feat: add new lc problems

Author: yanglbme

solution/1300-1399/1323.Maximum 69 Number/README_EN.md

@@ -29,7 +29,7 @@ tags:
 <pre>
 <strong>Input:</strong> num = 9669
 <strong>Output:</strong> 9969
-<strong>Explanation:</strong>
+<strong>Explanation:</strong> 
 Changing the first digit results in 6669.
 Changing the second digit results in 9969.
 Changing the third digit results in 9699.

solution/3600-3699/3643.Flip Square Submatrix Vertically/README.md

@@ -2,6 +2,8 @@
 comments: true
 difficulty: 简单
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3600-3699/3643.Flip%20Square%20Submatrix%20Vertically/README.md
+rating: 1234
+source: 第 462 场周赛 Q1
 ---
 
 <!-- problem:start -->

solution/3600-3699/3643.Flip Square Submatrix Vertically/README_EN.md

@@ -2,6 +2,8 @@
 comments: true
 difficulty: Easy
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3600-3699/3643.Flip%20Square%20Submatrix%20Vertically/README_EN.md
+rating: 1234
+source: Weekly Contest 462 Q1
 ---
 
 <!-- problem:start -->

solution/3600-3699/3644.Maximum K to Sort a Permutation/README.md

@@ -2,6 +2,8 @@
 comments: true
 difficulty: 中等
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3600-3699/3644.Maximum%20K%20to%20Sort%20a%20Permutation/README.md
+rating: 1775
+source: 第 462 场周赛 Q2
 ---
 
 <!-- problem:start -->

solution/3600-3699/3644.Maximum K to Sort a Permutation/README_EN.md

@@ -2,6 +2,8 @@
 comments: true
 difficulty: Medium
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3600-3699/3644.Maximum%20K%20to%20Sort%20a%20Permutation/README_EN.md
+rating: 1775
+source: Weekly Contest 462 Q2
 ---
 
 <!-- problem:start -->

solution/3600-3699/3645.Maximum Total from Optimal Activation Order/README.md

@@ -2,6 +2,8 @@
 comments: true
 difficulty: 中等
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3600-3699/3645.Maximum%20Total%20from%20Optimal%20Activation%20Order/README.md
+rating: 2018
+source: 第 462 场周赛 Q3
 ---
 
 <!-- problem:start -->

solution/3600-3699/3645.Maximum Total from Optimal Activation Order/README_EN.md

@@ -2,6 +2,8 @@
 comments: true
 difficulty: Medium
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3600-3699/3645.Maximum%20Total%20from%20Optimal%20Activation%20Order/README_EN.md
+rating: 2018
+source: Weekly Contest 462 Q3
 ---
 
 <!-- problem:start -->

solution/3600-3699/3646.Next Special Palindrome Number/README.md

@@ -2,6 +2,8 @@
 comments: true
 difficulty: 困难
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3600-3699/3646.Next%20Special%20Palindrome%20Number/README.md
+rating: 2445
+source: 第 462 场周赛 Q4
 ---
 
 <!-- problem:start -->

solution/3600-3699/3646.Next Special Palindrome Number/README_EN.md

@@ -2,6 +2,8 @@
 comments: true
 difficulty: Hard
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3600-3699/3646.Next%20Special%20Palindrome%20Number/README_EN.md
+rating: 2445
+source: Weekly Contest 462 Q4
 ---
 
 <!-- problem:start -->

solution/3600-3699/3648.Minimum Sensors to Cover Grid/README.md

@@ -0,0 +1,99 @@
+---
+comments: true
+difficulty: 中等
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3600-3699/3648.Minimum%20Sensors%20to%20Cover%20Grid/README.md
+---
+
+<!-- problem:start -->
+
+# [3648. 覆盖网格的最少传感器数目](https://leetcode.cn/problems/minimum-sensors-to-cover-grid)
+
+[English Version](/solution/3600-3699/3648.Minimum%20Sensors%20to%20Cover%20Grid/README_EN.md)
+
+## 题目描述
+
+<!-- description:start -->
+
+<p>给你一个 <code>n × m</code> 的网格和一个整数 <code>k</code>。</p>
+
+<p>一个放置在单元格 <code>(r, c)</code> 的传感器可以覆盖所有与 <code>(r, c)</code> 的&nbsp;<strong>切比雪夫距离</strong><strong>不超过</strong> <code>k</code> 的单元格。</p>
+
+<p>两个单元格 <code>(r<sub>1</sub>, c<sub>1</sub>)</code> 和 <code>(r<sub>2</sub>, c<sub>2</sub>)</code> 之间的&nbsp;<strong>切比雪夫距离&nbsp;</strong>为 <code>max(|r<sub>1</sub> − r<sub>2</sub>|,|c<sub>1</sub> − c<sub>2</sub>|)</code>。</p>
+
+<p>你的任务是返回覆盖整个网格所需传感器的&nbsp;<strong>最少&nbsp;</strong>数量。</p>
+
+<p>&nbsp;</p>
+
+<p><strong class="example">示例 1:</strong></p>
+
+<div class="example-block">
+<p><strong>输入:</strong> <span class="example-io">n = 5, m = 5, k = 1</span></p>
+
+<p><strong>输出:</strong> <span class="example-io">4</span></p>
+
+<p><strong>解释:</strong></p>
+
+<p>在位置 <code>(0, 3)</code>、<code>(1, 0)</code>、<code>(3, 3)</code> 和 <code>(4, 1)</code> 放置传感器可以确保网格中的每个单元格都被覆盖。因此，答案是 4。</p>
+</div>
+
+<p><strong class="example">示例 2:</strong></p>
+
+<div class="example-block">
+<p><strong>输入:</strong> <span class="example-io">n = 2, m = 2, k = 2</span></p>
+
+<p><strong>输出:</strong> <span class="example-io">1</span></p>
+
+<p><strong>解释:</strong></p>
+
+<p>当 <code>k = 2</code> 时，无论传感器放在哪个位置，单个传感器都可以覆盖整个 <code>2 * 2</code> 的网格。因此，答案是 1。</p>
+</div>
+
+<p>&nbsp;</p>
+
+<p><strong>提示:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= n &lt;= 10<sup>3</sup></code></li>
+	<li><code>1 &lt;= m &lt;= 10<sup>3</sup></code></li>
+	<li><code>0 &lt;= k &lt;= 10<sup>3</sup></code></li>
+</ul>
+
+<!-- description:end -->
+
+## 解法
+
+<!-- solution:start -->
+
+### 方法一
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+
+```
+
+#### Java
+
+```java
+
+```
+
+#### C++
+
+```cpp
+
+```
+
+#### Go
+
+```go
+
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->