feat: add new lc problems

main.py

@@ -149,11 +149,12 @@ def get_paths(dirs: str, m: int):
                     break
                 j = content.find(end)
                 codes = content[i + len(start) : j].strip()
-                res = re.findall(r"```(.+?)\n(.+?)\n```", codes, re.DOTALL)
+                res = re.findall(r"```(.*?)\n(.*?)\n```", codes, re.S)
                 result = []
                 if res:
                     for lang, code in res:
                         name = mapping.get(lang)
+                        code = code or ''
                         # 需要将 code 缩进 4 个空格
                         code = code.replace("\n", "\n    ")
                         code_snippet = (

solution/3000-3099/3018.Maximum Number of Removal Queries That Can Be Processed I/README.md

@@ -0,0 +1,102 @@
+# [3018. Maximum Number of Removal Queries That Can Be Processed I](https://leetcode.cn/problems/maximum-number-of-removal-queries-that-can-be-processed-i)
+
+[English Version](/solution/3000-3099/3018.Maximum%20Number%20of%20Removal%20Queries%20That%20Can%20Be%20Processed%20I/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>You are given a <strong>0-indexed</strong> array <code>nums</code> and a <strong>0-indexed</strong> array <code>queries</code>.</p>
+
+<p>You can do the following operation at the beginning <strong>at most once</strong>:</p>
+
+<ul>
+	<li>Replace <code>nums</code> with a <span data-keyword="subsequence-array">subsequence</span> of <code>nums</code>.</li>
+</ul>
+
+<p>We start processing queries in the given order; for each query, we do the following:</p>
+
+<ul>
+	<li>If the first <strong>and</strong> the last element of <code>nums</code> is <strong>less than</strong> <code>queries[i]</code>, the processing of queries <strong>ends</strong>.</li>
+	<li>Otherwise, we choose either the first <strong>or</strong> the last element of <code>nums</code> if it is <strong>greater than or equal to</strong> <code>queries[i]</code>, and we <strong>remove</strong> the chosen element from <code>nums</code>.</li>
+</ul>
+
+<p>Return <em>the <strong>maximum</strong> number of queries that can be processed by doing the operation optimally.</em></p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [1,2,3,4,5], queries = [1,2,3,4,6]
+<strong>Output:</strong> 4
+<strong>Explanation:</strong> We don&#39;t do any operation and process the queries as follows:
+1- We choose and remove nums[0] since 1 &lt;= 1, then nums becomes [2,3,4,5].
+2- We choose and remove nums[0] since 2 &lt;= 2, then nums becomes [3,4,5].
+3- We choose and remove nums[0] since 3 &lt;= 3, then nums becomes [4,5].
+4- We choose and remove nums[0] since 4 &lt;= 4, then nums becomes [5].
+5- We can not choose any elements from nums since they are not greater than or equal to 5.
+Hence, the answer is 4.
+It can be shown that we can&#39;t process more than 4 queries.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [2,3,2], queries = [2,2,3]
+<strong>Output:</strong> 3
+<strong>Explanation:</strong> We don&#39;t do any operation and process the queries as follows:
+1- We choose and remove nums[0] since 2 &lt;= 2, then nums becomes [3,2].
+2- We choose and remove nums[1] since 2 &lt;= 2, then nums becomes [3].
+3- We choose and remove nums[0] since 3 &lt;= 3, then nums becomes [].
+Hence, the answer is 3.
+It can be shown that we can&#39;t process more than 3 queries.
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [3,4,3], queries = [4,3,2]
+<strong>Output:</strong> 2
+<strong>Explanation:</strong> First we replace nums with the subsequence of nums [4,3].
+Then we can process the queries as follows:
+1- We choose and remove nums[0] since 4 &lt;= 4, then nums becomes [3].
+2- We choose and remove nums[0] since 3 &lt;= 3, then nums becomes [].
+3- We can not process any more queries since nums is empty.
+Hence, the answer is 2.
+It can be shown that we can&#39;t process more than 2 queries.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length &lt;= 1000</code></li>
+	<li><code>1 &lt;= queries.length &lt;= 1000</code></li>
+	<li><code>1 &lt;= nums[i], queries[i] &lt;= 10<sup>9</sup></code></li>
+</ul>
+
+## 解法
+
+### 方法一
+
+<!-- tabs:start -->
+
+```python
+
+```
+
+```java
+
+```
+
+```cpp
+
+```
+
+```go
+
+```
+
+<!-- tabs:end -->
+
+<!-- end -->

solution/3000-3099/3018.Maximum Number of Removal Queries That Can Be Processed I/README_EN.md

@@ -0,0 +1,100 @@
+# [3018. Maximum Number of Removal Queries That Can Be Processed I](https://leetcode.com/problems/maximum-number-of-removal-queries-that-can-be-processed-i)
+
+[中文文档](/solution/3000-3099/3018.Maximum%20Number%20of%20Removal%20Queries%20That%20Can%20Be%20Processed%20I/README.md)
+
+## Description
+
+<p>You are given a <strong>0-indexed</strong> array <code>nums</code> and a <strong>0-indexed</strong> array <code>queries</code>.</p>
+
+<p>You can do the following operation at the beginning <strong>at most once</strong>:</p>
+
+<ul>
+	<li>Replace <code>nums</code> with a <span data-keyword="subsequence-array">subsequence</span> of <code>nums</code>.</li>
+</ul>
+
+<p>We start processing queries in the given order; for each query, we do the following:</p>
+
+<ul>
+	<li>If the first <strong>and</strong> the last element of <code>nums</code> is <strong>less than</strong> <code>queries[i]</code>, the processing of queries <strong>ends</strong>.</li>
+	<li>Otherwise, we choose either the first <strong>or</strong> the last element of <code>nums</code> if it is <strong>greater than or equal to</strong> <code>queries[i]</code>, and we <strong>remove</strong> the chosen element from <code>nums</code>.</li>
+</ul>
+
+<p>Return <em>the <strong>maximum</strong> number of queries that can be processed by doing the operation optimally.</em></p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [1,2,3,4,5], queries = [1,2,3,4,6]
+<strong>Output:</strong> 4
+<strong>Explanation:</strong> We don&#39;t do any operation and process the queries as follows:
+1- We choose and remove nums[0] since 1 &lt;= 1, then nums becomes [2,3,4,5].
+2- We choose and remove nums[0] since 2 &lt;= 2, then nums becomes [3,4,5].
+3- We choose and remove nums[0] since 3 &lt;= 3, then nums becomes [4,5].
+4- We choose and remove nums[0] since 4 &lt;= 4, then nums becomes [5].
+5- We can not choose any elements from nums since they are not greater than or equal to 5.
+Hence, the answer is 4.
+It can be shown that we can&#39;t process more than 4 queries.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [2,3,2], queries = [2,2,3]
+<strong>Output:</strong> 3
+<strong>Explanation:</strong> We don&#39;t do any operation and process the queries as follows:
+1- We choose and remove nums[0] since 2 &lt;= 2, then nums becomes [3,2].
+2- We choose and remove nums[1] since 2 &lt;= 2, then nums becomes [3].
+3- We choose and remove nums[0] since 3 &lt;= 3, then nums becomes [].
+Hence, the answer is 3.
+It can be shown that we can&#39;t process more than 3 queries.
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [3,4,3], queries = [4,3,2]
+<strong>Output:</strong> 2
+<strong>Explanation:</strong> First we replace nums with the subsequence of nums [4,3].
+Then we can process the queries as follows:
+1- We choose and remove nums[0] since 4 &lt;= 4, then nums becomes [3].
+2- We choose and remove nums[0] since 3 &lt;= 3, then nums becomes [].
+3- We can not process any more queries since nums is empty.
+Hence, the answer is 2.
+It can be shown that we can&#39;t process more than 2 queries.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length &lt;= 1000</code></li>
+	<li><code>1 &lt;= queries.length &lt;= 1000</code></li>
+	<li><code>1 &lt;= nums[i], queries[i] &lt;= 10<sup>9</sup></code></li>
+</ul>
+
+## Solutions
+
+### Solution 1
+
+<!-- tabs:start -->
+
+```python
+
+```
+
+```java
+
+```
+
+```cpp
+
+```
+
+```go
+
+```
+
+<!-- tabs:end -->
+
+<!-- end -->

solution/README.md

@@ -3028,6 +3028,7 @@
 |  3015  |  [按距离统计房屋对数目 I](/solution/3000-3099/3015.Count%20the%20Number%20of%20Houses%20at%20a%20Certain%20Distance%20I/README.md)  |    |  中等  |  第 381 场周赛  |
 |  3016  |  [输入单词需要的最少按键次数 II](/solution/3000-3099/3016.Minimum%20Number%20of%20Pushes%20to%20Type%20Word%20II/README.md)  |    |  中等  |  第 381 场周赛  |
 |  3017  |  [按距离统计房屋对数目 II](/solution/3000-3099/3017.Count%20the%20Number%20of%20Houses%20at%20a%20Certain%20Distance%20II/README.md)  |    |  困难  |  第 381 场周赛  |
+|  3018  |  [Maximum Number of Removal Queries That Can Be Processed I](/solution/3000-3099/3018.Maximum%20Number%20of%20Removal%20Queries%20That%20Can%20Be%20Processed%20I/README.md)  |    |  困难  |  🔒  |
 
 ## 版权
 

solution/README_EN.md

@@ -3026,6 +3026,7 @@ Press <kbd>Control</kbd> + <kbd>F</kbd>(or <kbd>Command</kbd> + <kbd>F</kbd> on
 |  3015  |  [Count the Number of Houses at a Certain Distance I](/solution/3000-3099/3015.Count%20the%20Number%20of%20Houses%20at%20a%20Certain%20Distance%20I/README_EN.md)  |    |  Medium  |  Weekly Contest 381  |
 |  3016  |  [Minimum Number of Pushes to Type Word II](/solution/3000-3099/3016.Minimum%20Number%20of%20Pushes%20to%20Type%20Word%20II/README_EN.md)  |    |  Medium  |  Weekly Contest 381  |
 |  3017  |  [Count the Number of Houses at a Certain Distance II](/solution/3000-3099/3017.Count%20the%20Number%20of%20Houses%20at%20a%20Certain%20Distance%20II/README_EN.md)  |    |  Hard  |  Weekly Contest 381  |
+|  3018  |  [Maximum Number of Removal Queries That Can Be Processed I](/solution/3000-3099/3018.Maximum%20Number%20of%20Removal%20Queries%20That%20Can%20Be%20Processed%20I/README_EN.md)  |    |  Hard  |  🔒  |
 
 ## Copyright