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authoredFeb 23, 2025··
feat: add weekly contest 438 (#4101)
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‎solution/1200-1299/1261.Find Elements in a Contaminated Binary Tree/README.md

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

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

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

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

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---
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comments: true
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difficulty: 简单
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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
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---
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<!-- problem:start -->
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# [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)
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[English Version](/solution/3400-3499/3461.Check%20If%20Digits%20Are%20Equal%20in%20String%20After%20Operations%20I/README_EN.md)
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## 题目描述
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<!-- description:start -->
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<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>
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<ul>
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<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>
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<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>
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</ul>
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<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>
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<p>&nbsp;</p>
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<p><strong class="example">示例 1：</strong></p>
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<div class="example-block">
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<p><strong>输入：</strong> <span class="example-io">s = "3902"</span></p>
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<p><strong>输出：</strong> <span class="example-io">true</span></p>
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<p><strong>解释：</strong></p>
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<ul>
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<li>一开始，<code>s = "3902"</code></li>
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<li>第一次操作：
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<ul>
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<li><code>(s[0] + s[1]) % 10 = (3 + 9) % 10 = 2</code></li>
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<li><code>(s[1] + s[2]) % 10 = (9 + 0) % 10 = 9</code></li>
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<li><code>(s[2] + s[3]) % 10 = (0 + 2) % 10 = 2</code></li>
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<li><code>s</code> 变为 <code>"292"</code></li>
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</ul>
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</li>
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<li>第二次操作：
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<ul>
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<li><code>(s[0] + s[1]) % 10 = (2 + 9) % 10 = 1</code></li>
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<li><code>(s[1] + s[2]) % 10 = (9 + 2) % 10 = 1</code></li>
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<li><code>s</code> 变为 <code>"11"</code></li>
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</ul>
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</li>
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<li>由于 <code>"11"</code> 中的数字相同，输出为 <code>true</code>。</li>
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</ul>
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</div>
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<p><strong class="example">示例 2：</strong></p>
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<div class="example-block">
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<p><strong>输入：</strong> <span class="example-io">s = "34789"</span></p>
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<p><strong>输出：</strong> <span class="example-io">false</span></p>
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<p><strong>解释：</strong></p>
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<ul>
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<li>一开始，<code>s = "34789"</code>。</li>
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<li>第一次操作后，<code>s = "7157"</code>。</li>
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<li>第二次操作后，<code>s = "862"</code>。</li>
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<li>第三次操作后，<code>s = "48"</code>。</li>
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<li>由于 <code>'4' != '8'</code>，输出为 <code>false</code>。</li>
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</ul>
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<p>&nbsp;</p>
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</div>
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<p><strong>提示：</strong></p>
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<ul>
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<li><code>3 &lt;= s.length &lt;= 100</code></li>
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<li><code>s</code> 仅由数字组成。</li>
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</ul>
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<!-- description:end -->
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## 解法
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<!-- solution:start -->
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### 方法一
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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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#### C++
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```cpp
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```
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#### Go
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```go
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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<!-- problem:end -->

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

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---
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comments: true
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difficulty: Easy
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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
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---
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<!-- problem:start -->
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# [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)
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[中文文档](/solution/3400-3499/3461.Check%20If%20Digits%20Are%20Equal%20in%20String%20After%20Operations%20I/README.md)
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## Description
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<!-- description:start -->
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<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>
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<ul>
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<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>
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<li>Replace <code>s</code> with the sequence of newly calculated digits, <em>maintaining the order</em> in which they are computed.</li>
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</ul>
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<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>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<div class="example-block">
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<p><strong>Input:</strong> <span class="example-io">s = &quot;3902&quot;</span></p>
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<p><strong>Output:</strong> <span class="example-io">true</span></p>
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<p><strong>Explanation:</strong></p>
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<ul>
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<li>Initially, <code>s = &quot;3902&quot;</code></li>
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<li>First operation:
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<ul>
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<li><code>(s[0] + s[1]) % 10 = (3 + 9) % 10 = 2</code></li>
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<li><code>(s[1] + s[2]) % 10 = (9 + 0) % 10 = 9</code></li>
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<li><code>(s[2] + s[3]) % 10 = (0 + 2) % 10 = 2</code></li>
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<li><code>s</code> becomes <code>&quot;292&quot;</code></li>
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</ul>
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</li>
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<li>Second operation:
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<ul>
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<li><code>(s[0] + s[1]) % 10 = (2 + 9) % 10 = 1</code></li>
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<li><code>(s[1] + s[2]) % 10 = (9 + 2) % 10 = 1</code></li>
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<li><code>s</code> becomes <code>&quot;11&quot;</code></li>
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</ul>
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</li>
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<li>Since the digits in <code>&quot;11&quot;</code> are the same, the output is <code>true</code>.</li>
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</ul>
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</div>
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<p><strong class="example">Example 2:</strong></p>
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<div class="example-block">
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<p><strong>Input:</strong> <span class="example-io">s = &quot;34789&quot;</span></p>
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<p><strong>Output:</strong> <span class="example-io">false</span></p>
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<p><strong>Explanation:</strong></p>
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<ul>
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<li>Initially, <code>s = &quot;34789&quot;</code>.</li>
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<li>After the first operation, <code>s = &quot;7157&quot;</code>.</li>
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<li>After the second operation, <code>s = &quot;862&quot;</code>.</li>
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<li>After the third operation, <code>s = &quot;48&quot;</code>.</li>
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<li>Since <code>&#39;4&#39; != &#39;8&#39;</code>, the output is <code>false</code>.</li>
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</ul>
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</div>
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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>3 &lt;= s.length &lt;= 100</code></li>
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<li><code>s</code> consists of only digits.</li>
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</ul>
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<!-- description:end -->
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## Solutions
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<!-- solution:start -->
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### Solution 1
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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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#### C++
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```cpp
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```
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#### Go
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```go
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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<!-- problem:end -->

‎solution/3400-3499/3462.Maximum Sum With at Most K Elements/README.md

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---
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comments: true
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difficulty: 中等
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edit_url: https://github.com/doocs/leetcode/edit/main/solution/3400-3499/3462.Maximum%20Sum%20With%20at%20Most%20K%20Elements/README.md
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---
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<!-- problem:start -->
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# [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)
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[English Version](/solution/3400-3499/3462.Maximum%20Sum%20With%20at%20Most%20K%20Elements/README_EN.md)
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## 题目描述
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<!-- description:start -->
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<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>
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<ul data-spread="false">
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<li>
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<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>
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</li>
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</ul>
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<p data-pm-slice="1 1 []">返回最大总和。</p>
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<p>&nbsp;</p>
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<p><b>示例 1：</b></p>
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<div class="example-block">
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<p><span class="example-io"><b>输入：</b>grid = [[1,2],[3,4]], limits = [1,2], k = 2</span></p>
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<p><span class="example-io"><b>输出：</b>7</span></p>
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<p><b>解释：</b></p>
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<ul>
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<li>从第 2 行提取至多 2 个元素，取出 4 和 3 。</li>
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<li>至多提取 2 个元素时的最大总和&nbsp;<code>4 + 3 = 7</code>&nbsp;。</li>
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</ul>
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</div>
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<p><b>示例 2：</b></p>
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<div class="example-block">
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<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>
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<p><span class="example-io"><b>输出：</b></span><span class="example-io">21</span></p>
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<p><b>解释：</b></p>
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<ul>
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<li>从第 1&nbsp;行提取至多 2 个元素，取出 7 。</li>
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<li>从第 2 行提取至多 2 个元素，取出&nbsp;8 和 6 。</li>
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<li>至多提取 3&nbsp;个元素时的最大总和 <code>7 + 8 + 6 = 21</code>&nbsp;。</li>
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</ul>
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</div>
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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 == grid.length == limits.length</code></li>
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<li><code>m == grid[i].length</code></li>
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<li><code>1 &lt;= n, m &lt;= 500</code></li>
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<li><code>0 &lt;= grid[i][j] &lt;= 10<sup>5</sup></code></li>
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<li><code>0 &lt;= limits[i] &lt;= m</code></li>
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<li><code>0 &lt;= k &lt;= min(n * m, sum(limits))</code></li>
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</ul>
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<!-- description:end -->
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## 解法
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<!-- solution:start -->
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### 方法一
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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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#### C++
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```cpp
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```
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#### Go
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```go
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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<!-- problem:end -->

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