feat: add new lc problems

solution/3500-3599/3540.Minimum Time to Visit All Houses/README.md

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+---
+comments: true
+difficulty: 中等
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3500-3599/3540.Minimum%20Time%20to%20Visit%20All%20Houses/README.md
+---
+
+<!-- problem:start -->
+
+# [3540. 访问所有房屋的最短时间 🔒](https://leetcode.cn/problems/minimum-time-to-visit-all-houses)
+
+[English Version](/solution/3500-3599/3540.Minimum%20Time%20to%20Visit%20All%20Houses/README_EN.md)
+
+## 题目描述
+
+<!-- description:start -->
+
+<p>给定两个整数数组&nbsp;<code>forward</code> 和&nbsp;<code>backward</code>，长度都为&nbsp;<code>n</code>。同时给定另一个整数数组&nbsp;<code>queries</code>。</p>
+
+<p>有&nbsp;<code>n</code>&nbsp;个排列为环形的房屋。房屋通过道路以特殊方式相连：</p>
+
+<ul>
+	<li>对于所有的&nbsp;<code>0 &lt;= i &lt;= n - 2</code>，房屋&nbsp;<code>i</code>&nbsp;通过一条长度为&nbsp;<code>forward[i]</code>&nbsp;米的道路连接到房屋&nbsp;<code>i + 1</code>。另外，房屋&nbsp;<code>n - 1</code>&nbsp;通过一条长度为&nbsp;<code>forward[n - 1]</code>&nbsp;米的道路连接回房屋 0，形成一个环。</li>
+	<li>对于所有的 <code>1 &lt;= i &lt;= n - 1</code>，房屋&nbsp;<code>i</code>&nbsp;通过一条长度为&nbsp;<code>backward[i]</code>&nbsp;米的道路连接到房屋&nbsp;<code>i - 1</code>。另外，房屋&nbsp;0 通过一条长度为&nbsp;<code>backward[n - 1]</code>&nbsp;米的道路连接回房屋&nbsp;<code>n - 1</code>，形成一个环。</li>
+</ul>
+
+<p>你可以以 <strong>1</strong> 米每秒的速度行走。从房屋&nbsp;0 开始，找到按照&nbsp;<code>queries</code>&nbsp;指定的顺序访问每所房屋的 <strong>最小</strong> 时间。</p>
+
+<p>返回访问房屋所需的 <strong>最短</strong> 总时间。</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>forward = [1,4,4], backward = [4,1,2], queries = [1,2,0,2]</span></p>
+
+<p><b>输出：</b>12</p>
+
+<p><b>解释：</b></p>
+
+<p>路径如下：<code><u>0</u><sup>(0)</sup> → <u>1</u><sup>(1)</sup> → <u>2</u><sup>(5)</sup> <u>→</u> 1<sup>(7)</sup> <u>→</u> <u>0</u><sup>(8)</sup> <u>→</u> <u>2</u><sup>(12)</sup></code>。</p>
+
+<p><strong>注意：</strong>使用的&nbsp;<code>node<sup>(total time)</sup></code>&nbsp;符号，<code>→</code>&nbsp;表示前向道路，<code><u>→</u></code>&nbsp;表示反向道路。</p>
+</div>
+
+<p><strong class="example">示例 2：</strong></p>
+
+<div class="example-block">
+<p><strong>输入：</strong><span class="example-io">forward = [1,1,1,1], backward = [2,2,2,2], queries = [1,2,3,0]</span></p>
+
+<p><span class="example-io"><b>输出：</b>4</span></p>
+
+<p><strong>解释：</strong></p>
+
+<p>经过路径是&nbsp;<code><u>0</u> → <u>1</u> → <u>2</u> →​​​​​​​ <u>3</u> → <u>0</u></code>。每一步都在前向方向，需要 1 秒。</p>
+</div>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>2 &lt;= n &lt;= 10<sup>5</sup></code></li>
+	<li><code>n == forward.length == backward.length</code></li>
+	<li><code>1 &lt;= forward[i], backward[i] &lt;= 10<sup>5</sup></code></li>
+	<li><code>1 &lt;= queries.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>0 &lt;= queries[i] &lt; n</code></li>
+	<li><code>queries[i] != queries[i + 1]</code></li>
+	<li><code>queries[0]</code>&nbsp;非 0。</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/3500-3599/3540.Minimum Time to Visit All Houses/README_EN.md

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+---
+comments: true
+difficulty: Medium
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3500-3599/3540.Minimum%20Time%20to%20Visit%20All%20Houses/README_EN.md
+---
+
+<!-- problem:start -->
+
+# [3540. Minimum Time to Visit All Houses 🔒](https://leetcode.com/problems/minimum-time-to-visit-all-houses)
+
+[中文文档](/solution/3500-3599/3540.Minimum%20Time%20to%20Visit%20All%20Houses/README.md)
+
+## Description
+
+<!-- description:start -->
+
+<p>You are given two integer arrays <code>forward</code> and <code>backward</code>, both of size <code>n</code>. You are also given another integer array <code>queries</code>.</p>
+
+<p>There are <code>n</code> houses <em>arranged in a circle</em>. The houses are connected via roads in a special arrangement:</p>
+
+<ul>
+	<li>For all <code>0 &lt;= i &lt;= n - 2</code>, house <code>i</code> is connected to house <code>i + 1</code> via a road with length <code>forward[i]</code> meters. Additionally, house <code>n - 1</code> is connected back to house 0 via a road with length <code>forward[n - 1]</code> meters, completing the circle.</li>
+	<li>For all <code>1 &lt;= i &lt;= n - 1</code>, house <code>i</code> is connected to house <code>i - 1</code> via a road with length <code>backward[i]</code> meters. Additionally, house 0 is connected back to house <code>n - 1</code> via a road with length <code>backward[0]</code> meters, completing the circle.</li>
+</ul>
+
+<p>You can walk at a pace of <strong>one</strong> meter per second. Starting from house 0, find the <strong>minimum</strong> time taken to visit each house in the order specified by <code>queries</code>.</p>
+
+<p>Return the <strong>minimum</strong> total time taken to visit the houses.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">forward = [1,4,4], backward = [4,1,2], queries = [1,2,0,2]</span></p>
+
+<p><strong>Output:</strong> 12</p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>The path followed is <code><u>0</u><sup>(0)</sup> &rarr; <u>1</u><sup>(1)</sup> &rarr;​​​​​​​ <u>2</u><sup>(5)</sup> <u>&rarr;</u> 1<sup>(7)</sup> <u>&rarr;</u>​​​​​​​ <u>0</u><sup>(8)</sup> <u>&rarr;</u> <u>2</u><sup>(12)</sup></code>.</p>
+
+<p><strong>Note:</strong> The notation used is <code>node<sup>(total time)</sup></code>, <code>&rarr;</code> represents forward road, and <code><u>&rarr;</u></code> represents backward road.</p>
+</div>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">forward = [1,1,1,1], backward = [2,2,2,2], queries = [1,2,3,0]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">4</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>The path travelled is <code><u>0</u> &rarr;​​​​​​​ <u>1</u> &rarr;​​​​​​​ <u>2</u> &rarr;​​​​​​​ <u>3</u> &rarr; <u>0</u></code>. Each step is in the forward direction and requires 1 second.</p>
+</div>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>2 &lt;= n &lt;= 10<sup>5</sup></code></li>
+	<li><code>n == forward.length == backward.length</code></li>
+	<li><code>1 &lt;= forward[i], backward[i] &lt;= 10<sup>5</sup></code></li>
+	<li><code>1 &lt;= queries.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>0 &lt;= queries[i] &lt; n</code></li>
+	<li><code>queries[i] != queries[i + 1]</code></li>
+	<li><code>queries[0]</code> is not 0.</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/3500-3599/3541.Find Most Frequent Vowel and Consonant/README.md

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+---
+comments: true
+difficulty: 简单
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3500-3599/3541.Find%20Most%20Frequent%20Vowel%20and%20Consonant/README.md
+---
+
+<!-- problem:start -->
+
+# [3541. 找到频率最高的元音和辅音](https://leetcode.cn/problems/find-most-frequent-vowel-and-consonant)
+
+[English Version](/solution/3500-3599/3541.Find%20Most%20Frequent%20Vowel%20and%20Consonant/README_EN.md)
+
+## 题目描述
+
+<!-- description:start -->
+
+<p>给你一个由小写英文字母（<code>'a'</code> 到 <code>'z'</code>）组成的字符串 <code>s</code>。你的任务是找出出现频率&nbsp;<strong>最高&nbsp;</strong>的元音（<code>'a'</code>、<code>'e'</code>、<code>'i'</code>、<code>'o'</code>、<code>'u'</code> 中的一个）和出现频率<strong>最高</strong>的辅音（除元音以外的所有字母），并返回这两个频率之和。</p>
+
+<p><strong>注意</strong>：如果有多个元音或辅音具有相同的最高频率，可以任选其中一个。如果字符串中没有元音或没有辅音，则其频率视为 0。</p>
+一个字母 <code>x</code> 的&nbsp;<strong>频率&nbsp;</strong>是它在字符串中出现的次数。
+
+<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 = "successes"</span></p>
+
+<p><strong>输出:</strong> <span class="example-io">6</span></p>
+
+<p><strong>解释:</strong></p>
+
+<ul>
+	<li>元音有：<code>'u'</code> 出现 1 次，<code>'e'</code> 出现 2 次。最大元音频率 = 2。</li>
+	<li>辅音有：<code>'s'</code> 出现 4 次，<code>'c'</code> 出现 2 次。最大辅音频率 = 4。</li>
+	<li>输出为 <code>2 + 4 = 6</code>。</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">s = "aeiaeia"</span></p>
+
+<p><strong>输出:</strong> <span class="example-io">3</span></p>
+
+<p><strong>解释:</strong></p>
+
+<ul>
+	<li>元音有：<code>'a'</code> 出现 3 次，<code>'e'</code> 出现 2 次，<code>'i'</code> 出现 2 次。最大元音频率 = 3。</li>
+	<li><code>s</code> 中没有辅音。因此，最大辅音频率 = 0。</li>
+	<li>输出为 <code>3 + 0 = 3</code>。</li>
+</ul>
+</div>
+
+<p>&nbsp;</p>
+
+<p><strong>提示:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= s.length &lt;= 100</code></li>
+	<li><code>s</code> 只包含小写英文字母</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 -->