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feat: add weekly contest 392 (#2549)
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36 changes: 23 additions & 13 deletions solution/1300-1399/1364.Number of Trusted Contacts of a Customer/README.md
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<p>顾客表：<code>Customers</code></p>

<pre>+---------------+---------+
<pre>
+---------------+---------+
| Column Name | Type |
+---------------+---------+
| customer_id | int |
| customer_name | varchar |
| email | varchar |
+---------------+---------+
customer_id 是这张表的主键。
customer_id 是这张表具有唯一值的列。
此表的每一行包含了某在线商店顾客的姓名和电子邮件。
</pre>

<p>&nbsp;</p>

<p>联系方式表：<code>Contacts</code></p>

<pre>+---------------+---------+
<pre>
+---------------+---------+
| Column Name | Type |
+---------------+---------+
| user_id | id |
| contact_name | varchar |
| contact_email | varchar |
+---------------+---------+
(user_id, contact_email) 是这张表的主键。
(user_id, contact_email) 是这张表的主键（具有唯一值的列的组合）。
此表的每一行表示编号为 user_id 的顾客的某位联系人的姓名和电子邮件。
此表包含每位顾客的联系人信息，但顾客的联系人不一定存在于顾客表中。
</pre>
Expand All @@ -41,33 +43,40 @@ customer_id 是这张表的主键。

<p>发票表：<code>Invoices</code></p>

<pre>+--------------+---------+
<pre>
+--------------+---------+
| Column Name | Type |
+--------------+---------+
| invoice_id | int |
| price | int |
| user_id | int |
+--------------+---------+
invoice_id 是这张表的主键。
invoice_id 是这张表具有唯一值的列。
此表的每一行分别表示编号为 user_id 的顾客拥有有一张编号为 invoice_id、价格为 price 的发票。
</pre>

<p>&nbsp;</p>

<p>为每张发票 <code>invoice_id</code> 编写一个SQL查询以查找以下内容：</p>
<p>为每张发票 <code>invoice_id</code> 编写一个查询方案以查找以下内容：</p>

<ul>
<li><code>customer_name</code>：与发票相关的顾客名称。</li>
<li><code>price</code>：发票的价格。</li>
<li><code>contacts_cnt</code>：该顾客的联系人数量。</li>
<li><code>trusted_contacts_cnt</code>：可信联系人的数量：既是该顾客的联系人又是商店顾客的联系人数量（即：可信联系人的电子邮件存在于客户表中）。</li>
<li><code>contacts_cnt</code>：该顾客的联系人数量</li>
<li><code>trusted_contacts_cnt</code>：可信联系人的数量：既是该顾客的联系人又是商店顾客的联系人数量（即：可信联系人的电子邮件存在于 <meta charset="UTF-8" />&nbsp;<code>Customers</code>&nbsp;表中）。</li>
</ul>

<p>将查询的结果按照&nbsp;<code>invoice_id</code>&nbsp;排序。</p>
<p>返回结果按照&nbsp;<code>invoice_id</code>&nbsp;<strong>排序</strong>。</p>

<p>结果的格式如下例所示。</p>

<p>&nbsp;</p>

<p>查询结果的格式如下例所示：</p>
<p><strong>示例 1：</strong></p>

<pre><code>Customers</code> table:
<pre>
<strong>输入：</strong>
<code>Customers</code> table:
+-------------+---------------+--------------------+
| customer_id | customer_name | email |
+-------------+---------------+--------------------+
Expand Down Expand Up @@ -98,7 +107,7 @@ Invoices table:
| 55 | 500 | 13 |
| 44 | 60 | 6 |
+------------+-------+---------+
Result table:
<strong>输出：</strong>
+------------+---------------+-------+--------------+----------------------+
| invoice_id | customer_name | price | contacts_cnt | trusted_contacts_cnt |
+------------+---------------+-------+--------------+----------------------+
Expand All @@ -109,6 +118,7 @@ Result table:
| 88 | Alice | 200 | 3 | 2 |
| 99 | Bob | 300 | 2 | 0 |
+------------+---------------+-------+--------------+----------------------+
<strong>解释：</strong>
Alice 有三位联系人，其中两位(Bob 和 John)是可信联系人。
Bob 有两位联系人, 他们中的任何一位都不是可信联系人。
Alex 只有一位联系人(Alice)，并是一位可信联系人。
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10 changes: 5 additions & 5 deletions solution/1300-1399/1364.Number of Trusted Contacts of a Customer/README_EN.md
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| customer_name | varchar |
| email | varchar |
+---------------+---------+
customer_id is the primary key for this table.
customer_id is the column of unique values for this table.
Each row of this table contains the name and the email of a customer of an online shop.
</pre>

Expand All @@ -32,7 +32,7 @@ Each row of this table contains the name and the email of a customer of an onlin
| contact_name | varchar |
| contact_email | varchar |
+---------------+---------+
(user_id, contact_email) is the primary key for this table.
(user_id, contact_email) is the primary key (combination of columns with unique values) for this table.
Each row of this table contains the name and email of one contact of customer with user_id.
This table contains information about people each customer trust. The contact may or may not exist in the Customers table.
</pre>
Expand All @@ -49,13 +49,13 @@ This table contains information about people each customer trust. The contact ma
| price | int |
| user_id | int |
+--------------+---------+
invoice_id is the primary key for this table.
invoice_id is the column of unique values for this table.
Each row of this table indicates that user_id has an invoice with invoice_id and a price.
</pre>

<p>&nbsp;</p>

<p>Write an SQL query to find the following for each <code>invoice_id</code>:</p>
<p>Write a solution to find the following for each <code>invoice_id</code>:</p>

<ul>
<li><code>customer_name</code>: The name of the customer the invoice is related to.</li>
Expand All @@ -66,7 +66,7 @@ Each row of this table indicates that user_id has an invoice with invoice_id and

<p>Return the result table <strong>ordered</strong> by <code>invoice_id</code>.</p>

<p>The query result format is in the following example.</p>
<p>The result format is in the following example.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
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2 changes: 1 addition & 1 deletion solution/1900-1999/1918.Kth Smallest Subarray Sum/README.md
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# [1918. 第 K 小的子数组和·](https://leetcode.cn/problems/kth-smallest-subarray-sum)
# [1918. 第 K 小的子数组和](https://leetcode.cn/problems/kth-smallest-subarray-sum)

[English Version](/solution/1900-1999/1918.Kth%20Smallest%20Subarray%20Sum/README_EN.md)

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42 changes: 9 additions & 33 deletions solution/2800-2899/2826.Sorting Three Groups/README.md
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<!-- 这里写题目描述 -->

<p>给你一个下标从 <strong>0</strong>&nbsp;开始长度为 <code>n</code>&nbsp;的整数数组&nbsp;<code>nums</code>&nbsp;。<br />
<br />
从&nbsp;<code>0</code>&nbsp;到&nbsp;<code>n - 1</code>&nbsp;的数字被分为编号从&nbsp;<code>1</code>&nbsp;到&nbsp;<code>3</code>&nbsp;的三个组，数字&nbsp;<code>i</code>&nbsp;属于组&nbsp;<code>nums[i]</code>&nbsp;。注意，有的组可能是&nbsp;<strong>空的</strong>&nbsp;。<br />
<br />
你可以执行以下操作任意次：</p>

<ul>
<li>选择数字&nbsp;<code>x</code>&nbsp;并改变它的组。更正式的，你可以将&nbsp;<code>nums[x]</code>&nbsp;改为数字&nbsp;<code>1</code>&nbsp;到&nbsp;<code>3</code>&nbsp;中的任意一个。</li>
</ul>

<p>你将按照以下过程构建一个新的数组&nbsp;<code>res</code>&nbsp;：</p>

<ol>
<li>将每个组中的数字分别排序。</li>
<li>将组&nbsp;<code>1</code>&nbsp;，<code>2</code>&nbsp;和&nbsp;<code>3</code>&nbsp;中的元素&nbsp;<strong>依次</strong>&nbsp;连接以得到&nbsp;<code>res</code>&nbsp;。</li>
</ol>

<p>如果得到的&nbsp;<code>res</code>&nbsp;是 <strong>非递减</strong>顺序的，那么我们称数组&nbsp;<code>nums</code>&nbsp;是 <strong>美丽数组</strong>&nbsp;。</p>

<p>请你返回将<em>&nbsp;</em><code>nums</code>&nbsp;变为&nbsp;<strong>美丽数组</strong>&nbsp;需要的最少步数。</p>
<p>给你一个整数数组&nbsp;<code>nums</code>&nbsp;。<code>nums</code>&nbsp;的每个元素是 1，2 或 3。在每次操作中，你可以删除&nbsp;<code>nums</code>&nbsp;中的一个元素。返回使 nums 成为 <strong>非递减</strong>&nbsp;顺序所需操作数的 <strong>最小值</strong>。</p>

<p>&nbsp;</p>

Expand All @@ -36,33 +17,26 @@
<pre>
<b>输入：</b>nums = [2,1,3,2,1]
<b>输出：</b>3
<b>解释：</b>以下三步操作是最优方案：
1. 将 nums[0] 变为 1 。
2. 将 nums[2] 变为 1 。
3. 将 nums[3] 变为 1 。
执行以上操作后，将每组中的数字排序，组 1 为 [0,1,2,3,4] ，组 2 和组 3 都为空。所以 res 等于 [0,1,2,3,4] ，它是非递减顺序的。
三步操作是最少需要的步数。
<b>解释：</b>
其中一个最优方案是删除 nums[0]，nums[2] 和 nums[3]。
</pre>

<p><strong class="example">示例 2：</strong></p>

<pre>
<b>输入：</b>nums = [1,3,2,1,3,3]
<b>输出：</b>2
<b>解释：</b>以下两步操作是最优方案：
1. 将 nums[1] 变为 1 。
2. 将 nums[2] 变为 1 。
执行以上操作后，将每组中的数字排序，组 1 为 [0,1,2,3] ，组 2 为空，组 3 为 [4,5] 。所以 res 等于 [0,1,2,3,4,5] ，它是非递减顺序的。
两步操作是最少需要的步数。
<b>解释：</b>
其中一个最优方案是删除 nums[1] 和 nums[2]。
</pre>

<p><strong class="example">示例 3：</strong></p>

<pre>
<b>输入：</b>nums = [2,2,2,2,3,3]
<b>输出：</b>0
<b>解释：</b>不需要执行任何操作。
组 1 为空，组 2 为 [0,1,2,3] ，组 3 为 [4,5] 。所以 res 等于 [0,1,2,3,4,5] ，它是非递减顺序的。
<b>解释：</b>
nums 已是非递减顺序的。
</pre>

<p>&nbsp;</p>
Expand All @@ -74,6 +48,8 @@
<li><code>1 &lt;= nums[i] &lt;= 3</code></li>
</ul>

<p><strong>进阶：</strong>你可以使用&nbsp;<code>O(n)</code>&nbsp;时间复杂度以内的算法解决吗？</p>

## 解法

### 方法一：动态规划
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2 changes: 1 addition & 1 deletion solution/3000-3099/3087.Find Trending Hashtags/README.md
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Expand Up @@ -23,7 +23,7 @@ tweet_id 是这张表的主键 (值互不相同的列)。
这张表的每一行都包含 user_id, tweet_id, tweet_date 和 tweet。
</pre>

<p>编写一个解决方案来找到&nbsp;<code>2024</code>&nbsp;年 <strong>二月&nbsp;</strong>的 <strong>前</strong>&nbsp;<code>3</code>&nbsp;热门话题 <strong>标签</strong>。</p>
<p>编写一个解决方案来找到&nbsp;<code>2024</code>&nbsp;年 <strong>二月&nbsp;</strong>的 <strong>前</strong>&nbsp;<code>3</code>&nbsp;热门话题 <strong>标签</strong>。每条推文只包含一个标签。</p>

<p>返回结果表，根据标签的数量和标签&nbsp;<strong>降序</strong> 排序。</p>

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38 changes: 20 additions & 18 deletions solution/3000-3099/3094.Guess the Number Using Bitwise Questions II/README.md
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# [3094. Guess the Number Using Bitwise Questions II](https://leetcode.cn/problems/guess-the-number-using-bitwise-questions-ii)
# [3094. 使用按位查询猜测数字 II](https://leetcode.cn/problems/guess-the-number-using-bitwise-questions-ii)

[English Version](/solution/3000-3099/3094.Guess%20the%20Number%20Using%20Bitwise%20Questions%20II/README_EN.md)

Expand All @@ -8,50 +8,52 @@

<!-- 这里写题目描述 -->

<p>There is a number <code>n</code> between <code>0</code> and <code>2<sup>30</sup> - 1</code> (both inclusive) that you have to find.</p>
<p>你需要找到一个在 <code>0</code> 和&nbsp;<code>2<sup>30</sup> - 1</code>&nbsp;（均包含）之间的数字 <code>n</code>。</p>

<p>There is a pre-defined API <code>int commonBits(int num)</code> that helps you with your mission. But here is the challenge, every time you call this function, <code>n</code> changes in some way. But keep in mind, that you have to find the <strong>initial value of </strong><code>n</code>.</p>
<p>有一个预定义的 API <code>int commonBits(int num)</code>&nbsp;能帮助你完成任务。但挑战是每次你调用这个函数，<code>n</code>&nbsp;都会以某种方式改变。但是记住，你需要找到的是<strong>&nbsp;</strong><code>n</code>&nbsp;的 <strong>初始值</strong>。</p>

<p><code>commonBits(int num)</code> acts as follows:</p>
<p><code>commonBits(int num)</code> 的操作如下：</p>

<ul>
<li>Calculate <code>count</code> which is the number of bits where both <code>n</code> and <code>num</code> have the same value in that position of their binary representation.</li>
<li>计算&nbsp;<code>n</code>&nbsp;和&nbsp;<code>num</code>&nbsp;的二进制表示中值相同的二进制位的位的数量&nbsp;<code>count</code>。</li>
<li><code>n = n XOR num</code></li>
<li>Return <code>count</code>.</li>
<li>返回&nbsp;<code>count</code>。</li>
</ul>

<p>Return <em>the number</em> <code>n</code>.</p>
<p>返回数字&nbsp;<code>n</code>。</p>

<p><strong>Note:</strong> In this world, all numbers are between <code>0</code> and <code>2<sup>30</sup> - 1</code> (both inclusive), thus for counting common bits, we see only the first 30 bits of those numbers.</p>
<p><strong>注意：</strong>在这个世界中，所有数字都在&nbsp;<code>0</code>&nbsp;和&nbsp;<code>2<sup>30</sup> - 1</code>&nbsp;之间（均包含），因此在计算公共二进制位时，我们只看那些数字的前 30 个二进制位。</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1: </strong></p>

<p><strong class="example">示例 1：</strong></p>

<div class="example-block" style="border-color: var(--border-tertiary); border-left-width: 2px; color: var(--text-secondary); font-size: .875rem; margin-bottom: 1rem; margin-top: 1rem; overflow: visible; padding-left: 1rem;">
<p><strong>Input: </strong> <span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;"> n = 31 </span></p>
<p><strong>输入：</strong><span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;">n = 31 </span></p>

<p><strong>Output: </strong> <span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;"> 31 </span></p>
<p><strong>输出：</strong><span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;">31 </span></p>

<p><strong>Explanation: </strong> It can be proven that it&#39;s possible to find 31 using the provided API.</p>
<p><strong>解释：</strong>可以证明，使用提供的 API 可以找到 31。</p>
</div>

<p><strong class="example">Example 2: </strong></p>
<p><strong class="example">示例 2：</strong></p>

<div class="example-block" style="border-color: var(--border-tertiary); border-left-width: 2px; color: var(--text-secondary); font-size: .875rem; margin-bottom: 1rem; margin-top: 1rem; overflow: visible; padding-left: 1rem;">
<p><strong>Input: </strong> <span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;"> n = 33 </span></p>
<p><strong>输入：</strong><span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;">n = 33 </span></p>

<p><strong>Output: </strong> <span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;"> 33 </span></p>
<p><strong>输出：</strong><span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;">33 </span></p>

<p><strong>Explanation: </strong> It can be proven that it&#39;s possible to find 33 using the provided API.</p>
<p><strong>解释：</strong>可以证明，使用提供的 API 可以找到 33。</p>
</div>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<p><strong>提示：</strong></p>

<ul>
<li><code>0 &lt;= n &lt;= 2<sup>30</sup> - 1</code></li>
<li><code>0 &lt;= num &lt;= 2<sup>30</sup> - 1</code></li>
<li>If you ask for some <code>num</code> out of the given range, the output wouldn&#39;t be reliable.</li>
<li>如果你查询的&nbsp;<code>num</code>&nbsp;超出了给定的范围，输出将会是不可靠的。</li>
</ul>

## 解法
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37 changes: 19 additions & 18 deletions solution/3100-3199/3103.Find Trending Hashtags II/README.md
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# [3103. Find Trending Hashtags II](https://leetcode.cn/problems/find-trending-hashtags-ii)
# [3103. 查找热门话题标签 II](https://leetcode.cn/problems/find-trending-hashtags-ii)

[English Version](/solution/3100-3199/3103.Find%20Trending%20Hashtags%20II/README_EN.md)

Expand All @@ -8,7 +8,7 @@

<!-- 这里写题目描述 -->

<p>Table: <code>Tweets</code></p>
<p>表：<code>Tweets</code></p>

<pre>
+-------------+---------+
Expand All @@ -19,23 +19,24 @@
| tweet_date | date |
| tweet | varchar |
+-------------+---------+
tweet_id is the primary key (column with unique values) for this table.
Each row of this table contains user_id, tweet_id, tweet_date and tweet.
tweet_id 是这张表的主键 (值互不相同的列)。
这张表的每一行都包含 user_id, tweet_id, tweet_date 和 tweet。
</pre>

<p>Write a solution to find the <strong>top</strong> <code>3</code> trending <strong>hashtags</strong> in <strong>February</strong> <code>2024</code>. Every tweet may contain <strong>several</strong> <strong>hashtags</strong>.</p>
<p>编写一个解决方案来找到&nbsp;<code>2024</code>&nbsp;年 <strong>二月&nbsp;</strong>的 <strong>前</strong>&nbsp;<code>3</code>&nbsp;热门话题 <strong>标签</strong>。每条推文可能含有 <strong>几个标签</strong>。</p>

<p>Return <em>the result table orderd by count of hashtag, hashtag in </em><strong>descending</strong><em> order.</em></p>
<p>返回结果表，根据标签的数量和标签&nbsp;<strong>降序</strong> 排序。</p>

<p>The result format is in the following example.</p>
<p>结果格式如下所示。</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<p><strong class="example">示例 1：</strong></p>

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

<p>Tweets table:</p>
<p>Tweets 表：</p>

<pre class="example-io">
+---------+----------+------------------------------------------------------------+------------+
Expand All @@ -45,13 +46,13 @@ Each row of this table contains user_id, tweet_id, tweet_date and tweet.
| 136 | 14 | Another #HappyDay with good vibes! #FeelGood | 2024-02-03 |
| 137 | 15 | Productivity peaks! #WorkLife #ProductiveDay | 2024-02-04 |
| 138 | 16 | Exploring new tech frontiers. #TechLife #Innovation | 2024-02-04 |
| 139 | 17 | Gratitude for today&#39;s moments. #HappyDay #Thankful | 2024-02-05 |
| 139 | 17 | Gratitude for today's moments. #HappyDay #Thankful | 2024-02-05 |
| 140 | 18 | Innovation drives us. #TechLife #FutureTech | 2024-02-07 |
| 141 | 19 | Connecting with nature&#39;s serenity. #Nature #Peaceful | 2024-02-09 |
| 141 | 19 | Connecting with nature's serenity. #Nature #Peaceful | 2024-02-09 |
+---------+----------+------------------------------------------------------------+------------+
</pre>

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

<pre class="example-io">
+-----------+-------+
Expand All @@ -64,15 +65,15 @@ Each row of this table contains user_id, tweet_id, tweet_date and tweet.

</pre>

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

<ul>
<li><strong>#HappyDay:</strong> Appeared in tweet IDs 13, 14, and 17, with a total count of 3 mentions.</li>
<li><strong>#TechLife:</strong> Appeared in tweet IDs 16 and 18, with a total count of 2 mentions.</li>
<li><strong>#WorkLife:</strong> Appeared in tweet ID 15, with a total count of 1 mention.</li>
<li><strong>#HappyDay：</strong>在 ID 为 13，14，17 的推文中出现，总共提及&nbsp;3 次。</li>
<li><strong>#TechLife：</strong>在 ID 为 16，18 的推文中出现，总共提及 2&nbsp;次。</li>
<li><strong>#WorkLife：</strong>在 ID 为 15 的推文中出现，总共提及 1&nbsp;次。</li>
</ul>

<p><b>Note:</b> Output table is sorted in descending order by count and hashtag respectively.</p>
<p><b>注意：</b>输出表分别按 count 和 hashtag 降序排序。</p>
</div>

## 解法
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