feat: add new lc problem

solution/3000-3099/3073.Maximum Increasing Triplet Value/README.md

@@ -0,0 +1,74 @@
+# [3073. 最大递增三元组](https://leetcode.cn/problems/maximum-increasing-triplet-value)
+
+[English Version](/solution/3000-3099/3073.Maximum%20Increasing%20Triplet%20Value/README_EN.md)
+
+<!-- tags: -->
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>给定一个数组&nbsp;<code>nums</code>，返回满足 <code>i &lt; j &lt; k</code> 且 <code>nums[i] &lt; nums[j] &lt; nums[k]</code>&nbsp;的三元组 <code>(i, j, k)</code> <em>&nbsp;</em>的 <strong>最大值</strong><i>。</i></p>
+
+<p>三元组&nbsp;<code>(i, j, k)</code> &nbsp;的&nbsp;<strong>值&nbsp;</strong>为&nbsp;<code>nums[i] - nums[j] + nums[k]</code>。</p>
+
+<div id="gtx-trans" style="position: absolute; left: 274px; top: 102px;">
+<div class="gtx-trans-icon">&nbsp;</div>
+</div>
+
+<p>&nbsp;</p>
+
+<p><strong class="example">示例 1: </strong></p>
+
+<pre>
+输入：nums = [5,6,9]
+输出：8
+解释：对于一个递增的三元组，我们只有一个选择，那就是选择所有三个元素。三元组的值为 5 - 6 + 9 = 8。
+</pre>
+
+<p><strong class="example">示例 2: </strong></p>
+
+<pre>
+输入：nums = [1,5,3,6]
+输出：4
+解释：只有两个递增三元组：
+(0, 1, 3)：这个三元组的值为 nums[0] - nums[1] + nums[3] = 1 - 5 + 6 = 2。
+(0, 2, 3)：这个三元组的值为 nums[0] - nums[2] + nums[3] = 1 - 3 + 6 = 4。
+因此答案是 4。
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示:</strong></p>
+
+<ul>
+	<li><code>3 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>1 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
+	<li>输入数据保证至少一个三元组满足给定条件。</li>
+</ul>
+
+## 解法
+
+### 方法一
+
+<!-- tabs:start -->
+
+```python
+
+```
+
+```java
+
+```
+
+```cpp
+
+```
+
+```go
+
+```
+
+<!-- tabs:end -->
+
+<!-- end -->

solution/3000-3099/3073.Maximum Increasing Triplet Value/README_EN.md

@@ -0,0 +1,77 @@
+# [3073. Maximum Increasing Triplet Value](https://leetcode.com/problems/maximum-increasing-triplet-value)
+
+[中文文档](/solution/3000-3099/3073.Maximum%20Increasing%20Triplet%20Value/README.md)
+
+<!-- tags: -->
+
+## Description
+
+<p>Given an array <code>nums</code>, return <em>the <strong>maximum value</strong> of a triplet</em> <code>(i, j, k)</code> <em>such that</em> <code>i &lt; j &lt; k</code> <em>and</em> <code>nums[i] &lt; nums[j] &lt; nums[k]</code>.</p>
+
+<p>The <strong>value</strong> of a triplet <code>(i, j, k)</code> is <code>nums[i] - nums[j] + nums[k]</code>.</p>
+
+<div id="gtx-trans" style="position: absolute; left: 274px; top: 102px;">
+<div class="gtx-trans-icon"> </div>
+</div>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1: </strong></p>
+
+<div class="example-block" style="border-color: var(--border-tertiary); border-left-width: 2px; color: var(--text-secondary); font-size: .875rem; margin-bottom: 1rem; margin-top: 1rem; overflow: visible; padding-left: 1rem;">
+<p><strong>Input: </strong> <span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;">nums = [5,6,9] </span></p>
+
+<p><strong>Output: </strong> <span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;">8 </span></p>
+
+<p><strong>Explanation: </strong> We only have one choice for an increasing triplet and that is choosing all three elements. The value of this triplet would be <code>5 - 6 + 9 = 8</code>.</p>
+</div>
+
+<p><strong class="example">Example 2: </strong></p>
+
+<div class="example-block" style="border-color: var(--border-tertiary); border-left-width: 2px; color: var(--text-secondary); font-size: .875rem; margin-bottom: 1rem; margin-top: 1rem; overflow: visible; padding-left: 1rem;">
+<p><strong>Input:</strong> <span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;"> nums = [1,5,3,6] </span></p>
+
+<p><strong>Output:</strong> <span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;"> 4 </span></p>
+
+<p><strong>Explanation: </strong> There are only two increasing triplets:</p>
+
+<p><code>(0, 1, 3)</code>: The value of this triplet is <code>nums[0] - nums[1] + nums[3] = 1 - 5 + 6 = 2</code>.</p>
+
+<p><code>(0, 2, 3)</code>: The value of this triplet is <code>nums[0] - nums[2] + nums[3] = 1 - 3 + 6 = 4</code>.</p>
+
+<p>Thus the answer would be <code>4</code>.</p>
+</div>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>3 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>1 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
+	<li>The input is generated such that at least one triplet meets the given condition.</li>
+</ul>
+
+## Solutions
+
+### Solution 1
+
+<!-- tabs:start -->
+
+```python
+
+```
+
+```java
+
+```
+
+```cpp
+
+```
+
+```go
+
+```
+
+<!-- tabs:end -->
+
+<!-- end -->

solution/README.md

@@ -3059,30 +3059,31 @@
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+|  3049  |  [标记所有下标的最早秒数 II](/solution/3000-3099/3049.Earliest%20Second%20to%20Mark%20Indices%20II/README.md)  |  `贪心`,`数组`,`二分查找`,`堆（优先队列）`  |  困难  |  第 386 场周赛  |
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+|  3051  |  [Find Candidates for Data Scientist Position](/solution/3000-3099/3051.Find%20Candidates%20for%20Data%20Scientist%20Position/README.md)  |  `数据库`  |  简单  |  🔒  |
+|  3052  |  [Maximize Items](/solution/3000-3099/3052.Maximize%20Items/README.md)  |  `数据库`  |  困难  |  🔒  |
+|  3053  |  [Classifying Triangles by Lengths](/solution/3000-3099/3053.Classifying%20Triangles%20by%20Lengths/README.md)  |  `数据库`  |  简单  |  🔒  |
+|  3054  |  [Binary Tree Nodes](/solution/3000-3099/3054.Binary%20Tree%20Nodes/README.md)  |  `数据库`  |  中等  |  🔒  |
+|  3055  |  [Top Percentile Fraud](/solution/3000-3099/3055.Top%20Percentile%20Fraud/README.md)  |  `数据库`  |  中等  |  🔒  |
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+|  3058  |  [Friends With No Mutual Friends](/solution/3000-3099/3058.Friends%20With%20No%20Mutual%20Friends/README.md)  |  `数据库`  |  中等  |  🔒  |
+|  3059  |  [Find All Unique Email Domains](/solution/3000-3099/3059.Find%20All%20Unique%20Email%20Domains/README.md)  |  `数据库`  |  简单  |  🔒  |
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+|  3066  |  [超过阈值的最少操作数 II](/solution/3000-3099/3066.Minimum%20Operations%20to%20Exceed%20Threshold%20Value%20II/README.md)  |  `数组`,`模拟`,`堆（优先队列）`  |  中等  |  第 125 场双周赛  |
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+|  3070  |  [元素和小于等于 k 的子矩阵的数目](/solution/3000-3099/3070.Count%20Submatrices%20with%20Top-Left%20Element%20and%20Sum%20Less%20Than%20k/README.md)  |  `数组`,`矩阵`,`前缀和`  |  中等  |  第 387 场周赛  |
+|  3071  |  [在矩阵上写出字母 Y 所需的最少操作次数](/solution/3000-3099/3071.Minimum%20Operations%20to%20Write%20the%20Letter%20Y%20on%20a%20Grid/README.md)  |  `数组`,`哈希表`,`计数`,`矩阵`  |  中等  |  第 387 场周赛  |
+|  3072  |  [将元素分配到两个数组中 II](/solution/3000-3099/3072.Distribute%20Elements%20Into%20Two%20Arrays%20II/README.md)  |  `树状数组`,`线段树`,`数组`,`模拟`  |  困难  |  第 387 场周赛  |
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 ## 版权