chore: update lc problems (#1806)

Author: yanglbme

solution/2800-2899/2899.Last Visited Integers/README.md

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+# [2899. 上一个遍历的整数](https://leetcode.cn/problems/last-visited-integers)
+
+[English Version](/solution/2800-2899/2899.Last%20Visited%20Integers/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>给你一个下标从 <strong>0</strong>&nbsp;开始的字符串数组&nbsp;<code>words</code>&nbsp;，其中&nbsp;<code>words[i]</code>&nbsp;要么是一个字符串形式的正整数，要么是字符串&nbsp;<code>"prev"</code>&nbsp;。</p>
+
+<p>我们从数组的开头开始遍历，对于 <code>words</code>&nbsp;中的每个&nbsp;<code>"prev"</code>&nbsp;字符串，找到 <code>words</code>&nbsp;中的 <strong>上一个遍历的整数</strong>&nbsp;，定义如下：</p>
+
+<ul>
+	<li><code>k</code>&nbsp;表示到当前位置为止的连续&nbsp;<code>"prev"</code>&nbsp;字符串数目（包含当前字符串），令下标从&nbsp;<strong>0</strong>&nbsp;开始的&nbsp;<strong>整数</strong> 数组&nbsp;<code>nums</code>&nbsp;表示目前为止遍历过的所有整数，同时用&nbsp;<code>nums_reverse</code>&nbsp;表示&nbsp;<code>nums</code>&nbsp;反转得到的数组，那么当前 <code>"prev"</code>&nbsp;对应的 <strong>上一个遍历的整数</strong>&nbsp;是&nbsp;<code>nums_reverse</code>&nbsp;数组中下标为 <code>(k - 1)</code>&nbsp;的整数。</li>
+	<li>如果&nbsp;<code>k</code>&nbsp;比目前为止遍历过的整数数目 <strong>更多</strong>&nbsp;，那么上一个遍历的整数为&nbsp;<code>-1</code>&nbsp;。</li>
+</ul>
+
+<p>请你返回一个整数数组，包含所有上一个遍历的整数。</p>
+
+<p>&nbsp;</p>
+
+<p><strong class="example">示例 1：</strong></p>
+
+<pre>
+<b>输入：</b><code>words</code> = ["1","2","prev","prev","prev"]
+<b>输出：</b>[2,1,-1]
+<b>解释：</b>
+对于下标为 2 处的 "prev" ，上一个遍历的整数是 2 ，因为连续 "prev" 数目为 1 ，同时在数组 reverse_nums 中，第一个元素是 2 。
+对于下标为 3 处的 "prev" ，上一个遍历的整数是 1 ，因为连续 "prev" 数目为 2 ，同时在数组 reverse_nums 中，第二个元素是 1 。
+对于下标为 4 处的 "prev" ，上一个遍历的整数是 -1 ，因为连续 "prev" 数目为 3 ，但总共只遍历过 2 个整数。
+</pre>
+
+<p><strong class="example">示例 2：</strong></p>
+
+<pre>
+<b>输入：</b><code>words</code> = ["1","prev","2","prev","prev"]
+<b>输出：</b>[1,2,1]
+<strong>解释：</strong>
+对于下标为 1 处的 "prev" ，上一个遍历的整数是 1 。
+对于下标为 3 处的 "prev" ，上一个遍历的整数是 2 。
+对于下标为 4 处的 "prev" ，上一个遍历的整数是 1 ，因为连续 "prev"<strong>&nbsp;</strong>数目为 2 ，同时在数组 reverse_nums 中，第二个元素是 1 。
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= words.length &lt;= 100</code></li>
+	<li><code>words[i] == "prev"</code>&nbsp;或&nbsp;<code>1 &lt;= int(words[i]) &lt;= 100</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
+
+```
+
+### **C++**
+
+```cpp
+
+```
+
+### **Go**
+
+```go
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

solution/2800-2899/2899.Last Visited Integers/README_EN.md

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+# [2899. Last Visited Integers](https://leetcode.com/problems/last-visited-integers)
+
+[中文文档](/solution/2800-2899/2899.Last%20Visited%20Integers/README.md)
+
+## Description
+
+<p>Given a <strong>0-indexed</strong> array of strings <code>words</code> where <code>words[i]</code> is either a positive integer represented as a string or the string <code>&quot;prev&quot;</code>.</p>
+
+<p>Start iterating from the beginning of the array; for every <code>&quot;prev&quot;</code> string seen in <code>words</code>, find the <strong>last visited integer</strong> in <code>words</code> which is defined as follows:</p>
+
+<ul>
+	<li>Let <code>k</code> be the number of consecutive <code>&quot;prev&quot;</code> strings seen so far (containing the current string). Let <code>nums</code> be the <strong>0-indexed </strong>array of <strong>integers</strong> seen so far and <code>nums_reverse</code> be the reverse of <code>nums</code>, then the integer at <code>(k - 1)<sup>th</sup></code> index of <code>nums_reverse</code> will be the <strong>last visited integer</strong> for this <code>&quot;prev&quot;</code>.</li>
+	<li>If <code>k</code> is <strong>greater</strong> than the total visited integers, then the last visited integer will be <code>-1</code>.</li>
+</ul>
+
+<p>Return <em>an integer array containing the last visited integers.</em></p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> words = [&quot;1&quot;,&quot;2&quot;,&quot;prev&quot;,&quot;prev&quot;,&quot;prev&quot;]
+<strong>Output:</strong> [2,1,-1]
+<strong>Explanation:</strong> 
+For &quot;prev&quot; at index = 2, last visited integer will be 2 as here the number of consecutive &quot;prev&quot; strings is 1, and in the array reverse_nums, 2 will be the first element.
+For &quot;prev&quot; at index = 3, last visited integer will be 1 as there are a total of two consecutive &quot;prev&quot; strings including this &quot;prev&quot; which are visited, and 1 is the second last visited integer.
+For &quot;prev&quot; at index = 4, last visited integer will be -1 as there are a total of three consecutive &quot;prev&quot; strings including this &quot;prev&quot; which are visited, but the total number of integers visited is two.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> words = [&quot;1&quot;,&quot;prev&quot;,&quot;2&quot;,&quot;prev&quot;,&quot;prev&quot;]
+<strong>Output:</strong> [1,2,1]
+<strong>Explanation:</strong>
+For &quot;prev&quot; at index = 1, last visited integer will be 1.
+For &quot;prev&quot; at index = 3, last visited integer will be 2.
+For &quot;prev&quot; at index = 4, last visited integer will be 1 as there are a total of two consecutive &quot;prev&quot; strings including this &quot;prev&quot; which are visited, and 1 is the second last visited integer.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= words.length &lt;= 100</code></li>
+	<li><code>words[i] == &quot;prev&quot;</code> or <code>1 &lt;= int(words[i]) &lt;= 100</code></li>
+</ul>
+
+## Solutions
+
+<!-- tabs:start -->
+
+### **Python3**
+
+```python
+
+```
+
+### **Java**
+
+```java
+
+```
+
+### **C++**
+
+```cpp
+
+```
+
+### **Go**
+
+```go
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

solution/2900-2999/2900.Longest Unequal Adjacent Groups Subsequence I/README.md

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+# [2900. 最长相邻不相等子序列 I](https://leetcode.cn/problems/longest-unequal-adjacent-groups-subsequence-i)
+
+[English Version](/solution/2900-2999/2900.Longest%20Unequal%20Adjacent%20Groups%20Subsequence%20I/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>给你一个整数&nbsp;<code>n</code>&nbsp;和一个下标从&nbsp;<strong>0</strong>&nbsp;开始的字符串数组&nbsp;<code>words</code>&nbsp;，和一个下标从 <strong>0</strong>&nbsp;开始的 <strong>二进制</strong>&nbsp;数组&nbsp;<code>groups</code>&nbsp;，两个数组长度都是&nbsp;<code>n</code>&nbsp;。</p>
+
+<p>你需要从下标&nbsp;<code>[0, 1, ..., n - 1]</code>&nbsp;中选出一个&nbsp;<strong>最长子序列</strong>&nbsp;，将这个子序列记作长度为 <code>k</code> 的&nbsp;<code>[i<sub>0</sub>, i<sub>1</sub>, ..., i<sub>k - 1</sub>]</code>&nbsp;，对于所有满足&nbsp;<code>0 &lt; j + 1 &lt; k</code>&nbsp;的&nbsp;<code>j</code>&nbsp;都有&nbsp;<code>groups[i<sub>j</sub>] != groups[i<sub>j + 1</sub>]</code>&nbsp;。</p>
+
+<p>请你返回一个字符串数组，它是下标子序列&nbsp;<strong>依次</strong>&nbsp;对应&nbsp;<code>words</code>&nbsp;数组中的字符串连接形成的字符串数组。如果有多个答案，返回任意一个。</p>
+
+<p><strong>子序列</strong>&nbsp;指的是从原数组中删掉一些（也可能一个也不删掉）元素，剩余元素不改变相对位置得到的新的数组。</p>
+
+<p><b>注意：</b><code>words</code>&nbsp;中的字符串长度可能 <strong>不相等</strong>&nbsp;。</p>
+
+<p>&nbsp;</p>
+
+<p><strong class="example">示例 1：</strong></p>
+
+<pre>
+<b>输入：</b>n = 3, words = ["e","a","b"], groups = [0,0,1]
+<b>输出：</b>["e","b"]
+<strong>解释：</strong>一个可行的子序列是 [0,2] ，因为 groups[0] != groups[2] 。
+所以一个可行的答案是 [words[0],words[2]] = ["e","b"] 。
+另一个可行的子序列是 [1,2] ，因为 groups[1] != groups[2] 。
+得到答案为 [words[1],words[2]] = ["a","b"] 。
+这也是一个可行的答案。
+符合题意的最长子序列的长度为 2 。</pre>
+
+<p><strong class="example">示例 2：</strong></p>
+
+<pre>
+<b>输入：</b>n = 4, words = ["a","b","c","d"], groups = [1,0,1,1]
+<b>输出：</b>["a","b","c"]
+<b>解释：</b>一个可行的子序列为 [0,1,2] 因为 groups[0] != groups[1] 且 groups[1] != groups[2] 。
+所以一个可行的答案是 [words[0],words[1],words[2]] = ["a","b","c"] 。
+另一个可行的子序列为 [0,1,3] 因为 groups[0] != groups[1] 且 groups[1] != groups[3] 。
+得到答案为 [words[0],words[1],words[3]] = ["a","b","d"] 。
+这也是一个可行的答案。
+符合题意的最长子序列的长度为 3 。</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= n == words.length == groups.length &lt;= 100</code></li>
+	<li><code>1 &lt;= words[i].length &lt;= 10</code></li>
+	<li><code>0 &lt;= groups[i] &lt; 2</code></li>
+	<li><code>words</code>&nbsp;中的字符串 <strong>互不相同</strong>&nbsp;。</li>
+	<li><code>words[i]</code>&nbsp;只包含小写英文字母。</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
+
+```
+
+### **C++**
+
+```cpp
+
+```
+
+### **Go**
+
+```go
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->

solution/2900-2999/2900.Longest Unequal Adjacent Groups Subsequence I/README_EN.md

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+# [2900. Longest Unequal Adjacent Groups Subsequence I](https://leetcode.com/problems/longest-unequal-adjacent-groups-subsequence-i)
+
+[中文文档](/solution/2900-2999/2900.Longest%20Unequal%20Adjacent%20Groups%20Subsequence%20I/README.md)
+
+## Description
+
+<p>You are given an integer <code>n</code>, a <strong>0-indexed</strong> string array <code>words</code>, and a <strong>0-indexed</strong> <strong>binary</strong> array <code>groups</code>, both arrays having length <code>n</code>.</p>
+
+<p>You need to select the <strong>longest</strong> <strong>subsequence</strong> from an array of indices <code>[0, 1, ..., n - 1]</code>, such that for the subsequence denoted as <code>[i<sub>0</sub>, i<sub>1</sub>, ..., i<sub>k - 1</sub>]</code> having length <code>k</code>, <code>groups[i<sub>j</sub>] != groups[i<sub>j + 1</sub>]</code>, for each <code>j</code> where <code>0 &lt; j + 1 &lt; k</code>.</p>
+
+<p>Return <em>a string array containing the words corresponding to the indices <strong>(in order)</strong> in the selected subsequence</em>. If there are multiple answers, return<em> any of them</em>.</p>
+
+<p>A <strong>subsequence</strong> of an array is a new array that is formed from the original array by deleting some (possibly none) of the elements without disturbing the relative positions of the remaining elements.</p>
+
+<p><strong>Note:</strong> strings in <code>words</code> may be <strong>unequal</strong> in length.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> n = 3, words = [&quot;e&quot;,&quot;a&quot;,&quot;b&quot;], groups = [0,0,1]
+<strong>Output:</strong> [&quot;e&quot;,&quot;b&quot;]
+<strong>Explanation: </strong>A subsequence that can be selected is [0,2] because groups[0] != groups[2].
+So, a valid answer is [words[0],words[2]] = [&quot;e&quot;,&quot;b&quot;].
+Another subsequence that can be selected is [1,2] because groups[1] != groups[2].
+This results in [words[1],words[2]] = [&quot;a&quot;,&quot;b&quot;].
+It is also a valid answer.
+It can be shown that the length of the longest subsequence of indices that satisfies the condition is 2.</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> n = 4, words = [&quot;a&quot;,&quot;b&quot;,&quot;c&quot;,&quot;d&quot;], groups = [1,0,1,1]
+<strong>Output:</strong> [&quot;a&quot;,&quot;b&quot;,&quot;c&quot;]
+<strong>Explanation:</strong> A subsequence that can be selected is [0,1,2] because groups[0] != groups[1] and groups[1] != groups[2].
+So, a valid answer is [words[0],words[1],words[2]] = [&quot;a&quot;,&quot;b&quot;,&quot;c&quot;].
+Another subsequence that can be selected is [0,1,3] because groups[0] != groups[1] and groups[1] != groups[3].
+This results in [words[0],words[1],words[3]] = [&quot;a&quot;,&quot;b&quot;,&quot;d&quot;].
+It is also a valid answer.
+It can be shown that the length of the longest subsequence of indices that satisfies the condition is 3.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= n == words.length == groups.length &lt;= 100</code></li>
+	<li><code>1 &lt;= words[i].length &lt;= 10</code></li>
+	<li><code>0 &lt;= groups[i] &lt; 2</code></li>
+	<li><code>words</code> consists of <strong>distinct</strong> strings.</li>
+	<li><code>words[i]</code> consists of lowercase English letters.</li>
+</ul>
+
+## Solutions
+
+<!-- tabs:start -->
+
+### **Python3**
+
+```python
+
+```
+
+### **Java**
+
+```java
+
+```
+
+### **C++**
+
+```cpp
+
+```
+
+### **Go**
+
+```go
+
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->