chore: update lc problems

Author: yanglbme

solution/3400-3499/3482.Analyze Organization Hierarchy/README.md

@@ -135,7 +135,7 @@ employee_id 是这张表的唯一主键。
 
 <ul>
 	<li>结果先以层级升序排序</li>
-	<li>在同一层级内，员工按预算降序排序，然后按姓名降序排序</li>
+	<li>在同一层级内，员工按预算降序排序，然后按姓名升序排序</li>
 </ul>
 </div>
 

solution/3400-3499/3496.Maximize Score After Pair Deletions/README.md

@@ -2,6 +2,9 @@
 comments: true
 difficulty: 中等
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3400-3499/3496.Maximize%20Score%20After%20Pair%20Deletions/README.md
+tags:
+    - 贪心
+    - 数组
 ---
 
 <!-- problem:start -->

solution/3400-3499/3496.Maximize Score After Pair Deletions/README_EN.md

@@ -2,6 +2,9 @@
 comments: true
 difficulty: Medium
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3400-3499/3496.Maximize%20Score%20After%20Pair%20Deletions/README_EN.md
+tags:
+    - Greedy
+    - Array
 ---
 
 <!-- problem:start -->

solution/3400-3499/3498.Reverse Degree of a String/README.md

@@ -2,6 +2,9 @@
 comments: true
 difficulty: 简单
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3400-3499/3498.Reverse%20Degree%20of%20a%20String/README.md
+tags:
+    - 字符串
+    - 模拟
 ---
 
 <!-- problem:start -->

solution/3400-3499/3498.Reverse Degree of a String/README_EN.md

@@ -2,6 +2,9 @@
 comments: true
 difficulty: Easy
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3400-3499/3498.Reverse%20Degree%20of%20a%20String/README_EN.md
+tags:
+    - String
+    - Simulation
 ---
 
 <!-- problem:start -->

solution/3400-3499/3499.Maximize Active Section with Trade I/README.md

@@ -2,6 +2,9 @@
 comments: true
 difficulty: 中等
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3400-3499/3499.Maximize%20Active%20Section%20with%20Trade%20I/README.md
+tags:
+    - 字符串
+    - 枚举
 ---
 
 <!-- problem:start -->

solution/3400-3499/3499.Maximize Active Section with Trade I/README_EN.md

@@ -2,6 +2,9 @@
 comments: true
 difficulty: Medium
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3400-3499/3499.Maximize%20Active%20Section%20with%20Trade%20I/README_EN.md
+tags:
+    - String
+    - Enumeration
 ---
 
 <!-- problem:start -->

solution/3500-3599/3500.Minimum Cost to Divide Array Into Subarrays/README.md

@@ -2,6 +2,10 @@
 comments: true
 difficulty: 困难
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3500-3599/3500.Minimum%20Cost%20to%20Divide%20Array%20Into%20Subarrays/README.md
+tags:
+    - 数组
+    - 动态规划
+    - 前缀和
 ---
 
 <!-- problem:start -->

solution/3500-3599/3500.Minimum Cost to Divide Array Into Subarrays/README_EN.md

@@ -2,6 +2,10 @@
 comments: true
 difficulty: Hard
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3500-3599/3500.Minimum%20Cost%20to%20Divide%20Array%20Into%20Subarrays/README_EN.md
+tags:
+    - Array
+    - Dynamic Programming
+    - Prefix Sum
 ---
 
 <!-- problem:start -->

solution/3500-3599/3501.Maximize Active Section with Trade II/README.md

@@ -2,6 +2,11 @@
 comments: true
 difficulty: 困难
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3500-3599/3501.Maximize%20Active%20Section%20with%20Trade%20II/README.md
+tags:
+    - 线段树
+    - 数组
+    - 字符串
+    - 二分查找
 ---
 
 <!-- problem:start -->

solution/3500-3599/3501.Maximize Active Section with Trade II/README_EN.md

@@ -2,6 +2,11 @@
 comments: true
 difficulty: Hard
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3500-3599/3501.Maximize%20Active%20Section%20with%20Trade%20II/README_EN.md
+tags:
+    - Segment Tree
+    - Array
+    - String
+    - Binary Search
 ---
 
 <!-- problem:start -->

solution/3500-3599/3502.Minimum Cost to Reach Every Position/README.md

@@ -2,6 +2,8 @@
 comments: true
 difficulty: 简单
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3500-3599/3502.Minimum%20Cost%20to%20Reach%20Every%20Position/README.md
+tags:
+    - 数组
 ---
 
 <!-- problem:start -->

solution/3500-3599/3502.Minimum Cost to Reach Every Position/README_EN.md

@@ -2,6 +2,8 @@
 comments: true
 difficulty: Easy
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3500-3599/3502.Minimum%20Cost%20to%20Reach%20Every%20Position/README_EN.md
+tags:
+    - Array
 ---
 
 <!-- problem:start -->

solution/3500-3599/3503.Longest Palindrome After Substring Concatenation I/README.md

@@ -2,6 +2,11 @@
 comments: true
 difficulty: 中等
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3500-3599/3503.Longest%20Palindrome%20After%20Substring%20Concatenation%20I/README.md
+tags:
+    - 双指针
+    - 字符串
+    - 动态规划
+    - 枚举
 ---
 
 <!-- problem:start -->

solution/3500-3599/3503.Longest Palindrome After Substring Concatenation I/README_EN.md

@@ -2,6 +2,11 @@
 comments: true
 difficulty: Medium
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3500-3599/3503.Longest%20Palindrome%20After%20Substring%20Concatenation%20I/README_EN.md
+tags:
+    - Two Pointers
+    - String
+    - Dynamic Programming
+    - Enumeration
 ---
 
 <!-- problem:start -->

solution/3500-3599/3504.Longest Palindrome After Substring Concatenation II/README.md

@@ -2,6 +2,10 @@
 comments: true
 difficulty: 困难
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3500-3599/3504.Longest%20Palindrome%20After%20Substring%20Concatenation%20II/README.md
+tags:
+    - 双指针
+    - 字符串
+    - 动态规划
 ---
 
 <!-- problem:start -->

solution/3500-3599/3504.Longest Palindrome After Substring Concatenation II/README_EN.md

@@ -2,6 +2,10 @@
 comments: true
 difficulty: Hard
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3500-3599/3504.Longest%20Palindrome%20After%20Substring%20Concatenation%20II/README_EN.md
+tags:
+    - Two Pointers
+    - String
+    - Dynamic Programming
 ---
 
 <!-- problem:start -->

solution/3500-3599/3505.Minimum Operations to Make Elements Within K Subarrays Equal/README.md

@@ -2,6 +2,13 @@
 comments: true
 difficulty: 困难
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3500-3599/3505.Minimum%20Operations%20to%20Make%20Elements%20Within%20K%20Subarrays%20Equal/README.md
+tags:
+    - 数组
+    - 哈希表
+    - 数学
+    - 动态规划
+    - 滑动窗口
+    - 堆（优先队列）
 ---
 
 <!-- problem:start -->

solution/3500-3599/3505.Minimum Operations to Make Elements Within K Subarrays Equal/README_EN.md

@@ -2,6 +2,13 @@
 comments: true
 difficulty: Hard
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3500-3599/3505.Minimum%20Operations%20to%20Make%20Elements%20Within%20K%20Subarrays%20Equal/README_EN.md
+tags:
+    - Array
+    - Hash Table
+    - Math
+    - Dynamic Programming
+    - Sliding Window
+    - Heap (Priority Queue)
 ---
 
 <!-- problem:start -->

solution/3500-3599/3506.Find Time Required to Eliminate Bacterial Strains/README.md

@@ -0,0 +1,121 @@
+---
+comments: true
+difficulty: 困难
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3500-3599/3506.Find%20Time%20Required%20to%20Eliminate%20Bacterial%20Strains/README.md
+---
+
+<!-- problem:start -->
+
+# [3506. 查找消除细菌菌株所需时间 II 🔒](https://leetcode.cn/problems/find-time-required-to-eliminate-bacterial-strains)
+
+[English Version](/solution/3500-3599/3506.Find%20Time%20Required%20to%20Eliminate%20Bacterial%20Strains/README_EN.md)
+
+## 题目描述
+
+<!-- description:start -->
+
+<p>给定一个整数数组&nbsp;<code>timeReq</code>&nbsp;和一个整数&nbsp;<code>splitTime</code>。</p>
+
+<p>在人体微观世界中，免疫系统面临着一项非凡的挑战：对抗快速繁殖的细菌群落，这对身体的生存构成威胁。</p>
+
+<p>最初，只部署一个 <strong>白细胞</strong>（<strong>WBC</strong>）来消除细菌。然而，单独的白细胞很快意识到它无法跟上细菌的生长速度。</p>
+
+<p>WBC制定了一种巧妙的策略来对抗细菌：</p>
+
+<ul>
+	<li>第 <code>i</code> 个细菌菌株需要 <code>timeReq[i]</code> 个时间单位来被消除。</li>
+	<li>单个白细胞只能消除 <strong>一种</strong> 细菌菌株。之后，白细胞耗尽，无法执行任何其他任务。</li>
+	<li>一个白细胞可以将自身分裂为两个白细胞，但这需要&nbsp;<code>splitTime</code>&nbsp;单位时间。一旦分裂，两个白细胞就可以 <strong>并行</strong> 消灭细菌。</li>
+	<li>仅有一个白细胞可以针对一个单一细菌菌株工作。多个白细胞不能同时攻击一个菌株。</li>
+</ul>
+
+<p>您必须确定消除所有细菌菌株所需的 <strong>最短</strong> 时间。</p>
+
+<p><strong>注意</strong>，细菌菌株可以按任何顺序消除。</p>
+
+<p>&nbsp;</p>
+
+<p><strong class="example">示例 1：</strong></p>
+
+<div class="example-block">
+<p><span class="example-io"><b>输入：</b>timeReq = [10,4,5], splitTime = 2</span></p>
+
+<p><span class="example-io"><b>输出：</b>12</span></p>
+
+<p><b>解释：</b></p>
+
+<p>消除过程如下：</p>
+
+<ul>
+	<li>最初，有一个白细胞。经过 2 个时间单位后，白细胞分裂成 2 个白细胞。</li>
+	<li>其中一个白细胞在&nbsp;<code>t = 2 + 10 = 12</code>&nbsp;时间内消除菌株 0。另一个白细胞使用 2 个单位时间再次分裂。</li>
+	<li>2 个新的白细胞消灭细菌的时间是 <code>t = 2 + 2 + 4</code> 和&nbsp;<code>t = 2 + 2 + 5</code>。</li>
+</ul>
+</div>
+
+<p><strong class="example">示例 2：</strong></p>
+
+<div class="example-block">
+<p><span class="example-io"><b>输入：</b>timeReq = [10,4], splitTime = 5</span></p>
+
+<p><b>输出：</b>5</p>
+
+<p><strong>解释：</strong></p>
+
+<p>消除过程如下：</p>
+
+<ul>
+	<li>最初，有一个白细胞。经过 5 个时间单位后，白细胞分裂成 2 个白细胞。</li>
+	<li>2 个新的白细胞消灭细菌的时间是&nbsp;<code>t = 5 + 10</code> 和&nbsp;<code>t = 5 + 4</code>。</li>
+</ul>
+</div>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>2 &lt;= timeReq.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>1 &lt;= timeReq[i] &lt;= 10<sup>9</sup></code></li>
+	<li><code>1 &lt;= splitTime &lt;= 10<sup>9</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 -->