feat: add new lc problem

solution/3500-3599/3595.Once Twice/README.md

@@ -0,0 +1,110 @@
+---
+comments: true
+difficulty: 中等
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3500-3599/3595.Once%20Twice/README.md
+---
+
+<!-- problem:start -->
+
+# [3595. 一次或两次 🔒](https://leetcode.cn/problems/once-twice)
+
+[English Version](/solution/3500-3599/3595.Once%20Twice/README_EN.md)
+
+## 题目描述
+
+<!-- description:start -->
+
+<p>给定一个整数数组&nbsp;<code>nums</code>。在这个数组中：</p>
+
+<ul>
+	<li>
+	<p>有一个元素出现了 <strong>恰好 1&nbsp;</strong><strong>次</strong>。</p>
+	</li>
+	<li>
+	<p>有一个元素出现了 <strong>恰好 2&nbsp;</strong><strong>次</strong>。</p>
+	</li>
+	<li>
+	<p>其它所有元素都出现了 <strong>恰好 3 次</strong>。</p>
+	</li>
+</ul>
+
+<p>返回一个长度为 2 的整数数组，其中第一个元素是只出现 <strong>1 次</strong>&nbsp;的那个元素，第二个元素是只出现 <strong>2 次</strong>&nbsp;的那个元素。</p>
+
+<p>你的解决方案必须在 <strong>O(n) 时间</strong>&nbsp;与 <strong>O(1)</strong>&nbsp;空间中运行。</p>
+
+<p>&nbsp;</p>
+
+<p><strong class="example">示例 1：</strong></p>
+
+<div class="example-block">
+<p><strong>输入：</strong><span class="example-io">nums = [2,2,3,2,5,5,5,7,7]</span></p>
+
+<p><span class="example-io"><b>输出：</b>[3,7]</span></p>
+
+<p><strong>解释：</strong></p>
+
+<p>元素 3&nbsp;出现了 <strong>1 次</strong>，元素 7&nbsp;出现了 <strong>2 次</strong>。其余所有元素都出现了 <strong>3 次</strong>。</p>
+</div>
+
+<p><strong class="example">示例 2：</strong></p>
+
+<div class="example-block">
+<p><span class="example-io"><b>输入：</b>nums = [4,4,6,4,9,9,9,6,8]</span></p>
+
+<p><span class="example-io"><b>输出：</b>[8,6]</span></p>
+
+<p><strong>解释：</strong></p>
+
+<p>元素 8 出现了 <strong>1 次</strong>，元素 6 出现了 <strong>2 次</strong>。其余所有元素都出现了 <strong>3 次</strong>。</p>
+</div>
+
+<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>-2<sup>31</sup> &lt;= nums[i] &lt;= 2<sup>31</sup> - 1</code></li>
+	<li><code>nums.length</code>&nbsp;是 3 的倍数。</li>
+	<li>恰好有一个元素出现 1 次，一个元素出现 2 次，其余所有元素都出现了 3 次。</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/3500-3599/3595.Once Twice/README_EN.md

@@ -0,0 +1,108 @@
+---
+comments: true
+difficulty: Medium
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3500-3599/3595.Once%20Twice/README_EN.md
+---
+
+<!-- problem:start -->
+
+# [3595. Once Twice 🔒](https://leetcode.com/problems/once-twice)
+
+[中文文档](/solution/3500-3599/3595.Once%20Twice/README.md)
+
+## Description
+
+<!-- description:start -->
+
+<p>You are given an integer array <code>nums</code>. In this array:</p>
+
+<ul>
+	<li>
+	<p>Exactly one element appears <strong>once</strong>.</p>
+	</li>
+	<li>
+	<p>Exactly one element appears <strong>twice</strong>.</p>
+	</li>
+	<li>
+	<p>All other elements appear <strong>exactly three times</strong>.</p>
+	</li>
+</ul>
+
+<p>Return an integer array of length 2, where the first element is the one that appears <strong>once</strong>, and the second is the one that appears <strong>twice</strong>.</p>
+
+<p>Your solution must run in <strong>O(n)</strong> time and <strong>O(1)</strong> space.</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 = [2,2,3,2,5,5,5,7,7]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">[3,7]</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>The element 3 appears <b>once</b>, and the element 7 appears <b>twice</b>. The remaining elements each appear <b>three times</b>.</p>
+</div>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">nums = [4,4,6,4,9,9,9,6,8]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">[8,6]</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>The element 8 appears <b>once</b>, and the element 6 appears <b>twice</b>. The remaining elements each appear <b>three times</b>.</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>-2<sup>31</sup> &lt;= nums[i] &lt;= 2<sup>31</sup> - 1</code></li>
+	<li><code>nums.length</code> is a multiple of 3.</li>
+	<li>Exactly one element appears once, one element appears twice, and all other elements appear three times.</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/README.md

@@ -3605,6 +3605,7 @@
 |  3592  |  [硬币面值还原](/solution/3500-3599/3592.Inverse%20Coin%20Change/README.md)  |    |  中等  |  第 455 场周赛  |
 |  3593  |  [使叶子路径成本相等的最小增量](/solution/3500-3599/3593.Minimum%20Increments%20to%20Equalize%20Leaf%20Paths/README.md)  |    |  中等  |  第 455 场周赛  |
 |  3594  |  [所有人渡河所需的最短时间](/solution/3500-3599/3594.Minimum%20Time%20to%20Transport%20All%20Individuals/README.md)  |    |  困难  |  第 455 场周赛  |
+|  3595  |  [一次或两次](/solution/3500-3599/3595.Once%20Twice/README.md)  |    |  中等  |  🔒  |
 
 ## 版权
 

solution/README_EN.md

@@ -3603,6 +3603,7 @@ Press <kbd>Control</kbd> + <kbd>F</kbd>(or <kbd>Command</kbd> + <kbd>F</kbd> on
 |  3592  |  [Inverse Coin Change](/solution/3500-3599/3592.Inverse%20Coin%20Change/README_EN.md)  |    |  Medium  |  Weekly Contest 455  |
 |  3593  |  [Minimum Increments to Equalize Leaf Paths](/solution/3500-3599/3593.Minimum%20Increments%20to%20Equalize%20Leaf%20Paths/README_EN.md)  |    |  Medium  |  Weekly Contest 455  |
 |  3594  |  [Minimum Time to Transport All Individuals](/solution/3500-3599/3594.Minimum%20Time%20to%20Transport%20All%20Individuals/README_EN.md)  |    |  Hard  |  Weekly Contest 455  |
+|  3595  |  [Once Twice](/solution/3500-3599/3595.Once%20Twice/README_EN.md)  |    |  Medium  |  🔒  |
 
 ## Copyright