feat: add solutions to lc problems: No.2383~2386

* No.2383.Minimum Hours of Training to Win a Competition
* No.2384.Largest Palindromic Number
* No.2385.Amount of Time for Binary Tree to Be Infected
* No.2386.Find the K-Sum of an Array

Author: yanglbme

solution/1400-1499/1455.Check If a Word Occurs As a Prefix of Any Word in a Sentence/README.md

@@ -57,7 +57,7 @@
 
 将 $sentence$ 按空格分割为 $words$，然后遍历 $words$，检查 $words[i]$ 是否是 $searchWord$ 的前缀，是则返回 $i+1$。若遍历结束，所有单词都不满足，返回 $-1$。
 
-时间复杂度 $O(mn)$。其中 $m$ 是 $sentence$ 的长度，$n$ 是 $searchWord$ 的长度。
+时间复杂度 $O(mn)$。其中 $m$ 是 $sentence$ 的长度，而 $n$ 是 $searchWord$ 的长度。
 
 
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solution/2300-2399/2379.Minimum Recolors to Get K Consecutive Black Blocks/README.md

@@ -0,0 +1,87 @@
+# [2379. 得到 K 个黑块的最少涂色次数](https://leetcode.cn/problems/minimum-recolors-to-get-k-consecutive-black-blocks)
+
+[English Version](/solution/2300-2399/2379.Minimum%20Recolors%20to%20Get%20K%20Consecutive%20Black%20Blocks/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>给你一个长度为 <code>n</code>&nbsp;下标从 <strong>0</strong>&nbsp;开始的字符串&nbsp;<code>blocks</code>&nbsp;，<code>blocks[i]</code>&nbsp;要么是&nbsp;<code>'W'</code>&nbsp;要么是&nbsp;<code>'B'</code>&nbsp;，表示第&nbsp;<code>i</code>&nbsp;块的颜色。字符&nbsp;<code>'W'</code> 和&nbsp;<code>'B'</code>&nbsp;分别表示白色和黑色。</p>
+
+<p>给你一个整数&nbsp;<code>k</code>&nbsp;，表示想要&nbsp;<strong>连续</strong>&nbsp;黑色块的数目。</p>
+
+<p>每一次操作中，你可以选择一个白色块将它 <strong>涂成</strong>&nbsp;黑色块。</p>
+
+<p>请你返回至少出现 <strong>一次</strong>&nbsp;连续 <code>k</code>&nbsp;个黑色块的 <strong>最少</strong>&nbsp;操作次数。</p>
+
+<p>&nbsp;</p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre>
+<b>输入：</b>blocks = "WBBWWBBWBW", k = 7
+<b>输出：</b>3
+<strong>解释：</strong>
+一种得到 7 个连续黑色块的方法是把第 0 ，3 和 4 个块涂成黑色。
+得到 blocks = "BBBBBBBWBW" 。
+可以证明无法用少于 3 次操作得到 7 个连续的黑块。
+所以我们返回 3 。
+</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre>
+<b>输入：</b>blocks = "WBWBBBW", k = 2
+<b>输出：</b>0
+<strong>解释：</strong>
+不需要任何操作，因为已经有 2 个连续的黑块。
+所以我们返回 0 。
+</pre>
+
+<p>&nbsp;</p>
+
+<p><b>提示：</b></p>
+
+<ul>
+	<li><code>n == blocks.length</code></li>
+	<li><code>1 &lt;= n &lt;= 100</code></li>
+	<li><code>blocks[i]</code>&nbsp;要么是&nbsp;<code>'W'</code>&nbsp;，要么是&nbsp;<code>'B'</code> 。</li>
+	<li><code>1 &lt;= k &lt;= n</code></li>
+</ul>
+
+## 解法
+
+<!-- 这里可写通用的实现逻辑 -->
+
+<!-- tabs:start -->
+
+### **Python3**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```python
+
+```
+
+### **Java**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```java
+
+```
+
+### **TypeScript**
+
+```ts
+
+```
+
+### **...**
+
+```
+
+
+```
+
+<!-- tabs:end -->

solution/2300-2399/2379.Minimum Recolors to Get K Consecutive Black Blocks/README_EN.md

@@ -0,0 +1,77 @@
+# [2379. Minimum Recolors to Get K Consecutive Black Blocks](https://leetcode.com/problems/minimum-recolors-to-get-k-consecutive-black-blocks)
+
+[中文文档](/solution/2300-2399/2379.Minimum%20Recolors%20to%20Get%20K%20Consecutive%20Black%20Blocks/README.md)
+
+## Description
+
+<p>You are given a <strong>0-indexed</strong> string <code>blocks</code> of length <code>n</code>, where <code>blocks[i]</code> is either <code>&#39;W&#39;</code> or <code>&#39;B&#39;</code>, representing the color of the <code>i<sup>th</sup></code> block. The characters <code>&#39;W&#39;</code> and <code>&#39;B&#39;</code> denote the colors white and black, respectively.</p>
+
+<p>You are also given an integer <code>k</code>, which is the desired number of <strong>consecutive</strong> black blocks.</p>
+
+<p>In one operation, you can <strong>recolor</strong> a white block such that it becomes a black block.</p>
+
+<p>Return<em> the <strong>minimum</strong> number of operations needed such that there is at least <strong>one</strong> occurrence of </em><code>k</code><em> consecutive black blocks.</em></p>
+
+<p>&nbsp;</p>
+<p><strong>Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> blocks = &quot;WBBWWBBWBW&quot;, k = 7
+<strong>Output:</strong> 3
+<strong>Explanation:</strong>
+One way to achieve 7 consecutive black blocks is to recolor the 0th, 3rd, and 4th blocks
+so that blocks = &quot;BBBBBBBWBW&quot;. 
+It can be shown that there is no way to achieve 7 consecutive black blocks in less than 3 operations.
+Therefore, we return 3.
+</pre>
+
+<p><strong>Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> blocks = &quot;WBWBBBW&quot;, k = 2
+<strong>Output:</strong> 0
+<strong>Explanation:</strong>
+No changes need to be made, since 2 consecutive black blocks already exist.
+Therefore, we return 0.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>n == blocks.length</code></li>
+	<li><code>1 &lt;= n &lt;= 100</code></li>
+	<li><code>blocks[i]</code> is either <code>&#39;W&#39;</code> or <code>&#39;B&#39;</code>.</li>
+	<li><code>1 &lt;= k &lt;= n</code></li>
+</ul>
+
+## Solutions
+
+<!-- tabs:start -->
+
+### **Python3**
+
+```python
+
+```
+
+### **Java**
+
+```java
+
+```
+
+### **TypeScript**
+
+```ts
+
+```
+
+### **...**
+
+```
+
+
+```
+
+<!-- tabs:end -->

solution/2300-2399/2380.Time Needed to Rearrange a Binary String/README.md

@@ -0,0 +1,83 @@
+# [2380. 二进制字符串重新安排顺序需要的时间](https://leetcode.cn/problems/time-needed-to-rearrange-a-binary-string)
+
+[English Version](/solution/2300-2399/2380.Time%20Needed%20to%20Rearrange%20a%20Binary%20String/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>给你一个二进制字符串&nbsp;<code>s</code>&nbsp;。在一秒之中，<strong>所有</strong>&nbsp;子字符串&nbsp;<code>"01"</code> <strong>同时</strong>&nbsp;被替换成&nbsp;<code>"10"</code>&nbsp;。这个过程持续进行到没有&nbsp;<code>"01"</code>&nbsp;存在。</p>
+
+<p>请你返回完成这个过程所需要的秒数。</p>
+
+<p>&nbsp;</p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre>
+<b>输入：</b>s = "0110101"
+<b>输出：</b>4
+<b>解释：</b>
+一秒后，s 变成 "1011010" 。
+再过 1 秒后，s 变成 "1101100" 。
+第三秒过后，s 变成 "1110100" 。
+第四秒后，s 变成 "1111000" 。
+此时没有 "01" 存在，整个过程花费 4 秒。
+所以我们返回 4 。
+</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre>
+<b>输入：</b>s = "11100"
+<b>输出：</b>0
+<strong>解释：</strong>
+s 中没有 "01" 存在，整个过程花费 0 秒。
+所以我们返回 0 。
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= s.length &lt;= 1000</code></li>
+	<li><code>s[i]</code>&nbsp;要么是&nbsp;<code>'0'</code>&nbsp;，要么是&nbsp;<code>'1'</code> 。</li>
+</ul>
+
+## 解法
+
+<!-- 这里可写通用的实现逻辑 -->
+
+<!-- tabs:start -->
+
+### **Python3**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```python
+
+```
+
+### **Java**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```java
+
+```
+
+### **TypeScript**
+
+```ts
+
+```
+
+### **...**
+
+```
+
+
+```
+
+<!-- tabs:end -->

solution/2300-2399/2380.Time Needed to Rearrange a Binary String/README_EN.md

@@ -0,0 +1,73 @@
+# [2380. Time Needed to Rearrange a Binary String](https://leetcode.com/problems/time-needed-to-rearrange-a-binary-string)
+
+[中文文档](/solution/2300-2399/2380.Time%20Needed%20to%20Rearrange%20a%20Binary%20String/README.md)
+
+## Description
+
+<p>You are given a binary string <code>s</code>. In one second, <strong>all</strong> occurrences of <code>&quot;01&quot;</code> are <strong>simultaneously</strong> replaced with <code>&quot;10&quot;</code>. This process <strong>repeats</strong> until no occurrences of <code>&quot;01&quot;</code> exist.</p>
+
+<p>Return<em> the number of seconds needed to complete this process.</em></p>
+
+<p>&nbsp;</p>
+<p><strong>Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;0110101&quot;
+<strong>Output:</strong> 4
+<strong>Explanation:</strong> 
+After one second, s becomes &quot;1011010&quot;.
+After another second, s becomes &quot;1101100&quot;.
+After the third second, s becomes &quot;1110100&quot;.
+After the fourth second, s becomes &quot;1111000&quot;.
+No occurrence of &quot;01&quot; exists any longer, and the process needed 4 seconds to complete,
+so we return 4.
+</pre>
+
+<p><strong>Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;11100&quot;
+<strong>Output:</strong> 0
+<strong>Explanation:</strong>
+No occurrence of &quot;01&quot; exists in s, and the processes needed 0 seconds to complete,
+so we return 0.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= s.length &lt;= 1000</code></li>
+	<li><code>s[i]</code> is either <code>&#39;0&#39;</code> or <code>&#39;1&#39;</code>.</li>
+</ul>
+
+## Solutions
+
+<!-- tabs:start -->
+
+### **Python3**
+
+```python
+
+```
+
+### **Java**
+
+```java
+
+```
+
+### **TypeScript**
+
+```ts
+
+```
+
+### **...**
+
+```
+
+
+```
+
+<!-- tabs:end -->