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Commit bcdc734

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

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将 $sentence$ 按空格分割为 $words,ドル然后遍历 $words,ドル检查 $words[i]$ 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 $-1$。
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时间复杂度 $O(mn)$。其中 $m$ 是 $sentence$ 的长度,$n$ 是 $searchWord$ 的长度。
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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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# [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)
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[English Version](/solution/2300-2399/2379.Minimum%20Recolors%20to%20Get%20K%20Consecutive%20Black%20Blocks/README_EN.md)
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## 题目描述
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<!-- 这里写题目描述 -->
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<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>
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<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>
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<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>
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<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>
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<p>&nbsp;</p>
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<p><strong>示例 1:</strong></p>
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<pre>
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<b>输入:</b>blocks = "WBBWWBBWBW", k = 7
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<b>输出:</b>3
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<strong>解释:</strong>
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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。
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得到 blocks = "BBBBBBBWBW" 。
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可以证明无法用少于 3 次操作得到 7 个连续的黑块。
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所以我们返回 3 。
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</pre>
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<p><strong>示例 2:</strong></p>
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<pre>
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<b>输入:</b>blocks = "WBWBBBW", k = 2
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<b>输出:</b>0
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<strong>解释:</strong>
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不需要任何操作,因为已经有 2 个连续的黑块。
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所以我们返回 0 。
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</pre>
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<p>&nbsp;</p>
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<p><b>提示:</b></p>
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<ul>
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<li><code>n == blocks.length</code></li>
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<li><code>1 &lt;= n &lt;= 100</code></li>
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<li><code>blocks[i]</code>&nbsp;要么是&nbsp;<code>'W'</code>&nbsp;,要么是&nbsp;<code>'B'</code> 。</li>
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<li><code>1 &lt;= k &lt;= n</code></li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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```
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### **TypeScript**
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```ts
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```
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### **...**
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```
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```
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# [2379. Minimum Recolors to Get K Consecutive Black Blocks](https://leetcode.com/problems/minimum-recolors-to-get-k-consecutive-black-blocks)
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[中文文档](/solution/2300-2399/2379.Minimum%20Recolors%20to%20Get%20K%20Consecutive%20Black%20Blocks/README.md)
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## Description
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<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>
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<p>You are also given an integer <code>k</code>, which is the desired number of <strong>consecutive</strong> black blocks.</p>
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<p>In one operation, you can <strong>recolor</strong> a white block such that it becomes a black block.</p>
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<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>
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<p>&nbsp;</p>
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<p><strong>Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> blocks = &quot;WBBWWBBWBW&quot;, k = 7
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<strong>Output:</strong> 3
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<strong>Explanation:</strong>
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One way to achieve 7 consecutive black blocks is to recolor the 0th, 3rd, and 4th blocks
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so that blocks = &quot;BBBBBBBWBW&quot;.
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It can be shown that there is no way to achieve 7 consecutive black blocks in less than 3 operations.
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Therefore, we return 3.
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</pre>
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<p><strong>Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> blocks = &quot;WBWBBBW&quot;, k = 2
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<strong>Output:</strong> 0
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<strong>Explanation:</strong>
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No changes need to be made, since 2 consecutive black blocks already exist.
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Therefore, we return 0.
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</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>n == blocks.length</code></li>
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<li><code>1 &lt;= n &lt;= 100</code></li>
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<li><code>blocks[i]</code> is either <code>&#39;W&#39;</code> or <code>&#39;B&#39;</code>.</li>
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<li><code>1 &lt;= k &lt;= n</code></li>
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</ul>
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## Solutions
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<!-- tabs:start -->
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### **Python3**
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```python
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```
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### **Java**
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```java
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```
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### **TypeScript**
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```ts
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```
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### **...**
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```
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```
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# [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)
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[English Version](/solution/2300-2399/2380.Time%20Needed%20to%20Rearrange%20a%20Binary%20String/README_EN.md)
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## 题目描述
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<!-- 这里写题目描述 -->
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<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>
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<p>请你返回完成这个过程所需要的秒数。</p>
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<p>&nbsp;</p>
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<p><strong>示例 1:</strong></p>
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<pre>
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<b>输入:</b>s = "0110101"
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<b>输出:</b>4
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<b>解释:</b>
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一秒后,s 变成 "1011010" 。
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再过 1 秒后,s 变成 "1101100" 。
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第三秒过后,s 变成 "1110100" 。
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第四秒后,s 变成 "1111000" 。
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此时没有 "01" 存在,整个过程花费 4 秒。
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所以我们返回 4 。
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</pre>
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<p><strong>示例 2:</strong></p>
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<pre>
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<b>输入:</b>s = "11100"
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<b>输出:</b>0
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<strong>解释:</strong>
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s 中没有 "01" 存在,整个过程花费 0 秒。
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所以我们返回 0 。
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</pre>
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<p>&nbsp;</p>
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<p><strong>提示:</strong></p>
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<ul>
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<li><code>1 &lt;= s.length &lt;= 1000</code></li>
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<li><code>s[i]</code>&nbsp;要么是&nbsp;<code>'0'</code>&nbsp;,要么是&nbsp;<code>'1'</code> 。</li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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```
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### **TypeScript**
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```ts
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```
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### **...**
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```
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```
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# [2380. Time Needed to Rearrange a Binary String](https://leetcode.com/problems/time-needed-to-rearrange-a-binary-string)
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[中文文档](/solution/2300-2399/2380.Time%20Needed%20to%20Rearrange%20a%20Binary%20String/README.md)
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## Description
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<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>
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<p>Return<em> the number of seconds needed to complete this process.</em></p>
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<p>&nbsp;</p>
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<p><strong>Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> s = &quot;0110101&quot;
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<strong>Output:</strong> 4
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<strong>Explanation:</strong>
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After one second, s becomes &quot;1011010&quot;.
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After another second, s becomes &quot;1101100&quot;.
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After the third second, s becomes &quot;1110100&quot;.
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After the fourth second, s becomes &quot;1111000&quot;.
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No occurrence of &quot;01&quot; exists any longer, and the process needed 4 seconds to complete,
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so we return 4.
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</pre>
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<p><strong>Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> s = &quot;11100&quot;
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<strong>Output:</strong> 0
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<strong>Explanation:</strong>
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No occurrence of &quot;01&quot; exists in s, and the processes needed 0 seconds to complete,
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so we return 0.
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</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 &lt;= s.length &lt;= 1000</code></li>
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<li><code>s[i]</code> is either <code>&#39;0&#39;</code> or <code>&#39;1&#39;</code>.</li>
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</ul>
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## Solutions
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<!-- tabs:start -->
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### **Python3**
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```python
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```
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### **Java**
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```java
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```
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### **TypeScript**
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```ts
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```
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### **...**
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```
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```
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<!-- tabs:end -->

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