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feat: add new lc problems (doocs#2263)
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‎main.py

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

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

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

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

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

‎solution/README.md

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| 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 |
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| 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 |
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| 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 |
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| 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 | 🔒 |
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## 版权
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‎solution/README_EN.md

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| 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 |
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| 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 |
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| 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 |
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| 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 | 🔒 |
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## Copyright
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