feat: add weekly contest 448 (#4390)

Author: yanglbme

solution/0500-0599/0599.Minimum Index Sum of Two Lists/README.md

@@ -18,9 +18,13 @@ tags:
 
 <!-- description:start -->
 
-<p>假设 Andy 和 Doris 想在晚餐时选择一家餐厅，并且他们都有一个表示最喜爱餐厅的列表，每个餐厅的名字用字符串表示。</p>
+<p>给定两个字符串数组&nbsp;<code>list1</code> 和 <code>list2</code>，找到 <strong>索引和最小的公共字符串</strong>。</p>
 
-<p>你需要帮助他们用<strong>最少的索引和</strong>找出他们<strong>共同喜爱的餐厅</strong>。 如果答案不止一个，则输出所有答案并且不考虑顺序。 你可以假设答案总是存在。</p>
+<p><strong>公共字符串</strong>&nbsp;是同时出现在&nbsp;<code>list1</code> 和 <code>list2</code>&nbsp;中的字符串。</p>
+
+<p>具有 <strong>最小索引和的公共字符串</strong> 是指，如果它在 <code>list1[i]</code> 和 <code>list2[j]</code> 中出现，那么 <code>i + j</code> 应该是所有其他 <strong>公共字符串</strong> 中的最小值。</p>
+
+<p>返回所有 <strong>具有最小索引和的公共字符串</strong>。以 <strong>任何顺序</strong> 返回答案。</p>
 
 <p>&nbsp;</p>
 
@@ -29,17 +33,28 @@ tags:
 <pre>
 <strong>输入: </strong>list1 = ["Shogun", "Tapioca Express", "Burger King", "KFC"]，list2 = ["Piatti", "The Grill at Torrey Pines", "Hungry Hunter Steakhouse", "Shogun"]
 <strong>输出:</strong> ["Shogun"]
-<strong>解释:</strong> 他们唯一共同喜爱的餐厅是“Shogun”。
+<strong>解释:</strong> 唯一的公共字符串是 “Shogun”。
 </pre>
 
 <p><strong>示例 2:</strong></p>
 
 <pre>
 <strong>输入:</strong>list1 = ["Shogun", "Tapioca Express", "Burger King", "KFC"]，list2 = ["KFC", "Shogun", "Burger King"]
 <strong>输出:</strong> ["Shogun"]
-<strong>解释:</strong> 他们共同喜爱且具有最小索引和的餐厅是“Shogun”，它有最小的索引和1(0+1)。
+<strong>解释:</strong> 具有最小索引和的公共字符串是 “Shogun”，它有最小的索引和 = (0 + 1) = 1。
 </pre>
 
+<p><strong>示例 3：</strong></p>
+
+<pre>
+<strong>输入：</strong>list1 = ["happy","sad","good"], list2 = ["sad","happy","good"]
+<b>输出：</b>["sad","happy"]
+<b>解释：</b>有三个公共字符串：
+"happy" 索引和 = (0 + 1) = 1.
+"sad" 索引和 = (1 + 0) = 1.
+"good" 索引和 = (2 + 2) = 4.
+最小索引和的字符串是 "sad" 和 "happy"。</pre>
+
 <p>&nbsp;</p>
 
 <p><strong>提示:</strong></p>

solution/0800-0899/0819.Most Common Word/README_EN.md

@@ -23,6 +23,8 @@ tags:
 
 <p>The words in <code>paragraph</code> are <strong>case-insensitive</strong> and the answer should be returned in <strong>lowercase</strong>.</p>
 
+<p><strong>Note</strong> that words can not contain punctuation symbols.</p>
+
 <p>&nbsp;</p>
 <p><strong class="example">Example 1:</strong></p>
 

solution/0800-0899/0838.Push Dominoes/README.md

@@ -29,9 +29,9 @@ tags:
 <p>给你一个字符串 <code>dominoes</code> 表示这一行多米诺骨牌的初始状态，其中：</p>
 
 <ul>
- <li><code>dominoes[i] = 'L'</code>，表示第 <code>i</code> 张多米诺骨牌被推向左侧，</li>
- <li><code>dominoes[i] = 'R'</code>，表示第 <code>i</code> 张多米诺骨牌被推向右侧，</li>
- <li><code>dominoes[i] = '.'</code>，表示没有推动第 <code>i</code> 张多米诺骨牌。</li>
+	<li><code>dominoes[i] = 'L'</code>，表示第 <code>i</code> 张多米诺骨牌被推向左侧，</li>
+	<li><code>dominoes[i] = 'R'</code>，表示第 <code>i</code> 张多米诺骨牌被推向右侧，</li>
+	<li><code>dominoes[i] = '.'</code>，表示没有推动第 <code>i</code> 张多米诺骨牌。</li>
 </ul>
 
 <p>返回表示最终状态的字符串。</p>
@@ -57,9 +57,9 @@ tags:
 <p><strong>提示：</strong></p>
 
 <ul>
- <li><code>n == dominoes.length</code></li>
- <li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
- <li><code>dominoes[i]</code> 为 <code>'L'</code>、<code>'R'</code> 或 <code>'.'</code></li>
+	<li><code>n == dominoes.length</code></li>
+	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
+	<li><code>dominoes[i]</code> 为 <code>'L'</code>、<code>'R'</code> 或 <code>'.'</code></li>
 </ul>
 
 <!-- description:end -->

solution/0800-0899/0838.Push Dominoes/README_EN.md

@@ -29,9 +29,9 @@ tags:
 <p>You are given a string <code>dominoes</code> representing the initial state where:</p>
 
 <ul>
- <li><code>dominoes[i] = &#39;L&#39;</code>, if the <code>i<sup>th</sup></code> domino has been pushed to the left,</li>
- <li><code>dominoes[i] = &#39;R&#39;</code>, if the <code>i<sup>th</sup></code> domino has been pushed to the right, and</li>
- <li><code>dominoes[i] = &#39;.&#39;</code>, if the <code>i<sup>th</sup></code> domino has not been pushed.</li>
+	<li><code>dominoes[i] = &#39;L&#39;</code>, if the <code>i<sup>th</sup></code> domino has been pushed to the left,</li>
+	<li><code>dominoes[i] = &#39;R&#39;</code>, if the <code>i<sup>th</sup></code> domino has been pushed to the right, and</li>
+	<li><code>dominoes[i] = &#39;.&#39;</code>, if the <code>i<sup>th</sup></code> domino has not been pushed.</li>
 </ul>
 
 <p>Return <em>a string representing the final state</em>.</p>
@@ -56,9 +56,9 @@ tags:
 <p><strong>Constraints:</strong></p>
 
 <ul>
- <li><code>n == dominoes.length</code></li>
- <li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
- <li><code>dominoes[i]</code> is either <code>&#39;L&#39;</code>, <code>&#39;R&#39;</code>, or <code>&#39;.&#39;</code>.</li>
+	<li><code>n == dominoes.length</code></li>
+	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
+	<li><code>dominoes[i]</code> is either <code>&#39;L&#39;</code>, <code>&#39;R&#39;</code>, or <code>&#39;.&#39;</code>.</li>
 </ul>
 
 <!-- description:end -->

solution/1100-1199/1160.Find Words That Can Be Formed by Characters/README_EN.md

@@ -23,7 +23,7 @@ tags:
 
 <p>You are given an array of strings <code>words</code> and a string <code>chars</code>.</p>
 
-<p>A string is <strong>good</strong> if it can be formed by characters from <code>chars</code> (each character can only be used once).</p>
+<p>A string is <strong>good</strong> if it can be formed by characters from <code>chars</code> (each character can only be used once for <strong>each</strong> word in <code>words</code>).</p>
 
 <p>Return <em>the sum of lengths of all good strings in words</em>.</p>
 

solution/3400-3499/3401.Find Circular Gift Exchange Chains/README.md

@@ -8,15 +8,15 @@ tags:
 
 <!-- problem:start -->
 
-# [3401. Find Circular Gift Exchange Chains 🔒](https://leetcode.cn/problems/find-circular-gift-exchange-chains)
+# [3401. 寻找环形礼物交换链 🔒](https://leetcode.cn/problems/find-circular-gift-exchange-chains)
 
 [English Version](/solution/3400-3499/3401.Find%20Circular%20Gift%20Exchange%20Chains/README_EN.md)
 
 ## 题目描述
 
 <!-- description:start -->
 
-<p>Table: <code>SecretSanta</code></p>
+<p>表：<code>SecretSanta</code></p>
 
 <pre>
 +-------------+------+
@@ -26,31 +26,32 @@ tags:
 | receiver_id | int  |
 | gift_value  | int  |
 +-------------+------+
-(giver_id, receiver_id) is the unique key for this table.   
-Each row represents a record of a gift exchange between two employees, giver_id represents the employee who gives a gift, receiver_id represents the employee who receives the gift and gift_value represents the value of the gift given.  
+(giver_id, receiver_id) 是这张表的唯一主键。
+每一行表示两个员工之间的一次礼物交换记录，giver_id 表示给予礼物的员工，receiver_id 表示收到礼物的员工，gift_value 表示所给予礼物的价值。
 </pre>
 
-<p>Write a solution to find the <strong>total gift value</strong> and <strong>length</strong> of<strong> circular chains</strong> of Secret Santa gift exchanges:</p>
+<p>编写一个解决方案来找到 <strong>总礼物价值</strong>&nbsp;以及 <strong>环形礼物交换链的长度</strong>：</p>
 
-<p>A <strong>circular chain</strong> is defined as a series of exchanges where:</p>
+<p><strong>环形链</strong> 被定义为一系列交换，其中：</p>
 
 <ul>
-	<li>Each employee gives a gift to <strong>exactly one</strong> other employee.</li>
-	<li>Each employee receives a gift <strong>from exactly</strong> one other employee.</li>
-	<li>The exchanges form a continuous <strong>loop</strong> (e.g., employee A gives a gift to B, B gives to C, and C gives back to A).</li>
+	<li>每位员工都正好向另 <strong>一位</strong> 员工赠送一份礼物。</li>
+	<li>每位员工都正好从另 <strong>一位</strong> 员工那里收到一份礼物。</li>
+	<li>交换形成一个连续的循环（即 员工 A 给 B 一份礼物，B 给 C，C 再给 A）。</li>
 </ul>
 
-<p>Return <em>the result ordered by the chain length and total gift value of the chain in&nbsp;<strong>descending</strong> order</em>.&nbsp;</p>
+<p>返回结果以链的长度和总礼物价值 <strong>降序</strong>&nbsp;排序。</p>
 
-<p>The result format is in the following example.</p>
+<p>结果格式如下所示。</p>
 
 <p>&nbsp;</p>
-<p><strong class="example">Example:</strong></p>
+
+<p><strong class="example">示例：</strong></p>
 
 <div class="example-block">
-<p><strong>Input:</strong></p>
+<p><strong>输入：</strong></p>
 
-<p>SecretSanta table:</p>
+<p>SecretSanta 表：</p>
 
 <pre class="example-io">
 +----------+-------------+------------+
@@ -64,7 +65,7 @@ Each row represents a record of a gift exchange between two employees, giver_id
 +----------+-------------+------------+
 </pre>
 
-<p><strong>Output:</strong></p>
+<p><strong>输出：</strong></p>
 
 <pre class="example-io">
 +----------+--------------+------------------+
@@ -75,26 +76,26 @@ Each row represents a record of a gift exchange between two employees, giver_id
 +----------+--------------+------------------+
 </pre>
 
-<p><strong>Explanation:</strong></p>
+<p><strong>解释：</strong></p>
 
 <ul>
-	<li><strong>Chain 1</strong> involves employees 1, 2, and 3:
+	<li><strong>链 1</strong>&nbsp;包含员工 1，2 和 3：
 
     <ul>
-    	<li>Employee 1 gives a gift to 2, employee 2 gives a gift to 3, and employee 3 gives a gift to 1.</li>
-    	<li>Total gift value for this chain = 20 + 30 + 40 = 90.</li>
+    	<li>员工 1 给 2 一份礼物，员工&nbsp;2 给 3 一份礼物，员工 3 给 1 一份礼物。</li>
+    	<li>这个链的总礼物价值 = 20 + 30 + 40 = 90。</li>
     </ul>
     </li>
-    <li><strong>Chain 2</strong> involves employees 4 and 5:
+    <li><strong>链 2</strong> 包含员工 4 和 5：
     <ul>
-    	<li>Employee 4 gives a gift to 5, and employee 5 gives a gift to 4.</li>
-    	<li>Total gift value for this chain = 25 + 35 = 60.</li>
+    	<li>员工 4 给 5 一份礼物，员工 5 给 4&nbsp;一份礼物。</li>
+    	<li>这个链的总礼物价值 = 25 + 35 = 60。</li>
     </ul>
     </li>
 
 </ul>
 
-<p>The result table is ordered by the chain length and total gift value of the chain in descending order.</p>
+<p>结果表以链的长度和总礼物价值降序排序。</p>
 </div>
 
 <!-- description:end -->

solution/3500-3599/3528.Unit Conversion I/README.md

@@ -34,8 +34,8 @@ tags:
 <p><strong>解释：</strong></p>
 
 <ul>
- <li>使用 <code>conversions[0]</code>：将一个 0 类型单位转换为 2 个 1 类型单位。</li>
- <li>使用&nbsp;<code>conversions[0]</code>&nbsp;和&nbsp;<code>conversions[1]</code>&nbsp;将一个 0 类型单位转换为 6 个 2 类型单位。</li>
+	<li>使用 <code>conversions[0]</code>：将一个 0 类型单位转换为 2 个 1 类型单位。</li>
+	<li>使用&nbsp;<code>conversions[0]</code>&nbsp;和&nbsp;<code>conversions[1]</code>&nbsp;将一个 0 类型单位转换为 6 个 2 类型单位。</li>
 </ul>
 <img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/3500-3599/3528.Unit%20Conversion%20I/images/1745660099-FZhVTM-example1.png" style="width: 545px; height: 119px;" /></div>
 
@@ -49,13 +49,13 @@ tags:
 <p><strong>解释：</strong></p>
 
 <ul>
- <li>使用 <code>conversions[0]</code>&nbsp;将一个 0 类型单位转换为 2 个 1 类型单位。</li>
- <li>使用 <code>conversions[1]</code>&nbsp;将一个 0 类型单位转换为 3 个 2 类型单位。</li>
- <li>使用 <code>conversions[0]</code> 和 <code>conversions[2]</code>&nbsp;将一个 0 类型单位转换为 8 个 3 类型单位。</li>
- <li>使用 <code>conversions[0]</code> 和 <code>conversions[3]</code>&nbsp;将一个 0 类型单位转换为 10 个 4 类型单位。</li>
- <li>使用 <code>conversions[1]</code> 和 <code>conversions[4]</code>&nbsp;将一个 0 类型单位转换为 6 个 5 类型单位。</li>
- <li>使用 <code>conversions[0]</code>、<code>conversions[3]</code> 和 <code>conversions[5]</code>&nbsp;将一个 0 类型单位转换为 30 个 6 类型单位。</li>
- <li>使用 <code>conversions[1]</code>、<code>conversions[4]</code> 和 <code>conversions[6]</code>&nbsp;将一个 0 类型单位转换为 24 个 7 类型单位。</li>
+	<li>使用 <code>conversions[0]</code>&nbsp;将一个 0 类型单位转换为 2 个 1 类型单位。</li>
+	<li>使用 <code>conversions[1]</code>&nbsp;将一个 0 类型单位转换为 3 个 2 类型单位。</li>
+	<li>使用 <code>conversions[0]</code> 和 <code>conversions[2]</code>&nbsp;将一个 0 类型单位转换为 8 个 3 类型单位。</li>
+	<li>使用 <code>conversions[0]</code> 和 <code>conversions[3]</code>&nbsp;将一个 0 类型单位转换为 10 个 4 类型单位。</li>
+	<li>使用 <code>conversions[1]</code> 和 <code>conversions[4]</code>&nbsp;将一个 0 类型单位转换为 6 个 5 类型单位。</li>
+	<li>使用 <code>conversions[0]</code>、<code>conversions[3]</code> 和 <code>conversions[5]</code>&nbsp;将一个 0 类型单位转换为 30 个 6 类型单位。</li>
+	<li>使用 <code>conversions[1]</code>、<code>conversions[4]</code> 和 <code>conversions[6]</code>&nbsp;将一个 0 类型单位转换为 24 个 7 类型单位。</li>
 </ul>
 </div>
 
@@ -64,11 +64,11 @@ tags:
 <p><strong>提示：</strong></p>
 
 <ul>
- <li><code>2 &lt;= n &lt;= 10<sup>5</sup></code></li>
- <li><code>conversions.length == n - 1</code></li>
- <li><code>0 &lt;= sourceUnit<sub>i</sub>, targetUnit<sub>i</sub> &lt; n</code></li>
- <li><code>1 &lt;= conversionFactor<sub>i</sub> &lt;= 10<sup>9</sup></code></li>
- <li>保证单位&nbsp;0 可以通过&nbsp;<strong>唯一&nbsp;</strong>的转换路径（不需要反向转换）转换为任何其他单位。</li>
+	<li><code>2 &lt;= n &lt;= 10<sup>5</sup></code></li>
+	<li><code>conversions.length == n - 1</code></li>
+	<li><code>0 &lt;= sourceUnit<sub>i</sub>, targetUnit<sub>i</sub> &lt; n</code></li>
+	<li><code>1 &lt;= conversionFactor<sub>i</sub> &lt;= 10<sup>9</sup></code></li>
+	<li>保证单位&nbsp;0 可以通过&nbsp;<strong>唯一&nbsp;</strong>的转换路径（不需要反向转换）转换为任何其他单位。</li>
 </ul>
 
 <!-- description:end -->

solution/3500-3599/3528.Unit Conversion I/README_EN.md

@@ -33,8 +33,8 @@ tags:
 <p><strong>Explanation:</strong></p>
 
 <ul>
- <li>Convert a single unit of type 0 into 2 units of type 1 using <code>conversions[0]</code>.</li>
- <li>Convert a single unit of type 0 into 6 units of type 2 using <code>conversions[0]</code>, then <code>conversions[1]</code>.</li>
+	<li>Convert a single unit of type 0 into 2 units of type 1 using <code>conversions[0]</code>.</li>
+	<li>Convert a single unit of type 0 into 6 units of type 2 using <code>conversions[0]</code>, then <code>conversions[1]</code>.</li>
 </ul>
 <img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/3500-3599/3528.Unit%20Conversion%20I/images/example1.png" style="width: 545px; height: 118px;" /></div>
 
@@ -48,25 +48,25 @@ tags:
 <p><strong>Explanation:</strong></p>
 
 <ul>
- <li>Convert a single unit of type 0 into 2 units of type 1 using <code>conversions[0]</code>.</li>
- <li>Convert a single unit of type 0 into 3 units of type 2 using <code>conversions[1]</code>.</li>
- <li>Convert a single unit of type 0 into 8 units of type 3 using <code>conversions[0]</code>, then <code>conversions[2]</code>.</li>
- <li>Convert a single unit of type 0 into 10 units of type 4 using <code>conversions[0]</code>, then <code>conversions[3]</code>.</li>
- <li>Convert a single unit of type 0 into 6 units of type 5 using <code>conversions[1]</code>, then <code>conversions[4]</code>.</li>
- <li>Convert a single unit of type 0 into 30 units of type 6 using <code>conversions[0]</code>, <code>conversions[3]</code>, then <code>conversions[5]</code>.</li>
- <li>Convert a single unit of type 0 into 24 units of type 7 using <code>conversions[1]</code>, <code>conversions[4]</code>, then <code>conversions[6]</code>.</li>
+	<li>Convert a single unit of type 0 into 2 units of type 1 using <code>conversions[0]</code>.</li>
+	<li>Convert a single unit of type 0 into 3 units of type 2 using <code>conversions[1]</code>.</li>
+	<li>Convert a single unit of type 0 into 8 units of type 3 using <code>conversions[0]</code>, then <code>conversions[2]</code>.</li>
+	<li>Convert a single unit of type 0 into 10 units of type 4 using <code>conversions[0]</code>, then <code>conversions[3]</code>.</li>
+	<li>Convert a single unit of type 0 into 6 units of type 5 using <code>conversions[1]</code>, then <code>conversions[4]</code>.</li>
+	<li>Convert a single unit of type 0 into 30 units of type 6 using <code>conversions[0]</code>, <code>conversions[3]</code>, then <code>conversions[5]</code>.</li>
+	<li>Convert a single unit of type 0 into 24 units of type 7 using <code>conversions[1]</code>, <code>conversions[4]</code>, then <code>conversions[6]</code>.</li>
 </ul>
 </div>
 
 <p>&nbsp;</p>
 <p><strong>Constraints:</strong></p>
 
 <ul>
- <li><code>2 &lt;= n &lt;= 10<sup>5</sup></code></li>
- <li><code>conversions.length == n - 1</code></li>
- <li><code>0 &lt;= sourceUnit<sub>i</sub>, targetUnit<sub>i</sub> &lt; n</code></li>
- <li><code>1 &lt;= conversionFactor<sub>i</sub> &lt;= 10<sup>9</sup></code></li>
- <li>It is guaranteed that unit 0 can be converted into any other unit through a <strong>unique</strong> combination of conversions without using any conversions in the opposite direction.</li>
+	<li><code>2 &lt;= n &lt;= 10<sup>5</sup></code></li>
+	<li><code>conversions.length == n - 1</code></li>
+	<li><code>0 &lt;= sourceUnit<sub>i</sub>, targetUnit<sub>i</sub> &lt; n</code></li>
+	<li><code>1 &lt;= conversionFactor<sub>i</sub> &lt;= 10<sup>9</sup></code></li>
+	<li>It is guaranteed that unit 0 can be converted into any other unit through a <strong>unique</strong> combination of conversions without using any conversions in the opposite direction.</li>
 </ul>
 
 <!-- description:end -->

solution/3500-3599/3532.Path Existence Queries in a Graph I/README.md

@@ -2,6 +2,11 @@
 comments: true
 difficulty: 中等
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3500-3599/3532.Path%20Existence%20Queries%20in%20a%20Graph%20I/README.md
+tags:
+    - 并查集
+    - 图
+    - 数组
+    - 二分查找
 ---
 
 <!-- problem:start -->

solution/3500-3599/3532.Path Existence Queries in a Graph I/README_EN.md

@@ -2,6 +2,11 @@
 comments: true
 difficulty: Medium
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3500-3599/3532.Path%20Existence%20Queries%20in%20a%20Graph%20I/README_EN.md
+tags:
+    - Union Find
+    - Graph
+    - Array
+    - Binary Search
 ---
 
 <!-- problem:start -->

solution/3500-3599/3533.Concatenated Divisibility/README.md

@@ -2,6 +2,11 @@
 comments: true
 difficulty: 困难
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3500-3599/3533.Concatenated%20Divisibility/README.md
+tags:
+    - 位运算
+    - 数组
+    - 动态规划
+    - 状态压缩
 ---
 
 <!-- problem:start -->

solution/3500-3599/3533.Concatenated Divisibility/README_EN.md

@@ -2,6 +2,11 @@
 comments: true
 difficulty: Hard
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3500-3599/3533.Concatenated%20Divisibility/README_EN.md
+tags:
+    - Bit Manipulation
+    - Array
+    - Dynamic Programming
+    - Bitmask
 ---
 
 <!-- problem:start -->

solution/3500-3599/3534.Path Existence Queries in a Graph II/README.md

@@ -2,6 +2,12 @@
 comments: true
 difficulty: 困难
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3500-3599/3534.Path%20Existence%20Queries%20in%20a%20Graph%20II/README.md
+tags:
+    - 贪心
+    - 图
+    - 数组
+    - 二分查找
+    - 排序
 ---
 
 <!-- problem:start -->

solution/3500-3599/3534.Path Existence Queries in a Graph II/README_EN.md

@@ -2,6 +2,12 @@
 comments: true
 difficulty: Hard
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3500-3599/3534.Path%20Existence%20Queries%20in%20a%20Graph%20II/README_EN.md
+tags:
+    - Greedy
+    - Graph
+    - Array
+    - Binary Search
+    - Sorting
 ---
 
 <!-- problem:start -->