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feat: add solutions to lc problems: No.2108~2111
* No.2108.Find First Palindromic String in the Array * No.2109.Adding Spaces to a String * No.2110.Number of Smooth Descent Periods of a Stock * No.2111.Minimum Operations to Make the Array K-Increasing
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‎solution/1300-1399/1375.Number of Times Binary String Is Prefix-Aligned/README.md

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# [1375. 二进制字符串前缀一致的次数](https://leetcode-cn.com/problems/number-of-times-binary-string-is-prefix-aligned)
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[English Version](/solution/1300-1399/1375.Number%20of%20Times%20Binary%20String%20Is%20Prefix-Aligned/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> 、下标从 <strong>1</strong> 开始的二进制字符串，所有位最开始都是 <code>0</code> 。我们会按步翻转该二进制字符串的所有位（即，将 <code>0</code> 变为 <code>1</code>）。</p>
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<p>给你一个下标从 <strong>1</strong> 开始的整数数组 <code>flips</code> ，其中 <code>flips[i]</code> 表示对应下标 <code>i</code> 的位将会在第 <code>i</code> 步翻转。</p>
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<p>二进制字符串 <strong>前缀一致</strong> 需满足：在第 <code>i</code> 步之后，在 <strong>闭</strong> 区间&nbsp;<code>[1, i]</code> 内的所有位都是 1 ，而其他位都是 0 。</p>
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<p>返回二进制字符串在翻转过程中 <strong>前缀一致</strong> 的次数。</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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<strong>输入：</strong>flips = [3,2,4,1,5]
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<strong>输出：</strong>2
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<strong>解释：</strong>二进制字符串最开始是 "00000" 。
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执行第 1 步：字符串变为 "00100" ，不属于前缀一致的情况。
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执行第 2 步：字符串变为 "01100" ，不属于前缀一致的情况。
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执行第 3 步：字符串变为 "01110" ，不属于前缀一致的情况。
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执行第 4 步：字符串变为 "11110" ，属于前缀一致的情况。
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执行第 5 步：字符串变为 "11111" ，属于前缀一致的情况。
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在翻转过程中，前缀一致的次数为 2 ，所以返回 2 。
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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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<strong>输入：</strong>flips = [4,1,2,3]
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<strong>输出：</strong>1
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<strong>解释：</strong>二进制字符串最开始是 "0000" 。
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执行第 1 步：字符串变为 "0001" ，不属于前缀一致的情况。
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执行第 2 步：字符串变为 "1001" ，不属于前缀一致的情况。
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执行第 3 步：字符串变为 "1101" ，不属于前缀一致的情况。
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执行第 4 步：字符串变为 "1111" ，属于前缀一致的情况。
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在翻转过程中，前缀一致的次数为 1 ，所以返回 1 。</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>n == flips.length</code></li>
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<li><code>1 &lt;= n &lt;= 5 * 10<sup>4</sup></code></li>
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<li><code>flips</code> 是范围 <code>[1, n]</code> 中所有整数构成的一个排列</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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<!-- 这里可写当前语言的特殊实现逻辑 -->
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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 -->

‎solution/1300-1399/1375.Number of Times Binary String Is Prefix-Aligned/README_EN.md

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# [1375. Number of Times Binary String Is Prefix-Aligned](https://leetcode.com/problems/number-of-times-binary-string-is-prefix-aligned)
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[中文文档](/solution/1300-1399/1375.Number%20of%20Times%20Binary%20String%20Is%20Prefix-Aligned/README.md)
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## Description
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<p>You have a <strong>1-indexed</strong> binary string of length <code>n</code> where all the bits are <code>0</code> initially. We will flip all the bits of this binary string (i.e., change them from <code>0</code> to <code>1</code>) one by one. You are given a <strong>1-indexed</strong> integer array <code>flips</code> where <code>flips[i]</code> indicates that the bit at index <code>i</code> will be flipped in the <code>i<sup>th</sup></code> step.</p>
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<p>A binary string is <strong>prefix-aligned</strong> if, after the <code>i<sup>th</sup></code> step, all the bits in the <strong>inclusive</strong> range <code>[1, i]</code> are ones and all the other bits are zeros.</p>
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<p>Return <em>the number of times the binary string is <strong>prefix-aligned</strong> during the flipping 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> flips = [3,2,4,1,5]
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<strong>Output:</strong> 2
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<strong>Explanation:</strong> The binary string is initially &quot;00000&quot;.
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After applying step 1: The string becomes &quot;00100&quot;, which is not prefix-aligned.
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After applying step 2: The string becomes &quot;01100&quot;, which is not prefix-aligned.
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After applying step 3: The string becomes &quot;01110&quot;, which is not prefix-aligned.
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After applying step 4: The string becomes &quot;11110&quot;, which is prefix-aligned.
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After applying step 5: The string becomes &quot;11111&quot;, which is prefix-aligned.
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We can see that the string was prefix-aligned 2 times, so we return 2.
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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> flips = [4,1,2,3]
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<strong>Output:</strong> 1
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<strong>Explanation:</strong> The binary string is initially &quot;0000&quot;.
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After applying step 1: The string becomes &quot;0001&quot;, which is not prefix-aligned.
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After applying step 2: The string becomes &quot;1001&quot;, which is not prefix-aligned.
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After applying step 3: The string becomes &quot;1101&quot;, which is not prefix-aligned.
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After applying step 4: The string becomes &quot;1111&quot;, which is prefix-aligned.
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We can see that the string was prefix-aligned 1 time, so we return 1.
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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 == flips.length</code></li>
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<li><code>1 &lt;= n &lt;= 5 * 10<sup>4</sup></code></li>
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<li><code>flips</code> is a permutation of the integers in the range <code>[1, 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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‎solution/1500-1599/1578.Minimum Time to Make Rope Colorful/README.md

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# [1578. 使绳子变成彩色的最短时间](https://leetcode-cn.com/problems/minimum-time-to-make-rope-colorful)
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[English Version](/solution/1500-1599/1578.Minimum%20Time%20to%20Make%20Rope%20Colorful/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>Alice 把 <code>n</code> 个气球排列在一根绳子上。给你一个下标从 <strong>0</strong> 开始的字符串 <code>colors</code> ，其中 <code>colors[i]</code> 是第 <code>i</code> 个气球的颜色。</p>
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<p>Alice 想要把绳子装扮成 <strong>彩色</strong> ，且她不希望两个连续的气球涂着相同的颜色，所以她喊来 Bob 帮忙。Bob 可以从绳子上移除一些气球使绳子变成 <strong>彩色</strong> 。给你一个下标从 <strong>0</strong> 开始的整数数组 <code>neededTime</code> ，其中 <code>neededTime[i]</code> 是 Bob 从绳子上移除第 <code>i</code> 个气球需要的时间（以秒为单位）。</p>
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<p>返回 Bob 使绳子变成 <strong>彩色</strong> 需要的 <strong>最少时间</strong> 。</p>
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<p>&nbsp;</p>
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<p><strong>示例 1：</strong></p>
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<img alt="" src="https://assets.leetcode.com/uploads/2021/12/13/ballon1.jpg" style="width: 404px; height: 243px;" />
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<pre>
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<strong>输入：</strong>colors = "abaac", neededTime = [1,2,3,4,5]
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<strong>输出：</strong>3
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<strong>解释：</strong>在上图中，'a' 是蓝色，'b' 是红色且 'c' 是绿色。
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Bob 可以移除下标 2 的蓝色气球。这将花费 3 秒。
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移除后，不存在两个连续的气球涂着相同的颜色。总时间 = 3 。</pre>
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<p><strong>示例 2：</strong></p>
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<img alt="" src="https://assets.leetcode.com/uploads/2021/12/13/balloon2.jpg" style="width: 244px; height: 243px;" />
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<pre>
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<strong>输入：</strong>colors = "abc", neededTime = [1,2,3]
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<strong>输出：</strong>0
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<strong>解释：</strong>绳子已经是彩色的，Bob 不需要从绳子上移除任何气球。
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</pre>
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<p><strong>示例 3：</strong></p>
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<img alt="" src="https://assets.leetcode.com/uploads/2021/12/13/balloon3.jpg" style="width: 404px; height: 243px;" />
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<pre>
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<strong>输入：</strong>colors = "aabaa", neededTime = [1,2,3,4,1]
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<strong>输出：</strong>2
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<strong>解释：</strong>Bob 会移除下标 0 和下标 4 处的气球。这两个气球各需要 1 秒来移除。
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移除后，不存在两个连续的气球涂着相同的颜色。总时间 = 1 + 1 = 2 。
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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>n == colors.length == neededTime.length</code></li>
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<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
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<li><code>1 &lt;= neededTime[i] &lt;= 10<sup>4</sup></code></li>
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<li><code>colors</code> 仅由小写英文字母组成</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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<!-- 这里可写当前语言的特殊实现逻辑 -->
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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 -->

‎solution/1500-1599/1578.Minimum Time to Make Rope Colorful/README_EN.md

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# [1578. Minimum Time to Make Rope Colorful](https://leetcode.com/problems/minimum-time-to-make-rope-colorful)
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[中文文档](/solution/1500-1599/1578.Minimum%20Time%20to%20Make%20Rope%20Colorful/README.md)
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## Description
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<p>Alice has <code>n</code> balloons arranged on a rope. You are given a <strong>0-indexed</strong> string <code>colors</code> where <code>colors[i]</code> is the color of the <code>i<sup>th</sup></code> balloon.</p>
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<p>Alice wants the rope to be <strong>colorful</strong>. She does not want <strong>two consecutive balloons</strong> to be of the same color, so she asks Bob for help. Bob can remove some balloons from the rope to make it <strong>colorful</strong>. You are given a <strong>0-indexed</strong> integer array <code>neededTime</code> where <code>neededTime[i]</code> is the time (in seconds) that Bob needs to remove the <code>i<sup>th</sup></code> balloon from the rope.</p>
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<p>Return <em>the <strong>minimum time</strong> Bob needs to make the rope <strong>colorful</strong></em>.</p>
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<p>&nbsp;</p>
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<p><strong>Example 1:</strong></p>
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<img alt="" src="https://assets.leetcode.com/uploads/2021/12/13/ballon1.jpg" style="width: 404px; height: 243px;" />
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<pre>
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<strong>Input:</strong> colors = &quot;abaac&quot;, neededTime = [1,2,3,4,5]
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<strong>Output:</strong> 3
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<strong>Explanation:</strong> In the above image, &#39;a&#39; is blue, &#39;b&#39; is red, and &#39;c&#39; is green.
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Bob can remove the blue balloon at index 2. This takes 3 seconds.
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There are no longer two consecutive balloons of the same color. Total time = 3.</pre>
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<p><strong>Example 2:</strong></p>
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<img alt="" src="https://assets.leetcode.com/uploads/2021/12/13/balloon2.jpg" style="width: 244px; height: 243px;" />
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<pre>
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<strong>Input:</strong> colors = &quot;abc&quot;, neededTime = [1,2,3]
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<strong>Output:</strong> 0
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<strong>Explanation:</strong> The rope is already colorful. Bob does not need to remove any balloons from the rope.
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</pre>
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<p><strong>Example 3:</strong></p>
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<img alt="" src="https://assets.leetcode.com/uploads/2021/12/13/balloon3.jpg" style="width: 404px; height: 243px;" />
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<pre>
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<strong>Input:</strong> colors = &quot;aabaa&quot;, neededTime = [1,2,3,4,1]
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<strong>Output:</strong> 2
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<strong>Explanation:</strong> Bob will remove the ballons at indices 0 and 4. Each ballon takes 1 second to remove.
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There are no longer two consecutive balloons of the same color. Total time = 1 + 1 = 2.
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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 == colors.length == neededTime.length</code></li>
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<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
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<li><code>1 &lt;= neededTime[i] &lt;= 10<sup>4</sup></code></li>
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<li><code>colors</code> contains only lowercase English letters.</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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