) * 100 / (2 - 1) = 0.0
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- Student 8: percentage = (1 - 1) * 100 / (2 - 1) = 0.0
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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/2300-2399/2346.Compute the Rank as a Percentage/README_EN.md

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# [2346. Compute the Rank as a Percentage](https://leetcode.com/problems/compute-the-rank-as-a-percentage)
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[中文文档](/solution/2300-2399/2346.Compute%20the%20Rank%20as%20a%20Percentage/README.md)
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## Description
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<p>Table: <code>Students</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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| student_id | int |
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| department_id | int |
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| mark | int |
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+---------------+------+
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student_id is the primary key of this table.
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Each row of this table indicates a student&#39;s ID, the ID of the department in which the student enrolled, and their mark in the exam.
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</pre>
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<p>&nbsp;</p>
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<p>Write an SQL query that reports the rank of each student in their department as a percentage, where the rank as a percentage is computed using the following formula: <code>(student_rank_in_the_department - 1) * 100 / (the_number_of_students_in_the_department - 1)</code>. The <code>percentage</code> should be <strong>rounded to 2 decimal places</strong>. <code>student_rank_in_the_department</code> is determined by <strong>descending</strong><b> </b><code>mark</code>, such that the student with the highest <code>mark</code> is <code>rank 1</code>. If two students get the same mark, they also get the same rank.</p>
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<p>Return the result table in <strong>any order</strong>.</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>Example 1:</strong></p>
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<pre>
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<strong>Input:</strong>
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Students table:
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+------------+---------------+------+
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| student_id | department_id | mark |
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+------------+---------------+------+
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| 2 | 2 | 650 |
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| 8 | 2 | 650 |
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| 7 | 1 | 920 |
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| 1 | 1 | 610 |
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| 3 | 1 | 530 |
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+------------+---------------+------+
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<strong>Output:</strong>
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+------------+---------------+------------+
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| student_id | department_id | percentage |
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+------------+---------------+------------+
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| 7 | 1 | 0.0 |
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| 1 | 1 | 50.0 |
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| 3 | 1 | 100.0 |
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| 2 | 2 | 0.0 |
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| 8 | 2 | 0.0 |
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+------------+---------------+------------+
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<strong>Explanation:</strong>
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For Department 1:
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- Student 7: percentage = (1 - 1) * 100 / (3 - 1) = 0.0
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- Student 1: percentage = (2 - 1) * 100 / (3 - 1) = 50.0
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- Student 3: percentage = (3 - 1) * 100 / (3 - 1) = 100.0
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For Department 2:
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- Student 2: percentage = (1 - 1) * 100 / (2 - 1) = 0.0
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- Student 8: percentage = (1 - 1) * 100 / (2 - 1) = 0.0
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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/2300-2399/2347.Best Poker Hand/README.md

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# [2347. 最好的扑克手牌](https://leetcode.cn/problems/best-poker-hand)
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[English Version](/solution/2300-2399/2347.Best%20Poker%20Hand/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>给你一个整数数组&nbsp;<code>ranks</code>&nbsp;和一个字符数组&nbsp;<code>suit</code>&nbsp;。你有&nbsp;<code>5</code>&nbsp;张扑克牌，第&nbsp;<code>i</code>&nbsp;张牌大小为&nbsp;<code>ranks[i]</code>&nbsp;，花色为&nbsp;<code>suits[i]</code>&nbsp;。</p>
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<p>下述是从好到坏你可能持有的 <strong>手牌类型&nbsp;</strong>：</p>
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<ol>
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<li><code>"Flush"</code>：同花，五张相同花色的扑克牌。</li>
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<li><code>"Three of a Kind"</code>：三条，有 3 张大小相同的扑克牌。</li>
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<li><code>"Pair"</code>：对子，两张大小一样的扑克牌。</li>
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<li><code>"High Card"</code>：高牌，五张大小互不相同的扑克牌。</li>
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</ol>
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<p>请你返回一个字符串，表示给定的 5 张牌中，你能组成的 <strong>最好手牌类型</strong>&nbsp;。</p>
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<p><strong>注意：</strong>返回的字符串&nbsp;<strong>大小写</strong>&nbsp;需与题目描述相同。</p>
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<p>&nbsp;</p>
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<p><strong>示例 1：</strong></p>
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<pre><b>输入：</b>ranks = [13,2,3,1,9], suits = ["a","a","a","a","a"]
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<b>输出：</b>"Flush"
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<b>解释：</b>5 张扑克牌的花色相同，所以返回 "Flush" 。
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</pre>
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<p><strong>示例 2：</strong></p>
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<pre><b>输入：</b>ranks = [4,4,2,4,4], suits = ["d","a","a","b","c"]
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<b>输出：</b>"Three of a Kind"
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<b>解释：</b>第一、二和四张牌组成三张相同大小的扑克牌，所以得到 "Three of a Kind" 。
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注意我们也可以得到 "Pair" ，但是 "Three of a Kind" 是更好的手牌类型。
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有其他的 3 张牌也可以组成 "Three of a Kind" 手牌类型。</pre>
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<p><strong>示例 3：</strong></p>
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<pre><b>输入：</b>ranks = [10,10,2,12,9], suits = ["a","b","c","a","d"]
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<b>输出：</b>"Pair"
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<b>解释：</b>第一和第二张牌大小相同，所以得到 "Pair" 。
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我们无法得到 "Flush" 或者 "Three of a Kind" 。
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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>ranks.length == suits.length == 5</code></li>
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<li><code>1 &lt;= ranks[i] &lt;= 13</code></li>
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<li><code>'a' &lt;= suits[i] &lt;= 'd'</code></li>
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<li>任意两张扑克牌不会同时有相同的大小和花色。</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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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def bestHand(self, ranks: List[int], suits: List[str]) -> str:
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if len(set(suits)) == 1:
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return 'Flush'
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cnt = Counter(ranks)
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if any(v >= 3 for v in cnt.values()):
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return 'Three of a Kind'
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if any(v == 2 for v in cnt.values()):
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return 'Pair'
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return 'High Card'
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public String bestHand(int[] ranks, char[] suits) {
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Set<Character> s = new HashSet<>();
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for (char c : suits) {
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s.add(c);
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}
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if (s.size() == 1) {
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return "Flush";
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}
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int[] cnt = new int[20];
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for (int v : ranks) {
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++cnt[v];
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if (cnt[v] >= 3) {
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return "Three of a Kind";
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}
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}
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for (int v : cnt) {
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if (v == 2) {
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return "Pair";
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}
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}
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return "High Card";
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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string bestHand(vector<int>& ranks, vector<char>& suits) {
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unordered_set<char> s(suits.begin(), suits.end());
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if (s.size() == 1) return "Flush";
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vector<int> cnt(20);
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for (int v : ranks) if (++cnt[v] >= 3) return "Three of a Kind";
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for (int v : cnt) if (v == 2) return "Pair";
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return "High Card";
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}
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};
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```
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### **Go**
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```go
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func bestHand(ranks []int, suits []byte) string {
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s := map[byte]bool{}
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for _, v := range suits {
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s[v] = true
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}
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if len(s) == 1 {
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return "Flush"
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}
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cnt := make([]int, 20)
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for _, v := range ranks {
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cnt[v]++
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if cnt[v] >= 3 {
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return "Three of a Kind"
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}
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}
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for _, v := range cnt {
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if v == 2 {
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return "Pair"
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}
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}
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return "High Card"
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}
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```
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### **TypeScript**
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```ts
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```
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### **...**
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```
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```
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<!-- tabs:end -->

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