feat: add weekly contest 438

solution/1200-1299/1261.Find Elements in a Contaminated Binary Tree/README.md

@@ -57,8 +57,8 @@ tags:
 <strong>输出：</strong>
 [null,false,true]
 <strong>解释：</strong>
-FindElements findElements = new FindElements([-1,null,-1]);
-findElements.find(1); // return False
+FindElements findElements = new FindElements([-1,null,-1]); 
+findElements.find(1); // return False 
 findElements.find(2); // return True </pre>
 
 <p><strong class="example">示例 2：</strong></p>

solution/1200-1299/1261.Find Elements in a Contaminated Binary Tree/README_EN.md

@@ -54,8 +54,8 @@ tags:
 <strong>Output</strong>
 [null,false,true]
 <strong>Explanation</strong>
-FindElements findElements = new FindElements([-1,null,-1]);
-findElements.find(1); // return False
+FindElements findElements = new FindElements([-1,null,-1]); 
+findElements.find(1); // return False 
 findElements.find(2); // return True </pre>
 
 <p><strong class="example">Example 2:</strong></p>

solution/3400-3499/3461.Check If Digits Are Equal in String After Operations I/README.md

@@ -0,0 +1,123 @@
+---
+comments: true
+difficulty: 简单
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3400-3499/3461.Check%20If%20Digits%20Are%20Equal%20in%20String%20After%20Operations%20I/README.md
+---
+
+<!-- problem:start -->
+
+# [3461. 判断操作后字符串中的数字是否相等 I](https://leetcode.cn/problems/check-if-digits-are-equal-in-string-after-operations-i)
+
+[English Version](/solution/3400-3499/3461.Check%20If%20Digits%20Are%20Equal%20in%20String%20After%20Operations%20I/README_EN.md)
+
+## 题目描述
+
+<!-- description:start -->
+
+<p>给你一个由数字组成的字符串 <code>s</code>&nbsp;。重复执行以下操作，直到字符串恰好包含&nbsp;<strong>两个&nbsp;</strong>数字：</p>
+
+<ul>
+	<li>从第一个数字开始，对于 <code>s</code> 中的每一对连续数字，计算这两个数字的和&nbsp;<strong>模&nbsp;</strong>10。</li>
+	<li>用计算得到的新数字依次替换 <code>s</code>&nbsp;的每一个字符，并保持原本的顺序。</li>
+</ul>
+
+<p>如果 <code>s</code>&nbsp;最后剩下的两个数字 <strong>相同</strong> ，返回 <code>true</code>&nbsp;。否则，返回 <code>false</code>。</p>
+
+<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 = "3902"</span></p>
+
+<p><strong>输出：</strong> <span class="example-io">true</span></p>
+
+<p><strong>解释：</strong></p>
+
+<ul>
+	<li>一开始，<code>s = "3902"</code></li>
+	<li>第一次操作：
+	<ul>
+		<li><code>(s[0] + s[1]) % 10 = (3 + 9) % 10 = 2</code></li>
+		<li><code>(s[1] + s[2]) % 10 = (9 + 0) % 10 = 9</code></li>
+		<li><code>(s[2] + s[3]) % 10 = (0 + 2) % 10 = 2</code></li>
+		<li><code>s</code> 变为 <code>"292"</code></li>
+	</ul>
+	</li>
+	<li>第二次操作：
+	<ul>
+		<li><code>(s[0] + s[1]) % 10 = (2 + 9) % 10 = 1</code></li>
+		<li><code>(s[1] + s[2]) % 10 = (9 + 2) % 10 = 1</code></li>
+		<li><code>s</code> 变为 <code>"11"</code></li>
+	</ul>
+	</li>
+	<li>由于 <code>"11"</code> 中的数字相同，输出为 <code>true</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 = "34789"</span></p>
+
+<p><strong>输出：</strong> <span class="example-io">false</span></p>
+
+<p><strong>解释：</strong></p>
+
+<ul>
+	<li>一开始，<code>s = "34789"</code>。</li>
+	<li>第一次操作后，<code>s = "7157"</code>。</li>
+	<li>第二次操作后，<code>s = "862"</code>。</li>
+	<li>第三次操作后，<code>s = "48"</code>。</li>
+	<li>由于 <code>'4' != '8'</code>，输出为 <code>false</code>。</li>
+</ul>
+
+<p>&nbsp;</p>
+</div>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>3 &lt;= s.length &lt;= 100</code></li>
+	<li><code>s</code> 仅由数字组成。</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/3400-3499/3461.Check If Digits Are Equal in String After Operations I/README_EN.md

@@ -0,0 +1,121 @@
+---
+comments: true
+difficulty: Easy
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3400-3499/3461.Check%20If%20Digits%20Are%20Equal%20in%20String%20After%20Operations%20I/README_EN.md
+---
+
+<!-- problem:start -->
+
+# [3461. Check If Digits Are Equal in String After Operations I](https://leetcode.com/problems/check-if-digits-are-equal-in-string-after-operations-i)
+
+[中文文档](/solution/3400-3499/3461.Check%20If%20Digits%20Are%20Equal%20in%20String%20After%20Operations%20I/README.md)
+
+## Description
+
+<!-- description:start -->
+
+<p>You are given a string <code>s</code> consisting of digits. Perform the following operation repeatedly until the string has <strong>exactly</strong> two digits:</p>
+
+<ul>
+	<li>For each pair of consecutive digits in <code>s</code>, starting from the first digit, calculate a new digit as the sum of the two digits <strong>modulo</strong> 10.</li>
+	<li>Replace <code>s</code> with the sequence of newly calculated digits, <em>maintaining the order</em> in which they are computed.</li>
+</ul>
+
+<p>Return <code>true</code> if the final two digits in <code>s</code> are the <strong>same</strong>; otherwise, return <code>false</code>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">s = &quot;3902&quot;</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">true</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<ul>
+	<li>Initially, <code>s = &quot;3902&quot;</code></li>
+	<li>First operation:
+	<ul>
+		<li><code>(s[0] + s[1]) % 10 = (3 + 9) % 10 = 2</code></li>
+		<li><code>(s[1] + s[2]) % 10 = (9 + 0) % 10 = 9</code></li>
+		<li><code>(s[2] + s[3]) % 10 = (0 + 2) % 10 = 2</code></li>
+		<li><code>s</code> becomes <code>&quot;292&quot;</code></li>
+	</ul>
+	</li>
+	<li>Second operation:
+	<ul>
+		<li><code>(s[0] + s[1]) % 10 = (2 + 9) % 10 = 1</code></li>
+		<li><code>(s[1] + s[2]) % 10 = (9 + 2) % 10 = 1</code></li>
+		<li><code>s</code> becomes <code>&quot;11&quot;</code></li>
+	</ul>
+	</li>
+	<li>Since the digits in <code>&quot;11&quot;</code> are the same, the output is <code>true</code>.</li>
+</ul>
+</div>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">s = &quot;34789&quot;</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">false</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<ul>
+	<li>Initially, <code>s = &quot;34789&quot;</code>.</li>
+	<li>After the first operation, <code>s = &quot;7157&quot;</code>.</li>
+	<li>After the second operation, <code>s = &quot;862&quot;</code>.</li>
+	<li>After the third operation, <code>s = &quot;48&quot;</code>.</li>
+	<li>Since <code>&#39;4&#39; != &#39;8&#39;</code>, the output is <code>false</code>.</li>
+</ul>
+</div>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>3 &lt;= s.length &lt;= 100</code></li>
+	<li><code>s</code> consists of only digits.</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/3400-3499/3462.Maximum Sum With at Most K Elements/README.md

@@ -0,0 +1,111 @@
+---
+comments: true
+difficulty: 中等
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3400-3499/3462.Maximum%20Sum%20With%20at%20Most%20K%20Elements/README.md
+---
+
+<!-- problem:start -->
+
+# [3462. 提取至多 K 个元素的最大总和](https://leetcode.cn/problems/maximum-sum-with-at-most-k-elements)
+
+[English Version](/solution/3400-3499/3462.Maximum%20Sum%20With%20at%20Most%20K%20Elements/README_EN.md)
+
+## 题目描述
+
+<!-- description:start -->
+
+<p data-pm-slice="1 3 []">给你一个大小为 <code>n x m</code>&nbsp;的二维矩阵&nbsp;<code>grid</code>&nbsp;，以及一个长度为 <code>n</code>&nbsp;的整数数组&nbsp;<code>limits</code>&nbsp;，和一个整数&nbsp;<code>k</code>&nbsp;。你的目标是从矩阵 <code>grid</code> 中提取出&nbsp;<strong>至多</strong> <code>k</code>&nbsp;个元素，并计算这些元素的最大总和，提取时需满足以下限制<b>：</b></p>
+
+<ul data-spread="false">
+	<li>
+	<p>从 <code>grid</code>&nbsp;的第 <code>i</code> 行提取的元素数量不超过 <code>limits[i]</code> 。</p>
+	</li>
+</ul>
+
+<p data-pm-slice="1 1 []">返回最大总和。</p>
+
+<p>&nbsp;</p>
+
+<p><b>示例 1：</b></p>
+
+<div class="example-block">
+<p><span class="example-io"><b>输入：</b>grid = [[1,2],[3,4]], limits = [1,2], k = 2</span></p>
+
+<p><span class="example-io"><b>输出：</b>7</span></p>
+
+<p><b>解释：</b></p>
+
+<ul>
+	<li>从第 2 行提取至多 2 个元素，取出 4 和 3 。</li>
+	<li>至多提取 2 个元素时的最大总和&nbsp;<code>4 + 3 = 7</code>&nbsp;。</li>
+</ul>
+</div>
+
+<p><b>示例 2：</b></p>
+
+<div class="example-block">
+<p><span class="example-io"><b>输入：</b></span><span class="example-io">grid = [[5,3,7],[8,2,6]], limits = [2,2], k = 3</span></p>
+
+<p><span class="example-io"><b>输出：</b></span><span class="example-io">21</span></p>
+
+<p><b>解释：</b></p>
+
+<ul>
+	<li>从第 1&nbsp;行提取至多 2 个元素，取出 7 。</li>
+	<li>从第 2 行提取至多 2 个元素，取出&nbsp;8 和 6 。</li>
+	<li>至多提取 3&nbsp;个元素时的最大总和 <code>7 + 8 + 6 = 21</code>&nbsp;。</li>
+</ul>
+</div>
+
+<p>&nbsp;</p>
+
+<p><b>提示：</b></p>
+
+<ul>
+	<li><code>n == grid.length == limits.length</code></li>
+	<li><code>m == grid[i].length</code></li>
+	<li><code>1 &lt;= n, m &lt;= 500</code></li>
+	<li><code>0 &lt;= grid[i][j] &lt;= 10<sup>5</sup></code></li>
+	<li><code>0 &lt;= limits[i] &lt;= m</code></li>
+	<li><code>0 &lt;= k &lt;= min(n * m, sum(limits))</code></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 -->