feat: add biweekly contest 132 (doocs#3080)

Author: yanglbme

solution/3100-3199/3171.Find Subarray With Bitwise OR Closest to K/README.md

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+---
+comments: true
+difficulty: 困难
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3100-3199/3171.Find%20Subarray%20With%20Bitwise%20OR%20Closest%20to%20K/README.md
+tags:
+    - 位运算
+    - 线段树
+    - 数组
+    - 二分查找
+---
+
+<!-- problem:start -->
+
+# [3171. 找到按位与最接近 K 的子数组](https://leetcode.cn/problems/find-subarray-with-bitwise-or-closest-to-k)
+
+[English Version](/solution/3100-3199/3171.Find%20Subarray%20With%20Bitwise%20OR%20Closest%20to%20K/README_EN.md)
+
+## 题目描述
+
+<!-- description:start -->
+
+<p>给你一个数组&nbsp;<code>nums</code>&nbsp;和一个整数&nbsp;<code>k</code>&nbsp;。你需要找到&nbsp;<code>nums</code>&nbsp;的一个&nbsp;<span data-keyword="subarray-nonempty">子数组</span>&nbsp;，满足子数组中所有元素按位与运算&nbsp;<code>AND</code>&nbsp;的值与 <code>k</code>&nbsp;的 <strong>绝对差</strong>&nbsp;尽可能 <strong>小</strong>&nbsp;。换言之，你需要选择一个子数组&nbsp;<code>nums[l..r]</code>&nbsp;满足&nbsp;<code>|k - (nums[l] AND nums[l + 1] ... AND nums[r])|</code>&nbsp;最小。</p>
+
+<p>请你返回 <strong>最小</strong>&nbsp;的绝对差值。</p>
+
+<p><strong>子数组</strong>是数组中连续的&nbsp;<strong>非空</strong>&nbsp;元素序列。</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 = [1,2,4,5], k = 3</span></p>
+
+<p><span class="example-io"><b>输出：</b>1</span></p>
+
+<p><strong>解释：</strong></p>
+
+<p>子数组&nbsp;<code>nums[2..3]</code> 的按位&nbsp;<code>AND</code>&nbsp;运算值为 4 ，得到最小差值&nbsp;<code>|3 - 4| = 1</code>&nbsp;。</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 = [1,2,1,2], k = 2</span></p>
+
+<p><span class="example-io"><b>输出：</b>0</span></p>
+
+<p><strong>解释：</strong></p>
+
+<p>子数组&nbsp;<code>nums[1..1]</code> 的按位&nbsp;<code>AND</code>&nbsp;运算值为 2 ，得到最小差值&nbsp;<code>|2 - 2| = 0</code>&nbsp;。</p>
+</div>
+
+<p><strong class="example">示例 3：</strong></p>
+
+<div class="example-block">
+<p><span class="example-io"><b>输入：</b>nums = [1], k = 10</span></p>
+
+<p><span class="example-io"><b>输出：</b>9</span></p>
+
+<p><strong>解释：</strong></p>
+
+<p>只有一个子数组，按位&nbsp;<code>AND</code>&nbsp;运算值为 1 ，得到最小差值&nbsp;<code>|10 - 1| = 9</code>&nbsp;。</p>
+</div>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>1 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
+	<li><code>1 &lt;= k &lt;= 10<sup>9</sup></code></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/3100-3199/3171.Find Subarray With Bitwise OR Closest to K/README_EN.md

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+---
+comments: true
+difficulty: Hard
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3100-3199/3171.Find%20Subarray%20With%20Bitwise%20OR%20Closest%20to%20K/README_EN.md
+tags:
+    - Bit Manipulation
+    - Segment Tree
+    - Array
+    - Binary Search
+---
+
+<!-- problem:start -->
+
+# [3171. Find Subarray With Bitwise OR Closest to K](https://leetcode.com/problems/find-subarray-with-bitwise-or-closest-to-k)
+
+[中文文档](/solution/3100-3199/3171.Find%20Subarray%20With%20Bitwise%20OR%20Closest%20to%20K/README.md)
+
+## Description
+
+<!-- description:start -->
+
+<p>You are given an array <code>nums</code> and an integer <code>k</code>. You need to find a <span data-keyword="subarray-nonempty">subarray</span> of <code>nums</code> such that the <strong>absolute difference</strong> between <code>k</code> and the bitwise <code>OR</code> of the subarray elements is as<strong> small</strong> as possible. In other words, select a subarray <code>nums[l..r]</code> such that <code>|k - (nums[l] OR nums[l + 1] ... OR nums[r])|</code> is minimum.</p>
+
+<p>Return the <strong>minimum</strong> possible value of the absolute difference.</p>
+
+<p>A <strong>subarray</strong> is a contiguous <b>non-empty</b> sequence of elements within an array.</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 = [1,2,4,5], k = 3</span></p>
+
+<p><strong>Output:</strong> 0</p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>The subarray <code>nums[0..1]</code> has <code>OR</code> value 3, which gives the minimum absolute difference <code>|3 - 3| = 0</code>.</p>
+</div>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">nums = [1,3,1,3], k = 2</span></p>
+
+<p><strong>Output:</strong> 1</p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>The subarray <code>nums[1..1]</code> has <code>OR</code> value 3, which gives the minimum absolute difference <code>|3 - 2| = 1</code>.</p>
+</div>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">nums = [1], k = 10</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">9</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>There is a single subarray with <code>OR</code> value 1, which gives the minimum absolute difference <code>|10 - 1| = 9</code>.</p>
+</div>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>1 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
+	<li><code>1 &lt;= k &lt;= 10<sup>9</sup></code></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/3100-3199/3174.Clear Digits/README.md

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+---
+comments: true
+difficulty: 简单
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3100-3199/3174.Clear%20Digits/README.md
+---
+
+<!-- problem:start -->
+
+# [3174. 清除数字](https://leetcode.cn/problems/clear-digits)
+
+[English Version](/solution/3100-3199/3174.Clear%20Digits/README_EN.md)
+
+## 题目描述
+
+<!-- description:start -->
+
+<p>给你一个字符串&nbsp;<code>s</code>&nbsp;。</p>
+
+<p>你的任务是重复以下操作删除 <strong>所有</strong>&nbsp;数字字符：</p>
+
+<ul>
+	<li>删除 <strong>第一个数字字符</strong>&nbsp;以及它左边 <strong>最近</strong>&nbsp;的 <strong>非数字</strong>&nbsp;字符。</li>
+</ul>
+
+<p>请你返回删除所有数字字符以后剩下的字符串。</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>s = "abc"</span></p>
+
+<p><span class="example-io"><b>输出：</b>"abc"</span></p>
+
+<p><strong>解释：</strong></p>
+
+<p>字符串中没有数字。<!-- notionvc: ff07e34f-b1d6-41fb-9f83-5d0ba3c1ecde --></p>
+</div>
+
+<p><strong class="example">示例 2：</strong></p>
+
+<div class="example-block">
+<p><span class="example-io"><b>输入：</b>s = "cb34"</span></p>
+
+<p><span class="example-io"><b>输出：</b>""</span></p>
+
+<p><b>解释：</b></p>
+
+<p>一开始，我们对&nbsp;<code>s[2]</code>&nbsp;执行操作，<code>s</code>&nbsp;变为&nbsp;<code>"c4"</code>&nbsp;。</p>
+
+<p>然后对&nbsp;<code>s[1]</code>&nbsp;执行操作，<code>s</code>&nbsp;变为&nbsp;<code>""</code>&nbsp;。</p>
+</div>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= s.length &lt;= 100</code></li>
+	<li><code>s</code>&nbsp;只包含小写英文字母和数字字符。</li>
+	<li>输入保证所有数字都可以按以上操作被删除。</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 -->