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feat: add weekly contest 386 (doocs#2374)
 Co-authored-by: Doocs Bot <doocs-bot@outlook.com>
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# [3046. 分割数组](https://leetcode.cn/problems/split-the-array)
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[English Version](/solution/3000-3099/3046.Split%20the%20Array/README_EN.md)
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<!-- tags: -->
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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 <strong>偶数 </strong>的整数数组 <code>nums</code> 。你需要将这个数组分割成 <code>nums1</code> 和 <code>nums2</code> 两部分,要求:</p>
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<ul>
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<li><code>nums1.length == nums2.length == nums.length / 2</code> 。</li>
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<li><code>nums1</code> 应包含 <strong>互不相同</strong><strong> </strong>的元素。</li>
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<li><code>nums2</code>也应包含<strong> 互不相同</strong> 的元素。</li>
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</ul>
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<p>如果能够分割数组就返回 <code>true</code> ,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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<pre>
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<strong>输入:</strong>nums = [1,1,2,2,3,4]
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<strong>输出:</strong>true
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<strong>解释:</strong>分割 nums 的可行方案之一是 nums1 = [1,2,3] 和 nums2 = [1,2,4] 。
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</pre>
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<p><strong class="example">示例 2:</strong></p>
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<pre>
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<strong>输入:</strong>nums = [1,1,1,1]
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<strong>输出:</strong>false
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<strong>解释:</strong>分割 nums 的唯一可行方案是 nums1 = [1,1] 和 nums2 = [1,1] 。但 nums1 和 nums2 都不是由互不相同的元素构成。因此,返回 false 。
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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>1 &lt;= nums.length &lt;= 100</code></li>
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<li><code>nums.length % 2 == 0</code></li>
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<li><code>1 &lt;= nums[i] &lt;= 100</code></li>
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</ul>
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## 解法
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### 方法一
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<!-- tabs:start -->
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```python
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```
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```java
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```
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```cpp
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```
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```go
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```
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<!-- tabs:end -->
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<!-- end -->
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# [3046. Split the Array](https://leetcode.com/problems/split-the-array)
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[中文文档](/solution/3000-3099/3046.Split%20the%20Array/README.md)
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<!-- tags: -->
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## Description
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<p>You are given an integer array <code>nums</code> of <strong>even</strong> length. You have to split the array into two parts <code>nums1</code> and <code>nums2</code> such that:</p>
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<ul>
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<li><code>nums1.length == nums2.length == nums.length / 2</code>.</li>
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<li><code>nums1</code> should contain <strong>distinct </strong>elements.</li>
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<li><code>nums2</code> should also contain <strong>distinct</strong> elements.</li>
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</ul>
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<p>Return <code>true</code><em> if it is possible to split the array, and </em><code>false</code> <em>otherwise</em><em>.</em></p>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> nums = [1,1,2,2,3,4]
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<strong>Output:</strong> true
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<strong>Explanation:</strong> One of the possible ways to split nums is nums1 = [1,2,3] and nums2 = [1,2,4].
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</pre>
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<p><strong class="example">Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> nums = [1,1,1,1]
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<strong>Output:</strong> false
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<strong>Explanation:</strong> The only possible way to split nums is nums1 = [1,1] and nums2 = [1,1]. Both nums1 and nums2 do not contain distinct elements. Therefore, we return false.
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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>1 &lt;= nums.length &lt;= 100</code></li>
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<li><code>nums.length % 2 == 0 </code></li>
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<li><code>1 &lt;= nums[i] &lt;= 100</code></li>
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</ul>
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## Solutions
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### Solution 1
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<!-- tabs:start -->
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```python
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```
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```java
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```
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```cpp
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```
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```go
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```
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<!-- tabs:end -->
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<!-- end -->
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# [3047. 求交集区域内的最大正方形面积](https://leetcode.cn/problems/find-the-largest-area-of-square-inside-two-rectangles)
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[English Version](/solution/3000-3099/3047.Find%20the%20Largest%20Area%20of%20Square%20Inside%20Two%20Rectangles/README_EN.md)
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<!-- tags: -->
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## 题目描述
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<!-- 这里写题目描述 -->
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<p>在二维平面上存在 <code>n</code> 个矩形。给你两个下标从 <strong>0</strong> 开始的二维整数数组 <code>bottomLeft</code> 和 <code>topRight</code>,两个数组的大小都是 <code>n x 2</code> ,其中 <code>bottomLeft[i]</code> 和 <code>topRight[i]</code> 分别代表第 <code>i</code> 个矩形的<strong> 左下角 </strong>和 <strong>右上角 </strong>坐标。</p>
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<p>我们定义 <strong>向右 </strong>的方向为 x 轴正半轴(<strong>x 坐标增加</strong>),<strong>向左 </strong>的方向为 x 轴负半轴(<strong>x 坐标减少</strong>)。同样地,定义 <strong>向上 </strong>的方向为 y 轴正半轴(<strong>y 坐标增加</strong>)<strong>,向下 </strong>的方向为 y 轴负半轴(<strong>y 坐标减少</strong>)。</p>
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<p>你可以选择一个区域,该区域由两个矩形的 <strong>交集</strong>&nbsp;形成。你需要找出能够放入该区域 <strong>内 </strong>的<strong> 最大 </strong>正方形面积,并选择最优解。</p>
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<p>返回能够放入交集区域的正方形的 <strong>最大 </strong>可能面积,如果矩形之间不存在任何交集区域,则返回 <code>0</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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<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/3000-3099/3047.Find%20the%20Largest%20Area%20of%20Square%20Inside%20Two%20Rectangles/images/example12.png" style="width: 443px; height: 364px; padding: 10px; background: rgb(255, 255, 255); border-radius: 0.5rem;" />
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<pre>
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<strong>输入:</strong>bottomLeft = [[1,1],[2,2],[3,1]], topRight = [[3,3],[4,4],[6,6]]
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<strong>输出:</strong>1
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<strong>解释:</strong>边长为 1 的正方形可以放入矩形 0 和矩形 1 的交集区域,或矩形 1 和矩形 2 的交集区域。因此最大面积是边长 * 边长,即 1 * 1 = 1。
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可以证明,边长更大的正方形无法放入任何交集区域。
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</pre>
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<p><strong class="example">示例 2:</strong></p>
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<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/3000-3099/3047.Find%20the%20Largest%20Area%20of%20Square%20Inside%20Two%20Rectangles/images/rectanglesexample2.png" style="padding: 10px; background: rgb(255, 255, 255); border-radius: 0.5rem; width: 445px; height: 365px;" />
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<pre>
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<strong>输入:</strong>bottomLeft = [[1,1],[2,2],[1,2]], topRight = [[3,3],[4,4],[3,4]]
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<strong>输出:</strong>1
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<strong>解释:</strong>边长为 1 的正方形可以放入矩形 0 和矩形 1,矩形 1 和矩形 2,或所有三个矩形的交集区域。因此最大面积是边长 * 边长,即 1 * 1 = 1。
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可以证明,边长更大的正方形无法放入任何交集区域。
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请注意,区域可以由多于两个矩形的交集构成。
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</pre>
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<p><strong class="example">示例 3:</strong></p>
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<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/3000-3099/3047.Find%20the%20Largest%20Area%20of%20Square%20Inside%20Two%20Rectangles/images/rectanglesexample3.png" style="padding: 10px; background: rgb(255, 255, 255); border-radius: 0.5rem; width: 444px; height: 364px;" />
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<pre>
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<strong>输入:</strong>bottomLeft = [[1,1],[3,3],[3,1]], topRight = [[2,2],[4,4],[4,2]]
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<strong>输出:</strong>0
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<strong>解释:</strong>不存在相交的矩形,因此,返回 0 。
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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 == bottomLeft.length == topRight.length</code></li>
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<li><code>2 &lt;= n &lt;= 10<sup>3</sup></code></li>
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<li><code>bottomLeft[i].length == topRight[i].length == 2</code></li>
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<li><code>1 &lt;= bottomLeft[i][0], bottomLeft[i][1] &lt;= 10<sup>7</sup></code></li>
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<li><code>1 &lt;= topRight[i][0], topRight[i][1] &lt;= 10<sup>7</sup></code></li>
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<li><code>bottomLeft[i][0] &lt; topRight[i][0]</code></li>
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<li><code>bottomLeft[i][1] &lt; topRight[i][1]</code></li>
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</ul>
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## 解法
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### 方法一
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<!-- tabs:start -->
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```python
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```
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```java
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```
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```cpp
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```
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```go
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```
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<!-- tabs:end -->
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<!-- end -->
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# [3047. Find the Largest Area of Square Inside Two Rectangles](https://leetcode.com/problems/find-the-largest-area-of-square-inside-two-rectangles)
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[中文文档](/solution/3000-3099/3047.Find%20the%20Largest%20Area%20of%20Square%20Inside%20Two%20Rectangles/README.md)
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<!-- tags: -->
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## Description
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<p>There exist <code>n</code> rectangles in a 2D plane. You are given two <strong>0-indexed</strong> 2D integer arrays <code>bottomLeft</code> and <code>topRight</code>, both of size <code>n x 2</code>, where <code>bottomLeft[i]</code> and <code>topRight[i]</code> represent the <strong>bottom-left</strong> and <strong>top-right</strong> coordinates of the <code>i<sup>th</sup></code> rectangle respectively.</p>
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<p>You can select a region formed from the <strong>intersection</strong> of&nbsp;two of the given rectangles. You need to find the <strong>largest </strong>area of a <strong>square</strong> that can fit <strong>inside</strong> this region if you select the region optimally.</p>
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<p>Return <em>the <strong>largest </strong>possible area of a square, or </em><code>0</code><em> if there <strong>do not</strong> exist any intersecting regions between the rectangles</em>.</p>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/3000-3099/3047.Find%20the%20Largest%20Area%20of%20Square%20Inside%20Two%20Rectangles/images/example12.png" style="width: 443px; height: 364px; padding: 10px; background: rgb(255, 255, 255); border-radius: 0.5rem;" />
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<pre>
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<strong>Input:</strong> bottomLeft = [[1,1],[2,2],[3,1]], topRight = [[3,3],[4,4],[6,6]]
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<strong>Output:</strong> 1
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<strong>Explanation:</strong> A square with side length 1 can fit inside either the intersecting region of rectangle 0 and rectangle 1, or the intersecting region of rectangle 1 and rectangle 2. Hence the largest area is side * side which is 1 * 1 == 1.
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It can be shown that a square with a greater side length can not fit inside any intersecting region.
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</pre>
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<p><strong class="example">Example 2:</strong></p>
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<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/3000-3099/3047.Find%20the%20Largest%20Area%20of%20Square%20Inside%20Two%20Rectangles/images/rectanglesexample2.png" style="padding: 10px; background: rgb(255, 255, 255); border-radius: 0.5rem; width: 445px; height: 365px;" />
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<pre>
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<strong>Input:</strong> bottomLeft = [[1,1],[2,2],[1,2]], topRight = [[3,3],[4,4],[3,4]]
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<strong>Output:</strong> 1
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<strong>Explanation:</strong> A square with side length 1 can fit inside either the intersecting region of rectangle 0 and rectangle 1, the intersecting region of rectangle 1 and rectangle 2, or the intersection region of all 3 rectangles. Hence the largest area is side * side which is 1 * 1 == 1.
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It can be shown that a square with a greater side length can not fit inside any intersecting region.
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Note that the region can be formed by the intersection of more than 2 rectangles.
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</pre>
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<p><strong class="example">Example 3:</strong></p>
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<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/3000-3099/3047.Find%20the%20Largest%20Area%20of%20Square%20Inside%20Two%20Rectangles/images/rectanglesexample3.png" style="padding: 10px; background: rgb(255, 255, 255); border-radius: 0.5rem; width: 444px; height: 364px;" />
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<pre>
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<strong>Input:</strong> bottomLeft = [[1,1],[3,3],[3,1]], topRight = [[2,2],[4,4],[4,2]]
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<strong>Output:</strong> 0
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<strong>Explanation:</strong> No pair of rectangles intersect, hence, we return 0.
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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 == bottomLeft.length == topRight.length</code></li>
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<li><code>2 &lt;= n &lt;= 10<sup>3</sup></code></li>
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<li><code>bottomLeft[i].length == topRight[i].length == 2</code></li>
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<li><code>1 &lt;= bottomLeft[i][0], bottomLeft[i][1] &lt;= 10<sup>7</sup></code></li>
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<li><code>1 &lt;= topRight[i][0], topRight[i][1] &lt;= 10<sup>7</sup></code></li>
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<li><code>bottomLeft[i][0] &lt; topRight[i][0]</code></li>
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<li><code>bottomLeft[i][1] &lt; topRight[i][1]</code></li>
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</ul>
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## Solutions
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### Solution 1
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<!-- tabs:start -->
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```python
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```
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```java
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```
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```cpp
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```
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```go
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```
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<!-- tabs:end -->
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<!-- end -->
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