feat: add biweekly contest 123

solution/3000-3099/3024.Type of Triangle II/README.md

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+# [3024. 三角形类型 II](https://leetcode.cn/problems/type-of-triangle-ii)
+
+[English Version](/solution/3000-3099/3024.Type%20of%20Triangle%20II/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;开始长度为 <code>3</code>&nbsp;的整数数组&nbsp;<code>nums</code>&nbsp;，需要用它们来构造三角形。</p>
+
+<ul>
+	<li>如果一个三角形的所有边长度相等，那么这个三角形称为&nbsp;<strong>equilateral</strong>&nbsp;。</li>
+	<li>如果一个三角形恰好有两条边长度相等，那么这个三角形称为&nbsp;<strong>isosceles</strong>&nbsp;。</li>
+	<li>如果一个三角形三条边的长度互不相同，那么这个三角形称为&nbsp;<strong>scalene</strong>&nbsp;。</li>
+</ul>
+
+<p>如果这个数组无法构成一个三角形，请你返回字符串&nbsp;<code>"none"</code>&nbsp;，否则返回一个字符串表示这个三角形的类型。</p>
+
+<p>&nbsp;</p>
+
+<p><strong class="example">示例 1：</strong></p>
+
+<pre>
+<b>输入：</b>nums = [3,3,3]
+<b>输出：</b>"equilateral"
+<b>解释：</b>由于三条边长度相等，所以可以构成一个等边三角形，返回 "equilateral" 。
+</pre>
+
+<p><strong class="example">示例 2：</strong></p>
+
+<pre>
+<b>输入：</b>nums = [3,4,5]
+<b>输出：</b>"scalene"
+<b>解释：</b>
+nums[0] + nums[1] = 3 + 4 = 7 ，大于 nums[2] = 5<code>&nbsp;。</code>
+nums[0] + nums[2] = 3 + 5 = 8 ，大于 nums[1] = 4 。
+nums[1] + nums[2] = 4 + 5 = 9 ，大于 nums[0] = 3 。
+由于任意两边纸盒都大于第三边，所以可以构成一个三角形。
+因为三条边的长度互不相等，所以返回 "scalene" 。
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>nums.length == 3</code></li>
+	<li><code>1 &lt;= nums[i] &lt;= 100</code></li>
+</ul>
+
+## 解法
+
+### 方法一
+
+<!-- tabs:start -->
+
+```python
+
+```
+
+```java
+
+```
+
+```cpp
+
+```
+
+```go
+
+```
+
+<!-- tabs:end -->
+
+<!-- end -->

solution/3000-3099/3024.Type of Triangle II/README_EN.md

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+# [3024. Type of Triangle II](https://leetcode.com/problems/type-of-triangle-ii)
+
+[中文文档](/solution/3000-3099/3024.Type%20of%20Triangle%20II/README.md)
+
+## Description
+
+<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code> of size <code>3</code> which can form the sides of a triangle.</p>
+
+<ul>
+	<li>A triangle is called <strong>equilateral</strong> if it has all sides of equal length.</li>
+	<li>A triangle is called <strong>isosceles</strong> if it has exactly two sides of equal length.</li>
+	<li>A triangle is called <strong>scalene</strong> if all its sides are of different lengths.</li>
+</ul>
+
+<p>Return <em>a string representing</em> <em>the type of triangle that can be formed </em><em>or </em><code>&quot;none&quot;</code><em> if it <strong>cannot</strong> form a triangle.</em></p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [3,3,3]
+<strong>Output:</strong> &quot;equilateral&quot;
+<strong>Explanation:</strong> Since all the sides are of equal length, therefore, it will form an equilateral triangle.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [3,4,5]
+<strong>Output:</strong> &quot;scalene&quot;
+<strong>Explanation:</strong> 
+nums[0] + nums[1] = 3 + 4 = 7, which is greater than nums[2] = 5.
+nums[0] + nums[2] = 3 + 5 = 8, which is greater than nums[1] = 4.
+nums[1] + nums[2] = 4 + 5 = 9, which is greater than nums[0] = 3. 
+Since the sum of the two sides is greater than the third side for all three cases, therefore, it can form a triangle.
+As all the sides are of different lengths, it will form a scalene triangle.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>nums.length == 3</code></li>
+	<li><code>1 &lt;= nums[i] &lt;= 100</code></li>
+</ul>
+
+## Solutions
+
+### Solution 1
+
+<!-- tabs:start -->
+
+```python
+
+```
+
+```java
+
+```
+
+```cpp
+
+```
+
+```go
+
+```
+
+<!-- tabs:end -->
+
+<!-- end -->

solution/3000-3099/3025.Find the Number of Ways to Place People I/README.md

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+# [3025. 人员站位的方案数 I](https://leetcode.cn/problems/find-the-number-of-ways-to-place-people-i)
+
+[English Version](/solution/3000-3099/3025.Find%20the%20Number%20of%20Ways%20to%20Place%20People%20I/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>给你一个&nbsp;&nbsp;<code>n x 2</code>&nbsp;的二维数组 <code>points</code>&nbsp;，它表示二维平面上的一些点坐标，其中&nbsp;<code>points[i] = [x<sub>i</sub>, y<sub>i</sub>]</code>&nbsp;。</p>
+
+<p>我们定义 x 轴的正方向为 <strong>右</strong>&nbsp;（<strong>x 轴递增的方向</strong>），x 轴的负方向为 <strong>左</strong>&nbsp;（<strong>x 轴递减的方向</strong>）。类似的，我们定义 y 轴的正方向为 <strong>上</strong>&nbsp;（<strong>y 轴递增的方向</strong>），y 轴的负方向为 <strong>下</strong>&nbsp;（<strong>y 轴递减的方向</strong>）。</p>
+
+<p>你需要安排这 <code>n</code>&nbsp;个人的站位，这 <code>n</code>&nbsp;个人中包括&nbsp;liupengsay 和小羊肖恩&nbsp;。你需要确保每个点处&nbsp;<strong>恰好</strong>&nbsp;有&nbsp;<strong>一个</strong>&nbsp;人。同时，liupengsay 想跟小羊肖恩单独玩耍，所以&nbsp;liupengsay&nbsp;会以 liupengsay<b>&nbsp;</b>的坐标为 <strong>左上角</strong>&nbsp;，小羊肖恩的坐标为 <strong>右下角</strong>&nbsp;建立一个矩形的围栏（<strong>注意</strong>，围栏可能&nbsp;<strong>不</strong> 包含任何区域，也就是说围栏可能是一条线段）。如果围栏的 <strong>内部</strong>&nbsp;或者 <strong>边缘</strong>&nbsp;上有任何其他人，liupengsay 都会难过。</p>
+
+<p>请你在确保 liupengsay&nbsp;<strong>不会</strong> 难过的前提下，返回 liupengsay 和小羊肖恩可以选择的 <strong>点对</strong>&nbsp;数目。</p>
+
+<p><b>注意</b>，liupengsay 建立的围栏必须确保 liupengsay 的位置是矩形的左上角，小羊肖恩的位置是矩形的右下角。比方说，以&nbsp;<code>(1, 1)</code>&nbsp;，<code>(1, 3)</code>&nbsp;，<code>(3, 1)</code>&nbsp;和&nbsp;<code>(3, 3)</code>&nbsp;为矩形的四个角，给定下图的两个输入，liupengsay 都不能建立围栏，原因如下：</p>
+
+<ul>
+	<li>图一中，liupengsay 在&nbsp;<code>(3, 3)</code>&nbsp;且小羊肖恩在&nbsp;<code>(1, 1)</code>&nbsp;，liupengsay 的位置不是左上角且小羊肖恩的位置不是右下角。</li>
+	<li>图二中，liupengsay 在&nbsp;<code>(1, 3)</code>&nbsp;且小羊肖恩在&nbsp;<code>(1, 1)</code>&nbsp;，小羊肖恩的位置不是在围栏的右下角。</li>
+</ul>
+<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/3000-3099/3025.Find%20the%20Number%20of%20Ways%20to%20Place%20People%20I/images/example0alicebob-1.png" style="width: 750px; height: 308px;padding: 10px; background: #fff; border-radius: .5rem;" />
+<p>&nbsp;</p>
+
+<p><strong class="example">示例 1：</strong></p>
+
+<p><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/3000-3099/3025.Find%20the%20Number%20of%20Ways%20to%20Place%20People%20I/images/example1alicebob.png" style="width: 376px; height: 308px; padding: 10px; background: rgb(255, 255, 255); border-radius: 0.5rem;" /></p>
+
+<pre>
+<b>输入：</b>points = [[1,1],[2,2],[3,3]]
+<b>输出：</b>0
+<strong>解释：</strong>没有办法可以让 liupengsay 的围栏以 liupengsay 的位置为左上角且小羊肖恩的位置为右下角。所以我们返回 0 。
+</pre>
+
+<p><strong class="example">示例 2：</strong></p>
+
+<p><strong class="example"><a href="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/3000-3099/3025.Find%20the%20Number%20of%20Ways%20to%20Place%20People%20I/images/1706880313-YelabI-example2.jpeg"><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/3000-3099/3025.Find%20the%20Number%20of%20Ways%20to%20Place%20People%20I/images/1706880313-YelabI-example2.jpeg" style="width: 900px; height: 247px;" /></a></strong></p>
+
+<pre>
+<b>输入：</b>points = [[6,2],[4,4],[2,6]]
+<b>输出：</b>2
+<b>解释：</b>总共有 2 种方案安排 liupengsay 和小羊肖恩的位置，使得 liupengsay 不会难过：
+- liupengsay 站在 (4, 4) ，小羊肖恩站在 (6, 2) 。
+- liupengsay 站在 (2, 6) ，小羊肖恩站在 (4, 4) 。
+不能安排 liupengsay 站在 (2, 6) 且小羊肖恩站在 (6, 2) ，因为站在 (4, 4) 的人处于围栏内。
+</pre>
+
+<p><strong class="example">示例 3：</strong></p>
+
+<p><strong class="example"><a href="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/3000-3099/3025.Find%20the%20Number%20of%20Ways%20to%20Place%20People%20I/images/1706880311-mtPGYC-example3.jpeg"><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/3000-3099/3025.Find%20the%20Number%20of%20Ways%20to%20Place%20People%20I/images/1706880311-mtPGYC-example3.jpeg" style="width: 900px; height: 247px;" /></a></strong></p>
+
+<pre>
+<b>输入：</b>points = [[3,1],[1,3],[1,1]]
+<b>输出：</b>2
+<b>解释：</b>总共有 2 种方案安排 liupengsay 和小羊肖恩的位置，使得 liupengsay 不会难过：
+- liupengsay 站在 (1, 1) ，小羊肖恩站在 (3, 1) 。
+- liupengsay 站在 (1, 3) ，小羊肖恩站在 (1, 1) 。
+不能安排 liupengsay 站在 (1, 3) 且小羊肖恩站在 (3, 1) ，因为站在 (1, 1) 的人处于围栏内。
+注意围栏是可以不包含任何面积的，上图中第一和第二个围栏都是合法的。
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>2 &lt;= n &lt;= 50</code></li>
+	<li><code>points[i].length == 2</code></li>
+	<li><code>0 &lt;= points[i][0], points[i][1] &lt;= 50</code></li>
+	<li><code>points[i]</code>&nbsp;点对两两不同。</li>
+</ul>
+
+## 解法
+
+### 方法一
+
+<!-- tabs:start -->
+
+```python
+
+```
+
+```java
+
+```
+
+```cpp
+
+```
+
+```go
+
+```
+
+<!-- tabs:end -->
+
+<!-- end -->

solution/3000-3099/3025.Find the Number of Ways to Place People I/README_EN.md

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+# [3025. Find the Number of Ways to Place People I](https://leetcode.com/problems/find-the-number-of-ways-to-place-people-i)
+
+[中文文档](/solution/3000-3099/3025.Find%20the%20Number%20of%20Ways%20to%20Place%20People%20I/README.md)
+
+## Description
+
+<p>You are given a 2D array <code>points</code> of size <code>n x 2</code> representing integer coordinates of some points on a 2D-plane, where <code>points[i] = [x<sub>i</sub>, y<sub>i</sub>]</code>.</p>
+
+<p>We define the <strong>right</strong> direction as positive x-axis (<strong>increasing x-coordinate</strong>) and the <strong>left</strong> direction as negative x-axis (<strong>decreasing x-coordinate</strong>). Similarly, we define the <strong>up</strong> direction as positive y-axis (<strong>increasing y-coordinate</strong>) and the <strong>down</strong> direction as negative y-axis (<strong>decreasing y-coordinate</strong>)</p>
+
+<p>You have to place <code>n</code> people, including Chisato and Takina, at these points such that there is <strong>exactly one</strong> person at every point. Chisato wants to be alone with Takina, so Chisato will build a rectangular fence with Chisato&#39;s position as the <strong>upper left corner</strong> and Takina&#39;s position as the <strong>lower right corner</strong> of the fence (<strong>Note</strong> that the fence <strong>might not</strong> enclose any area, i.e. it can be a line). If any person other than Chisato and Takina is either <strong>inside</strong> the fence or <strong>on</strong> the fence, Chisato will be sad.</p>
+
+<p>Return <em>the number of <strong>pairs of points</strong> where you can place Chisato and Takina, such that Chisato <strong>does not</strong> become sad on building the fence</em>.</p>
+
+<p><strong>Note</strong> that Chisato can only build a fence with Chisato&#39;s position as the upper left corner, and Takina&#39;s position as the lower right corner. For example, Chisato cannot build either of the fences in the picture below with four corners <code>(1, 1)</code>, <code>(1, 3)</code>, <code>(3, 1)</code>, and <code>(3, 3)</code>, because:</p>
+
+<ul>
+	<li>With Chisato at <code>(3, 3)</code> and Takina at <code>(1, 1)</code>, Chisato&#39;s position is not the upper left corner and Takina&#39;s position is not the lower right corner of the fence.</li>
+	<li>With Chisato at <code>(1, 3)</code> and Takina at <code>(1, 1)</code>, Takina&#39;s position is not the lower right corner of the fence.</li>
+</ul>
+<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/3000-3099/3025.Find%20the%20Number%20of%20Ways%20to%20Place%20People%20I/images/example0alicebob-1.png" style="width: 750px; height: 308px;padding: 10px; background: #fff; border-radius: .5rem;" />
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/3000-3099/3025.Find%20the%20Number%20of%20Ways%20to%20Place%20People%20I/images/example1alicebob.png" style="width: 376px; height: 308px; padding: 10px; background: rgb(255, 255, 255); border-radius: 0.5rem;" />
+<pre>
+<strong>Input:</strong> points = [[1,1],[2,2],[3,3]]
+<strong>Output:</strong> 0
+<strong>Explanation:</strong> There is no way to place Chisato and Takina such that Chisato can build a fence with Chisato&#39;s position as the upper left corner and Takina&#39;s position as the lower right corner. Hence we return 0. 
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/3000-3099/3025.Find%20the%20Number%20of%20Ways%20to%20Place%20People%20I/images/example2chisatotakina.png" style="width: 1121px; height: 308px; padding: 10px; background: rgb(255, 255, 255); border-radius: 0.5rem;" />
+<pre>
+<strong>Input:</strong> points = [[6,2],[4,4],[2,6]]
+<strong>Output:</strong> 2
+<strong>Explanation:</strong> There are two ways to place Chisato and Takina such that Chisato will not be sad:
+- Place Chisato at (4, 4) and Takina at (6, 2).
+- Place Chisato at (2, 6) and Takina at (4, 4).
+You cannot place Chisato at (2, 6) and Takina at (6, 2) because the person at (4, 4) will be inside the fence.
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/3000-3099/3025.Find%20the%20Number%20of%20Ways%20to%20Place%20People%20I/images/chisatotakinaexample3.png" style="width: 1123px; height: 308px; padding: 10px; background: rgb(255, 255, 255); border-radius: 0.5rem;" />
+<pre>
+<strong>Input:</strong> points = [[3,1],[1,3],[1,1]]
+<strong>Output:</strong> 2
+<strong>Explanation:</strong> There are two ways to place Chisato and Takina such that Chisato will not be sad:
+- Place Chisato at (1, 1) and Takina at (3, 1).
+- Place Chisato at (1, 3) and Takina at (1, 1).
+You cannot place Chisato at (1, 3) and Takina at (3, 1) because the person at (1, 1) will be on the fence.
+Note that it does not matter if the fence encloses any area, the first and second fences in the image are valid.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>2 &lt;= n &lt;= 50</code></li>
+	<li><code>points[i].length == 2</code></li>
+	<li><code>0 &lt;= points[i][0], points[i][1] &lt;= 50</code></li>
+	<li>All <code>points[i]</code> are distinct.</li>
+</ul>
+
+## Solutions
+
+### Solution 1
+
+<!-- tabs:start -->
+
+```python
+
+```
+
+```java
+
+```
+
+```cpp
+
+```
+
+```go
+
+```
+
+<!-- tabs:end -->
+
+<!-- end -->