feat: add weekly contest 419 and biweekly contest 141

solution/3100-3199/3162.Find the Number of Good Pairs I/README.md

@@ -21,7 +21,7 @@ tags:
 
 <p>给你两个整数数组 <code>nums1</code> 和 <code>nums2</code>，长度分别为 <code>n</code> 和 <code>m</code>。同时给你一个<strong>正整数</strong> <code>k</code>。</p>
 
-<p>如果 <code>nums1[i]</code> 可以被 <code>nums2[j] * k</code> 整除，则称数对 <code>(i, j)</code> 为 <strong>优质数对</strong>（<code>0 &lt;= i &lt;= n - 1</code>, <code>0 &lt;= j &lt;= m - 1</code>）。</p>
+<p>如果 <code>nums1[i]</code> 可以除尽&nbsp;<code>nums2[j] * k</code>，则称数对 <code>(i, j)</code> 为 <strong>优质数对</strong>（<code>0 &lt;= i &lt;= n - 1</code>, <code>0 &lt;= j &lt;= m - 1</code>）。</p>
 
 <p>返回<strong> 优质数对 </strong>的总数。</p>
 

solution/3300-3399/3311.Construct 2D Grid Matching Graph Layout/README.md

@@ -80,7 +80,7 @@ tags:
 	<li><code>1 &lt;= edges.length &lt;= 10<sup>5</sup></code></li>
 	<li><code>edges[i] = [u<sub>i</sub>, v<sub>i</sub>]</code></li>
 	<li><code>0 &lt;= u<sub>i</sub> &lt; v<sub>i</sub> &lt; n</code></li>
-	<li>树中的边互不相同。</li>
+	<li>图中的边互不相同。</li>
 	<li>输入保证&nbsp;<code>edges</code>&nbsp;可以形成一个符合上述条件的二维矩阵。</li>
 </ul>
 

solution/3300-3399/3314.Construct the Minimum Bitwise Array I/README.md

@@ -0,0 +1,112 @@
+---
+comments: true
+difficulty: 简单
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3300-3399/3314.Construct%20the%20Minimum%20Bitwise%20Array%20I/README.md
+---
+
+<!-- problem:start -->
+
+# [3314. 构造最小位运算数组 I](https://leetcode.cn/problems/construct-the-minimum-bitwise-array-i)
+
+[English Version](/solution/3300-3399/3314.Construct%20the%20Minimum%20Bitwise%20Array%20I/README_EN.md)
+
+## 题目描述
+
+<!-- description:start -->
+
+<p>给你一个长度为 <code>n</code>&nbsp;的<span data-keyword="prime">质数</span>数组&nbsp;<code>nums</code>&nbsp;。你的任务是返回一个长度为 <code>n</code>&nbsp;的数组 <code>ans</code>&nbsp;，对于每个下标 <code>i</code>&nbsp;，以下<strong>&nbsp;条件</strong>&nbsp;均成立：</p>
+
+<ul>
+	<li><code>ans[i] OR (ans[i] + 1) == nums[i]</code></li>
+</ul>
+
+<p>除此以外，你需要 <strong>最小化</strong>&nbsp;结果数组里每一个&nbsp;<code>ans[i]</code>&nbsp;。</p>
+
+<p>如果没法找到符合 <strong>条件</strong>&nbsp;的&nbsp;<code>ans[i]</code>&nbsp;，那么&nbsp;<code>ans[i] = -1</code>&nbsp;。</p>
+
+<p><strong>质数</strong>&nbsp;指的是一个大于 1 的自然数，且它只有 1 和自己两个因数。</p>
+
+<p>&nbsp;</p>
+
+<p><strong class="example">示例 1：</strong></p>
+
+<div class="example-block">
+<p><span class="example-io"><b>输入：</b>nums = [2,3,5,7]</span></p>
+
+<p><span class="example-io"><b>输出：</b>[-1,1,4,3]</span></p>
+
+<p><b>解释：</b></p>
+
+<ul>
+	<li>对于&nbsp;<code>i = 0</code>&nbsp;，不存在&nbsp;<code>ans[0]</code>&nbsp;满足&nbsp;<code>ans[0] OR (ans[0] + 1) = 2</code>&nbsp;，所以&nbsp;<code>ans[0] = -1</code>&nbsp;。</li>
+	<li>对于&nbsp;<code>i = 1</code>&nbsp;，满足 <code>ans[1] OR (ans[1] + 1) = 3</code>&nbsp;的最小&nbsp;<code>ans[1]</code>&nbsp;为&nbsp;<code>1</code>&nbsp;，因为&nbsp;<code>1 OR (1 + 1) = 3</code>&nbsp;。</li>
+	<li>对于&nbsp;<code>i = 2</code>&nbsp;，满足 <code>ans[2] OR (ans[2] + 1) = 5</code>&nbsp;的最小 <code>ans[2]</code>&nbsp;为&nbsp;<code>4</code>&nbsp;，因为&nbsp;<code>4 OR (4 + 1) = 5</code>&nbsp;。</li>
+	<li>对于&nbsp;<code>i = 3</code>&nbsp;，满足&nbsp;<code>ans[3] OR (ans[3] + 1) = 7</code>&nbsp;的最小&nbsp;<code>ans[3]</code>&nbsp;为&nbsp;<code>3</code>&nbsp;，因为&nbsp;<code>3 OR (3 + 1) = 7</code>&nbsp;。</li>
+</ul>
+</div>
+
+<p><strong class="example">示例 2：</strong></p>
+
+<div class="example-block">
+<p><span class="example-io"><b>输入：</b>nums = [11,13,31]</span></p>
+
+<p><span class="example-io"><b>输出：</b>[9,12,15]</span></p>
+
+<p><b>解释：</b></p>
+
+<ul>
+	<li>对于&nbsp;<code>i = 0</code>&nbsp;，满足&nbsp;<code>ans[0] OR (ans[0] + 1) = 11</code> 的最小&nbsp;<code>ans[0]</code>&nbsp;为&nbsp;<code>9</code>&nbsp;，因为&nbsp;<code>9 OR (9 + 1) = 11</code>&nbsp;。</li>
+	<li>对于&nbsp;<code>i = 1</code>&nbsp;，满足&nbsp;<code>ans[1] OR (ans[1] + 1) = 13</code>&nbsp;的最小&nbsp;<code>ans[1]</code>&nbsp;为&nbsp;<code>12</code>&nbsp;，因为&nbsp;<code>12 OR (12 + 1) = 13</code>&nbsp;。</li>
+	<li>对于&nbsp;<code>i = 2</code>&nbsp;，满足&nbsp;<code>ans[2] OR (ans[2] + 1) = 31</code>&nbsp;的最小&nbsp;<code>ans[2]</code>&nbsp;为&nbsp;<code>15</code>&nbsp;，因为&nbsp;<code>15 OR (15 + 1) = 31</code>&nbsp;。</li>
+</ul>
+</div>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length &lt;= 100</code></li>
+	<li><code>2 &lt;= nums[i] &lt;= 1000</code></li>
+	<li><code>nums[i]</code>&nbsp;是一个质数。</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/3300-3399/3314.Construct the Minimum Bitwise Array I/README_EN.md

@@ -0,0 +1,108 @@
+---
+comments: true
+difficulty: Easy
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3300-3399/3314.Construct%20the%20Minimum%20Bitwise%20Array%20I/README_EN.md
+---
+
+<!-- problem:start -->
+
+# [3314. Construct the Minimum Bitwise Array I](https://leetcode.com/problems/construct-the-minimum-bitwise-array-i)
+
+[中文文档](/solution/3300-3399/3314.Construct%20the%20Minimum%20Bitwise%20Array%20I/README.md)
+
+## Description
+
+<!-- description:start -->
+
+<p>You are given an array <code>nums</code> consisting of <code>n</code> <span data-keyword="prime-number">prime</span> integers.</p>
+
+<p>You need to construct an array <code>ans</code> of length <code>n</code>, such that, for each index <code>i</code>, the bitwise <code>OR</code> of <code>ans[i]</code> and <code>ans[i] + 1</code> is equal to <code>nums[i]</code>, i.e. <code>ans[i] OR (ans[i] + 1) == nums[i]</code>.</p>
+
+<p>Additionally, you must <strong>minimize</strong> each value of <code>ans[i]</code> in the resulting array.</p>
+
+<p>If it is <em>not possible</em> to find such a value for <code>ans[i]</code> that satisfies the <strong>condition</strong>, then set <code>ans[i] = -1</code>.</p>
+
+<p>A <strong>prime number</strong> is a natural number greater than 1 with only two factors, 1 and itself.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">nums = [2,3,5,7]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">[-1,1,4,3]</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<ul>
+	<li>For <code>i = 0</code>, as there is no value for <code>ans[0]</code> that satisfies <code>ans[0] OR (ans[0] + 1) = 2</code>, so <code>ans[0] = -1</code>.</li>
+	<li>For <code>i = 1</code>, the smallest <code>ans[1]</code> that satisfies <code>ans[1] OR (ans[1] + 1) = 3</code> is <code>1</code>, because <code>1 OR (1 + 1) = 3</code>.</li>
+	<li>For <code>i = 2</code>, the smallest <code>ans[2]</code> that satisfies <code>ans[2] OR (ans[2] + 1) = 5</code> is <code>4</code>, because <code>4 OR (4 + 1) = 5</code>.</li>
+	<li>For <code>i = 3</code>, the smallest <code>ans[3]</code> that satisfies <code>ans[3] OR (ans[3] + 1) = 7</code> is <code>3</code>, because <code>3 OR (3 + 1) = 7</code>.</li>
+</ul>
+</div>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">nums = [11,13,31]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">[9,12,15]</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<ul>
+	<li>For <code>i = 0</code>, the smallest <code>ans[0]</code> that satisfies <code>ans[0] OR (ans[0] + 1) = 11</code> is <code>9</code>, because <code>9 OR (9 + 1) = 11</code>.</li>
+	<li>For <code>i = 1</code>, the smallest <code>ans[1]</code> that satisfies <code>ans[1] OR (ans[1] + 1) = 13</code> is <code>12</code>, because <code>12 OR (12 + 1) = 13</code>.</li>
+	<li>For <code>i = 2</code>, the smallest <code>ans[2]</code> that satisfies <code>ans[2] OR (ans[2] + 1) = 31</code> is <code>15</code>, because <code>15 OR (15 + 1) = 31</code>.</li>
+</ul>
+</div>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length &lt;= 100</code></li>
+	<li><code>2 &lt;= nums[i] &lt;= 1000</code></li>
+	<li><code>nums[i]</code> is a prime number.</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/3300-3399/3315.Construct the Minimum Bitwise Array II/README.md

@@ -0,0 +1,112 @@
+---
+comments: true
+difficulty: 中等
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3300-3399/3315.Construct%20the%20Minimum%20Bitwise%20Array%20II/README.md
+---
+
+<!-- problem:start -->
+
+# [3315. 构造最小位运算数组 II](https://leetcode.cn/problems/construct-the-minimum-bitwise-array-ii)
+
+[English Version](/solution/3300-3399/3315.Construct%20the%20Minimum%20Bitwise%20Array%20II/README_EN.md)
+
+## 题目描述
+
+<!-- description:start -->
+
+<p>给你一个长度为 <code>n</code>&nbsp;的<span data-keyword="prime">质数</span>数组&nbsp;<code>nums</code>&nbsp;。你的任务是返回一个长度为 <code>n</code>&nbsp;的数组 <code>ans</code>&nbsp;，对于每个下标 <code>i</code>&nbsp;，以下<strong>&nbsp;条件</strong>&nbsp;均成立：</p>
+
+<ul>
+	<li><code>ans[i] OR (ans[i] + 1) == nums[i]</code></li>
+</ul>
+
+<p>除此以外，你需要 <strong>最小化</strong>&nbsp;结果数组里每一个&nbsp;<code>ans[i]</code>&nbsp;。</p>
+
+<p>如果没法找到符合 <strong>条件</strong>&nbsp;的&nbsp;<code>ans[i]</code>&nbsp;，那么&nbsp;<code>ans[i] = -1</code>&nbsp;。</p>
+
+<p><strong>质数</strong>&nbsp;指的是一个大于 1 的自然数，且它只有 1 和自己两个因数。</p>
+
+<p>&nbsp;</p>
+
+<p><strong class="example">示例 1：</strong></p>
+
+<div class="example-block">
+<p><span class="example-io"><b>输入：</b>nums = [2,3,5,7]</span></p>
+
+<p><span class="example-io"><b>输出：</b>[-1,1,4,3]</span></p>
+
+<p><b>解释：</b></p>
+
+<ul>
+	<li>对于&nbsp;<code>i = 0</code>&nbsp;，不存在&nbsp;<code>ans[0]</code>&nbsp;满足&nbsp;<code>ans[0] OR (ans[0] + 1) = 2</code>&nbsp;，所以&nbsp;<code>ans[0] = -1</code>&nbsp;。</li>
+	<li>对于&nbsp;<code>i = 1</code>&nbsp;，满足 <code>ans[1] OR (ans[1] + 1) = 3</code>&nbsp;的最小&nbsp;<code>ans[1]</code>&nbsp;为&nbsp;<code>1</code>&nbsp;，因为&nbsp;<code>1 OR (1 + 1) = 3</code>&nbsp;。</li>
+	<li>对于&nbsp;<code>i = 2</code>&nbsp;，满足 <code>ans[2] OR (ans[2] + 1) = 5</code>&nbsp;的最小 <code>ans[2]</code>&nbsp;为&nbsp;<code>4</code>&nbsp;，因为&nbsp;<code>4 OR (4 + 1) = 5</code>&nbsp;。</li>
+	<li>对于&nbsp;<code>i = 3</code>&nbsp;，满足&nbsp;<code>ans[3] OR (ans[3] + 1) = 7</code>&nbsp;的最小&nbsp;<code>ans[3]</code>&nbsp;为&nbsp;<code>3</code>&nbsp;，因为&nbsp;<code>3 OR (3 + 1) = 7</code>&nbsp;。</li>
+</ul>
+</div>
+
+<p><strong class="example">示例 2：</strong></p>
+
+<div class="example-block">
+<p><span class="example-io"><b>输入：</b>nums = [11,13,31]</span></p>
+
+<p><span class="example-io"><b>输出：</b>[9,12,15]</span></p>
+
+<p><b>解释：</b></p>
+
+<ul>
+	<li>对于&nbsp;<code>i = 0</code>&nbsp;，满足&nbsp;<code>ans[0] OR (ans[0] + 1) = 11</code> 的最小&nbsp;<code>ans[0]</code>&nbsp;为&nbsp;<code>9</code>&nbsp;，因为&nbsp;<code>9 OR (9 + 1) = 11</code>&nbsp;。</li>
+	<li>对于&nbsp;<code>i = 1</code>&nbsp;，满足&nbsp;<code>ans[1] OR (ans[1] + 1) = 13</code>&nbsp;的最小&nbsp;<code>ans[1]</code>&nbsp;为&nbsp;<code>12</code>&nbsp;，因为&nbsp;<code>12 OR (12 + 1) = 13</code>&nbsp;。</li>
+	<li>对于&nbsp;<code>i = 2</code>&nbsp;，满足&nbsp;<code>ans[2] OR (ans[2] + 1) = 31</code>&nbsp;的最小&nbsp;<code>ans[2]</code>&nbsp;为&nbsp;<code>15</code>&nbsp;，因为&nbsp;<code>15 OR (15 + 1) = 31</code>&nbsp;。</li>
+</ul>
+</div>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length &lt;= 100</code></li>
+	<li><code>2 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
+	<li><code>nums[i]</code>&nbsp;是一个质数。</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 -->