feat: add weekly contest 440 (#4138)

Author: yanglbme

solution/2200-2299/2227.Encrypt and Decrypt Strings/README.md

@@ -61,11 +61,11 @@ tags:
 
 <strong>解释：</strong>
 Encrypter encrypter = new Encrypter([['a', 'b', 'c', 'd'], ["ei", "zf", "ei", "am"], ["abcd", "acbd", "adbc", "badc", "dacb", "cadb", "cbda", "abad"]);
-encrypter.encrypt("abcd"); // 返回 "eizfeiam"。
+encrypter.encrypt("abcd"); // 返回 "eizfeiam"。 
 &nbsp;                          // 'a' 映射为 "ei"，'b' 映射为 "zf"，'c' 映射为 "ei"，'d' 映射为 "am"。
-encrypter.decrypt("eizfeiam"); // return 2.
-                              // "ei" 可以映射为 'a' 或 'c'，"zf" 映射为 'b'，"am" 映射为 'd'。
-                              // 因此，解密后可以得到的字符串是 "abad"，"cbad"，"abcd" 和 "cbcd"。
+encrypter.decrypt("eizfeiam"); // return 2. 
+                              // "ei" 可以映射为 'a' 或 'c'，"zf" 映射为 'b'，"am" 映射为 'd'。 
+                              // 因此，解密后可以得到的字符串是 "abad"，"cbad"，"abcd" 和 "cbcd"。 
                               // 其中 2 个字符串，"abad" 和 "abcd"，在 dictionary 中出现，所以答案是 2 。
 </pre>
 

solution/2200-2299/2227.Encrypt and Decrypt Strings/README_EN.md

@@ -60,11 +60,11 @@ tags:
 
 <strong>Explanation</strong>
 Encrypter encrypter = new Encrypter([[&#39;a&#39;, &#39;b&#39;, &#39;c&#39;, &#39;d&#39;], [&quot;ei&quot;, &quot;zf&quot;, &quot;ei&quot;, &quot;am&quot;], [&quot;abcd&quot;, &quot;acbd&quot;, &quot;adbc&quot;, &quot;badc&quot;, &quot;dacb&quot;, &quot;cadb&quot;, &quot;cbda&quot;, &quot;abad&quot;]);
-encrypter.encrypt(&quot;abcd&quot;); // return &quot;eizfeiam&quot;.
+encrypter.encrypt(&quot;abcd&quot;); // return &quot;eizfeiam&quot;. 
 &nbsp;                          // &#39;a&#39; maps to &quot;ei&quot;, &#39;b&#39; maps to &quot;zf&quot;, &#39;c&#39; maps to &quot;ei&quot;, and &#39;d&#39; maps to &quot;am&quot;.
-encrypter.decrypt(&quot;eizfeiam&quot;); // return 2.
-                              // &quot;ei&quot; can map to &#39;a&#39; or &#39;c&#39;, &quot;zf&quot; maps to &#39;b&#39;, and &quot;am&quot; maps to &#39;d&#39;.
-                              // Thus, the possible strings after decryption are &quot;abad&quot;, &quot;cbad&quot;, &quot;abcd&quot;, and &quot;cbcd&quot;.
+encrypter.decrypt(&quot;eizfeiam&quot;); // return 2. 
+                              // &quot;ei&quot; can map to &#39;a&#39; or &#39;c&#39;, &quot;zf&quot; maps to &#39;b&#39;, and &quot;am&quot; maps to &#39;d&#39;. 
+                              // Thus, the possible strings after decryption are &quot;abad&quot;, &quot;cbad&quot;, &quot;abcd&quot;, and &quot;cbcd&quot;. 
                               // 2 of those strings, &quot;abad&quot; and &quot;abcd&quot;, appear in dictionary, so the answer is 2.
 </pre>
 

solution/3400-3499/3477.Fruits Into Baskets II/README.md

@@ -0,0 +1,115 @@
+---
+comments: true
+difficulty: 简单
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3400-3499/3477.Fruits%20Into%20Baskets%20II/README.md
+---
+
+<!-- problem:start -->
+
+# [3477. 将水果放入篮子 II](https://leetcode.cn/problems/fruits-into-baskets-ii)
+
+[English Version](/solution/3400-3499/3477.Fruits%20Into%20Baskets%20II/README_EN.md)
+
+## 题目描述
+
+<!-- description:start -->
+
+<p>给你两个长度为 <code>n</code>&nbsp;的整数数组，<code>fruits</code> 和 <code>baskets</code>，其中 <code>fruits[i]</code> 表示第 <code>i</code>&nbsp;种水果的 <strong>数量</strong>，<code>baskets[j]</code> 表示第 <code>j</code>&nbsp;个篮子的 <strong>容量</strong>。</p>
+
+<p>你需要对 <code>fruits</code> 数组从左到右按照以下规则放置水果：</p>
+
+<ul>
+	<li>每种水果必须放入第一个 <strong>容量大于等于</strong> 该水果数量的 <strong>最左侧可用篮子</strong> 中。</li>
+	<li>每个篮子只能装 <b>一种</b> 水果。</li>
+	<li>如果一种水果 <b>无法放入</b> 任何篮子，它将保持 <b>未放置</b>。</li>
+</ul>
+
+<p>返回所有可能分配完成后，剩余未放置的水果种类的数量。</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">fruits = [4,2,5], baskets = [3,5,4]</span></p>
+
+<p><strong>输出：</strong> <span class="example-io">1</span></p>
+
+<p><strong>解释：</strong></p>
+
+<ul>
+	<li><code>fruits[0] = 4</code> 放入 <code>baskets[1] = 5</code>。</li>
+	<li><code>fruits[1] = 2</code> 放入 <code>baskets[0] = 3</code>。</li>
+	<li><code>fruits[2] = 5</code> 无法放入 <code>baskets[2] = 4</code>。</li>
+</ul>
+
+<p>由于有一种水果未放置，我们返回 1。</p>
+</div>
+
+<p><strong class="example">示例 2</strong></p>
+
+<div class="example-block">
+<p><strong>输入：</strong> <span class="example-io">fruits = [3,6,1], baskets = [6,4,7]</span></p>
+
+<p><strong>输出：</strong> <span class="example-io">0</span></p>
+
+<p><strong>解释：</strong></p>
+
+<ul>
+	<li><code>fruits[0] = 3</code> 放入 <code>baskets[0] = 6</code>。</li>
+	<li><code>fruits[1] = 6</code> 无法放入 <code>baskets[1] = 4</code>（容量不足），但可以放入下一个可用的篮子 <code>baskets[2] = 7</code>。</li>
+	<li><code>fruits[2] = 1</code> 放入 <code>baskets[1] = 4</code>。</li>
+</ul>
+
+<p>由于所有水果都已成功放置，我们返回 0。</p>
+</div>
+
+<p>&nbsp;</p>
+
+<p><b>提示：</b></p>
+
+<ul>
+	<li><code>n == fruits.length == baskets.length</code></li>
+	<li><code>1 &lt;= n &lt;= 100</code></li>
+	<li><code>1 &lt;= fruits[i], baskets[i] &lt;= 1000</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/3400-3499/3477.Fruits Into Baskets II/README_EN.md

@@ -0,0 +1,113 @@
+---
+comments: true
+difficulty: Easy
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3400-3499/3477.Fruits%20Into%20Baskets%20II/README_EN.md
+---
+
+<!-- problem:start -->
+
+# [3477. Fruits Into Baskets II](https://leetcode.com/problems/fruits-into-baskets-ii)
+
+[中文文档](/solution/3400-3499/3477.Fruits%20Into%20Baskets%20II/README.md)
+
+## Description
+
+<!-- description:start -->
+
+<p>You are given two arrays of integers, <code>fruits</code> and <code>baskets</code>, each of length <code>n</code>, where <code>fruits[i]</code> represents the <strong>quantity</strong> of the <code>i<sup>th</sup></code> type of fruit, and <code>baskets[j]</code> represents the <strong>capacity</strong> of the <code>j<sup>th</sup></code> basket.</p>
+
+<p>From left to right, place the fruits according to these rules:</p>
+
+<ul>
+	<li>Each fruit type must be placed in the <strong>leftmost available basket</strong> with a capacity <strong>greater than or equal</strong> to the quantity of that fruit type.</li>
+	<li>Each basket can hold <b>only one</b> type of fruit.</li>
+	<li>If a fruit type <b>cannot be placed</b> in any basket, it remains <b>unplaced</b>.</li>
+</ul>
+
+<p>Return the number of fruit types that remain unplaced after all possible allocations are made.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">fruits = [4,2,5], baskets = [3,5,4]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">1</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<ul>
+	<li><code>fruits[0] = 4</code> is placed in <code>baskets[1] = 5</code>.</li>
+	<li><code>fruits[1] = 2</code> is placed in <code>baskets[0] = 3</code>.</li>
+	<li><code>fruits[2] = 5</code> cannot be placed in <code>baskets[2] = 4</code>.</li>
+</ul>
+
+<p>Since one fruit type remains unplaced, we return 1.</p>
+</div>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">fruits = [3,6,1], baskets = [6,4,7]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">0</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<ul>
+	<li><code>fruits[0] = 3</code> is placed in <code>baskets[0] = 6</code>.</li>
+	<li><code>fruits[1] = 6</code> cannot be placed in <code>baskets[1] = 4</code> (insufficient capacity) but can be placed in the next available basket, <code>baskets[2] = 7</code>.</li>
+	<li><code>fruits[2] = 1</code> is placed in <code>baskets[1] = 4</code>.</li>
+</ul>
+
+<p>Since all fruits are successfully placed, we return 0.</p>
+</div>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>n == fruits.length == baskets.length</code></li>
+	<li><code>1 &lt;= n &lt;= 100</code></li>
+	<li><code>1 &lt;= fruits[i], baskets[i] &lt;= 1000</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/3400-3499/3478.Choose K Elements With Maximum Sum/README.md

@@ -0,0 +1,107 @@
+---
+comments: true
+difficulty: 中等
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3400-3499/3478.Choose%20K%20Elements%20With%20Maximum%20Sum/README.md
+---
+
+<!-- problem:start -->
+
+# [3478. 选出和最大的 K 个元素](https://leetcode.cn/problems/choose-k-elements-with-maximum-sum)
+
+[English Version](/solution/3400-3499/3478.Choose%20K%20Elements%20With%20Maximum%20Sum/README_EN.md)
+
+## 题目描述
+
+<!-- description:start -->
+
+<p>给你两个整数数组，<code>nums1</code> 和 <code>nums2</code>，长度均为 <code>n</code>，以及一个正整数 <code>k</code> 。</p>
+
+<p>对从 <code>0</code> 到 <code>n - 1</code> 每个下标 <code>i</code> ，执行下述操作：</p>
+
+<ul>
+	<li>找出所有满足 <code>nums1[j]</code> 小于 <code>nums1[i]</code> 的下标 <code>j</code> 。</li>
+	<li>从这些下标对应的 <code>nums2[j]</code> 中选出 <strong>至多</strong> <code>k</code> 个，并 <strong>最大化</strong> 这些值的总和作为结果。</li>
+</ul>
+
+<p>返回一个长度为 <code>n</code> 的数组 <code>answer</code> ，其中 <code>answer[i]</code> 表示对应下标 <code>i</code> 的结果。</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">nums1 = [4,2,1,5,3], nums2 = [10,20,30,40,50], k = 2</span></p>
+
+<p><strong>输出：</strong><span class="example-io">[80,30,0,80,50]</span></p>
+
+<p><strong>解释：</strong></p>
+
+<ul>
+	<li>对于 <code>i = 0</code> ：满足 <code>nums1[j] &lt; nums1[0]</code> 的下标为 <code>[1, 2, 4]</code> ，选出其中值最大的两个，结果为 <code>50 + 30 = 80</code> 。</li>
+	<li>对于 <code>i = 1</code> ：满足 <code>nums1[j] &lt; nums1[1]</code> 的下标为 <code>[2]</code> ，只能选择这个值，结果为 <code>30</code> 。</li>
+	<li>对于 <code>i = 2</code> ：不存在满足 <code>nums1[j] &lt; nums1[2]</code> 的下标，结果为 <code>0</code> 。</li>
+	<li>对于 <code>i = 3</code> ：满足 <code>nums1[j] &lt; nums1[3]</code> 的下标为 <code>[0, 1, 2, 4]</code> ，选出其中值最大的两个，结果为 <code>50 + 30 = 80</code> 。</li>
+	<li>对于 <code>i = 4</code> ：满足 <code>nums1[j] &lt; nums1[4]</code> 的下标为 <code>[1, 2]</code> ，选出其中值最大的两个，结果为 <code>30 + 20 = 50</code> 。</li>
+</ul>
+</div>
+
+<p><strong class="example">示例 2：</strong></p>
+
+<div class="example-block">
+<p><strong>输入：</strong><span class="example-io">nums1 = [2,2,2,2], nums2 = [3,1,2,3], k = 1</span></p>
+
+<p><strong>输出：</strong><span class="example-io">[0,0,0,0]</span></p>
+
+<p><strong>解释：</strong>由于 <code>nums1</code> 中的所有元素相等，不存在满足条件 <code>nums1[j] &lt; nums1[i]</code>，所有位置的结果都是 0 。</p>
+</div>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>n == nums1.length == nums2.length</code></li>
+	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
+	<li><code>1 &lt;= nums1[i], nums2[i] &lt;= 10<sup>6</sup></code></li>
+	<li><code>1 &lt;= k &lt;= n</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 -->