feat: add weekly contest 443

solution/3400-3499/3499.Maximize Active Section with Trade I/Solution.java

@@ -3,7 +3,7 @@ public int maxActiveSectionsAfterTrade(String s) {
         int n = s.length();
         int ans = 0, i = 0;
         int pre = Integer.MIN_VALUE, mx = 0;
-        
+
         while (i < n) {
             int j = i + 1;
             while (j < n && s.charAt(j) == s.charAt(i)) {
@@ -18,7 +18,7 @@ public int maxActiveSectionsAfterTrade(String s) {
             }
             i = j;
         }
-        
+
         ans += mx;
         return ans;
     }

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

@@ -0,0 +1,112 @@
+---
+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
+---
+
+<!-- problem:start -->
+
+# [3502. 到达每个位置的最小费用](https://leetcode.cn/problems/minimum-cost-to-reach-every-position)
+
+[English Version](/solution/3500-3599/3502.Minimum%20Cost%20to%20Reach%20Every%20Position/README_EN.md)
+
+## 题目描述
+
+<!-- description:start -->
+
+<p data-end="438" data-start="104">给你一个长度为 <code>n</code> 的整数数组 <code data-end="119" data-start="113">cost</code> 。当前你位于位置 <code data-end="166" data-start="163">n</code>（队伍的末尾），队伍中共有 <code data-end="187" data-start="180">n + 1</code> 人，编号从 0 到 <code>n</code> 。</p>
+
+<p data-end="438" data-start="104">你希望在队伍中向前移动，但队伍中每个人都会收取一定的费用才能与你 <strong>交换</strong>位置。与编号 <code data-end="375" data-start="372">i</code> 的人交换位置的费用为 <code data-end="397" data-start="388">cost[i]</code> 。</p>
+
+<p data-end="487" data-start="440">你可以按照以下规则与他人交换位置：</p>
+
+<ul data-end="632" data-start="488">
+	<li data-end="572" data-start="488">如果对方在你前面，你 <strong>必须</strong> 支付 <code data-end="546" data-start="537">cost[i]</code> 费用与他们交换位置。</li>
+	<li data-end="632" data-start="573">如果对方在你后面，他们可以免费与你交换位置。</li>
+</ul>
+
+<p data-end="755" data-start="634">返回一个大小为 <code>n</code> 的数组 <code>answer</code>，其中 <code>answer[i]</code> 表示到达队伍中每个位置 <code>i</code> 所需的 <strong data-end="680" data-start="664">最小</strong> 总费用。</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">cost = [5,3,4,1,3,2]</span></p>
+
+<p><strong>输出:</strong> <span class="example-io">[5,3,3,1,1,1]</span></p>
+
+<p><strong>解释:</strong></p>
+
+<p>我们可以通过以下方式到达每个位置：</p>
+
+<ul>
+	<li><code>i = 0</code>。可以花费 5 费用与编号 0 的人交换位置。</li>
+	<li><span class="example-io"><code>i = 1</code>。可以花费 3 费用与编号 1 的人交换位置。</span></li>
+	<li><span class="example-io"><code>i = 2</code>。可以花费 3 费用与编号 1 的人交换位置，然后免费与编号 2 的人交换位置。</span></li>
+	<li><span class="example-io"><code>i = 3</code>。可以花费 1 费用与编号 3 的人交换位置。</span></li>
+	<li><span class="example-io"><code>i = 4</code>。可以花费 1 费用与编号 3 的人交换位置，然后免费与编号 4 的人交换位置。</span></li>
+	<li><span class="example-io"><code>i = 5</code>。可以花费 1 费用与编号 3 的人交换位置，然后免费与编号 5 的人交换位置。</span></li>
+</ul>
+</div>
+
+<p><strong class="example">示例 2：</strong></p>
+
+<div class="example-block">
+<p><strong>输入:</strong> <span class="example-io">cost = [1,2,4,6,7]</span></p>
+
+<p><strong>输出:</strong> <span class="example-io">[1,1,1,1,1]</span></p>
+
+<p><strong>解释:</strong></p>
+
+<p>可以花费 1 费用与编号 0 的人交换位置，然后可以免费到达队伍中的任何位置 <code>i</code>。</p>
+</div>
+
+<p>&nbsp;</p>
+
+<p><strong>提示</strong></p>
+
+<ul>
+	<li><code>1 &lt;= n == cost.length &lt;= 100</code></li>
+	<li><code>1 &lt;= cost[i] &lt;= 100</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 -->

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

@@ -0,0 +1,110 @@
+---
+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
+---
+
+<!-- problem:start -->
+
+# [3502. Minimum Cost to Reach Every Position](https://leetcode.com/problems/minimum-cost-to-reach-every-position)
+
+[中文文档](/solution/3500-3599/3502.Minimum%20Cost%20to%20Reach%20Every%20Position/README.md)
+
+## Description
+
+<!-- description:start -->
+
+<p data-end="438" data-start="104">You are given an integer array <code data-end="119" data-start="113">cost</code> of size <code data-end="131" data-start="128">n</code>. You are currently at position <code data-end="166" data-start="163">n</code> (at the end of the line) in a line of <code data-end="187" data-start="180">n + 1</code> people (numbered from 0 to <code data-end="218" data-start="215">n</code>).</p>
+
+<p data-end="438" data-start="104">You wish to move forward in the line, but each person in front of you charges a specific amount to <strong>swap</strong> places. The cost to swap with person <code data-end="375" data-start="372">i</code> is given by <code data-end="397" data-start="388">cost[i]</code>.</p>
+
+<p data-end="487" data-start="440">You are allowed to swap places with people as follows:</p>
+
+<ul data-end="632" data-start="488">
+	<li data-end="572" data-start="488">If they are in front of you, you <strong>must</strong> pay them <code data-end="546" data-start="537">cost[i]</code> to swap with them.</li>
+	<li data-end="632" data-start="573">If they are behind you, they can swap with you for free.</li>
+</ul>
+
+<p data-end="755" data-start="634">Return an array <code>answer</code> of size <code>n</code>, where <code>answer[i]</code> is the <strong data-end="680" data-start="664">minimum</strong> total cost to reach each position <code>i</code> in the line<font face="monospace">.</font></p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">cost = [5,3,4,1,3,2]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">[5,3,3,1,1,1]</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>We can get to each position in the following way:</p>
+
+<ul>
+	<li><code>i = 0</code>. We can swap with person 0 for a cost of 5.</li>
+	<li><span class="example-io"><code><font face="monospace">i = </font>1</code>. We can swap with person 1 for a cost of 3.</span></li>
+	<li><span class="example-io"><code>i = 2</code>. We can swap with person 1 for a cost of 3, then swap with person 2 for free.</span></li>
+	<li><span class="example-io"><code>i = 3</code>. We can swap with person 3 for a cost of 1.</span></li>
+	<li><span class="example-io"><code>i = 4</code>. We can swap with person 3 for a cost of 1, then swap with person 4 for free.</span></li>
+	<li><span class="example-io"><code>i = 5</code>. We can swap with person 3 for a cost of 1, then swap with person 5 for free.</span></li>
+</ul>
+</div>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">cost = [1,2,4,6,7]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">[1,1,1,1,1]</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>We can swap with person 0 for a cost of <span class="example-io">1, then we will be able to reach any position <code>i</code> for free.</span></p>
+</div>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= n == cost.length &lt;= 100</code></li>
+	<li><code>1 &lt;= cost[i] &lt;= 100</code></li>
+</ul>
+
+<!-- description:end -->
+
+## Solutions
+
+<!-- solution:start -->
+
+### Solution 1
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+
+```
+
+#### Java
+
+```java
+
+```
+
+#### C++
+
+```cpp
+
+```
+
+#### Go
+
+```go
+
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->

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

@@ -0,0 +1,124 @@
+---
+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
+---
+
+<!-- problem:start -->
+
+# [3503. 子字符串连接后的最长回文串 I](https://leetcode.cn/problems/longest-palindrome-after-substring-concatenation-i)
+
+[English Version](/solution/3500-3599/3503.Longest%20Palindrome%20After%20Substring%20Concatenation%20I/README_EN.md)
+
+## 题目描述
+
+<!-- description:start -->
+
+<p>给你两个字符串 <code>s</code> 和 <code>t</code>。</p>
+
+<p>你可以从 <code>s</code> 中选择一个子串（可以为空）以及从 <code>t</code> 中选择一个子串（可以为空），然后将它们<strong> 按顺序 </strong>连接，得到一个新的字符串。</p>
+
+<p>返回可以由上述方法构造出的<strong> 最长</strong> 回文串的长度。</p>
+
+<p><strong>回文串</strong> 是指正着读和反着读都相同的字符串。</p>
+
+<p><strong>子字符串 </strong>是指字符串中的一个连续字符序列。</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">s = "a", t = "a"</span></p>
+
+<p><strong>输出：</strong> <span class="example-io">2</span></p>
+
+<p><strong>解释：</strong></p>
+
+<p>从 <code>s</code> 中选择 <code>"a"</code>，从 <code>t</code> 中选择 <code>"a"</code>，拼接得到 <code>"aa"</code>，这是一个长度为 2 的回文串。</p>
+</div>
+
+<p><strong class="example">示例 2：</strong></p>
+
+<div class="example-block">
+<p><strong>输入：</strong> <span class="example-io">s = "abc", t = "def"</span></p>
+
+<p><strong>输出：</strong> <span class="example-io">1</span></p>
+
+<p><strong>解释：</strong></p>
+
+<p>由于两个字符串的所有字符都不同，最长的回文串只能是任意一个单独的字符，因此答案是 1。</p>
+</div>
+
+<p><strong class="example">示例 3：</strong></p>
+
+<div class="example-block">
+<p><strong>输入：</strong> <span class="example-io">s = "b", t = "aaaa"</span></p>
+
+<p><strong>输出：</strong> 4</p>
+
+<p><strong>解释：</strong></p>
+
+<p>可以选择 <code>"aaaa"</code> 作为回文串，其长度为 4。</p>
+</div>
+
+<p><strong class="example">示例 4：</strong></p>
+
+<div class="example-block">
+<p><strong>输入：</strong> <span class="example-io">s = "abcde", t = "ecdba"</span></p>
+
+<p><strong>输出：</strong> 5</p>
+
+<p><strong>解释：</strong></p>
+
+<p>从 <code>s</code> 中选择 <code>"abc"</code>，从 <code>t</code> 中选择 <code>"ba"</code>，拼接得到 <code>"abcba"</code>，这是一个长度为 5 的回文串。</p>
+</div>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= s.length, t.length &lt;= 30</code></li>
+	<li><code>s</code> 和 <code>t</code> 仅由小写英文字母组成。</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 -->