feat: update lc problems

Author: yanglbme

solution/0000-0099/0020.Valid Parentheses/README_EN.md

@@ -11,6 +11,7 @@
 <ol>
 	<li>Open brackets must be closed by the same type of brackets.</li>
 	<li>Open brackets must be closed in the correct order.</li>
+	<li>Every close bracket has a corresponding open bracket of the same type.</li>
 </ol>
 
 <p>&nbsp;</p>

solution/0000-0099/0031.Next Permutation/README_EN.md

@@ -7,7 +7,7 @@
 <p>A <strong>permutation</strong> of an array of integers is an arrangement of its members into a sequence or linear order.</p>
 
 <ul>
-	<li>For example, for <code>arr = [1,2,3]</code>, the following are considered permutations of <code>arr</code>: <code>[1,2,3]</code>, <code>[1,3,2]</code>, <code>[3,1,2]</code>, <code>[2,3,1]</code>.</li>
+	<li>For example, for <code>arr = [1,2,3]</code>, the following are all the permutations of <code>arr</code>: <code>[1,2,3], [1,3,2], [2, 1, 3], [2, 3, 1], [3,1,2], [3,2,1]</code>.</li>
 </ul>
 
 <p>The <strong>next permutation</strong> of an array of integers is the next lexicographically greater permutation of its integer. More formally, if all the permutations of the array are sorted in one container according to their lexicographical order, then the <strong>next permutation</strong> of that array is the permutation that follows it in the sorted container. If such arrangement is not possible, the array must be rearranged as the lowest possible order (i.e., sorted in ascending order).</p>

solution/0200-0299/0234.Palindrome Linked List/README_EN.md

@@ -4,7 +4,7 @@
 
 ## Description
 
-<p>Given the <code>head</code> of a singly linked list, return <code>true</code> if it is a palindrome.</p>
+<p>Given the <code>head</code> of a singly linked list, return <code>true</code><em> if it is a palindrome or </em><code>false</code><em> otherwise</em>.</p>
 
 <p>&nbsp;</p>
 <p><strong>Example 1:</strong></p>

solution/0300-0399/0326.Power of Three/README_EN.md

@@ -14,20 +14,23 @@
 <pre>
 <strong>Input:</strong> n = 27
 <strong>Output:</strong> true
+<strong>Explanation:</strong> 27 = 3<sup>3</sup>
 </pre>
 
 <p><strong>Example 2:</strong></p>
 
 <pre>
 <strong>Input:</strong> n = 0
 <strong>Output:</strong> false
+<strong>Explanation:</strong> There is no x where 3<sup>x</sup> = 0.
 </pre>
 
 <p><strong>Example 3:</strong></p>
 
 <pre>
-<strong>Input:</strong> n = 9
-<strong>Output:</strong> true
+<strong>Input:</strong> n = -1
+<strong>Output:</strong> false
+<strong>Explanation:</strong> There is no x where 3<sup>x</sup> = (-1).
 </pre>
 
 <p>&nbsp;</p>

solution/0400-0499/0493.Reverse Pairs/README_EN.md

@@ -10,12 +10,26 @@
 
 <p>&nbsp;</p>
 <p><strong>Example 1:</strong></p>
-<pre><strong>Input:</strong> nums = [1,3,2,3,1]
+
+<pre>
+<strong>Input:</strong> nums = [1,3,2,3,1]
 <strong>Output:</strong> 2
-</pre><p><strong>Example 2:</strong></p>
-<pre><strong>Input:</strong> nums = [2,4,3,5,1]
+<strong>Explanation:</strong> The reverse pairs are:
+(1, 4) --&gt; nums[1] = 3, nums[4] = 1, 3 &gt; 2 * 1
+(3, 4) --&gt; nums[3] = 3, nums[4] = 1, 3 &gt; 2 * 1
+</pre>
+
+<p><strong>Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [2,4,3,5,1]
 <strong>Output:</strong> 3
+<strong>Explanation:</strong> The reverse pairs are:
+(1, 4) --&gt; nums[1] = 4, nums[4] = 1, 4 &gt; 2 * 1
+(2, 4) --&gt; nums[2] = 3, nums[4] = 1, 3 &gt; 2 * 1
+(3, 4) --&gt; nums[3] = 3, nums[4] = 1, 5 &gt; 2 * 1
 </pre>
+
 <p>&nbsp;</p>
 <p><strong>Constraints:</strong></p>
 

solution/0500-0599/0569.Median Employee Salary/README_EN.md

@@ -20,7 +20,7 @@ Each row of this table indicates the company and the salary of one employee.
 
 <p>&nbsp;</p>
 
-<p>Write an SQL query to find the median salary of each company.</p>
+<p>Write an SQL query to find the rows that contain the median salary of each company. While calculating the median, when you sort the salaries of the company, break the ties by <code>id</code>.</p>
 
 <p>Return the result table in <strong>any order</strong>.</p>
 
@@ -63,6 +63,39 @@ Employee table:
 | 9  | B       | 1154   |
 | 14 | C       | 2645   |
 +----+---------+--------+
+<strong>Explanation:</strong> 
+For company A, the rows sorted are as follows:
++----+---------+--------+
+| id | company | salary |
++----+---------+--------+
+| 3  | A       | 15     |
+| 2  | A       | 341    |
+| 5  | A       | 451    | &lt;-- median
+| 6  | A       | 513    | &lt;-- median
+| 1  | A       | 2341   |
+| 4  | A       | 15314  |
++----+---------+--------+
+For company B, the rows sorted are as follows:
++----+---------+--------+
+| id | company | salary |
++----+---------+--------+
+| 8  | B       | 13     |
+| 7  | B       | 15     |
+| 12 | B       | 234    | &lt;-- median
+| 11 | B       | 1221   | &lt;-- median
+| 9  | B       | 1154   |
+| 10 | B       | 1345   |
++----+---------+--------+
+For company C, the rows sorted are as follows:
++----+---------+--------+
+| id | company | salary |
++----+---------+--------+
+| 17 | C       | 65     |
+| 13 | C       | 2345   |
+| 14 | C       | 2645   | &lt;-- median
+| 15 | C       | 2645   | 
+| 16 | C       | 2652   |
++----+---------+--------+
 </pre>
 
 <p>&nbsp;</p>

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

@@ -4,25 +4,41 @@
 
 ## Description
 
-<p>Suppose Andy and Doris want to choose a restaurant for dinner, and they both have a list of favorite restaurants represented by strings.</p>
+<p>Given two arrays of strings <code>list1</code> and <code>list2</code>, find the <strong>common strings with the least index sum</strong>.</p>
 
-<p>You need to help them find out their <b>common interest</b> with the <b>least list index sum</b>. If there is a choice tie between answers, output all of them with no order requirement. You could assume there always exists an answer.</p>
+<p>A <strong>common string</strong> is a string that appeared in both <code>list1</code> and <code>list2</code>.</p>
+
+<p>A <strong>common string with the least index sum</strong> is a common string such that if it appeared at <code>list1[i]</code> and <code>list2[j]</code> then <code>i + j</code> should be the minimum value among all the other <strong>common strings</strong>.</p>
+
+<p>Return <em>all the <strong>common strings with the least index sum</strong></em>. Return the answer in <strong>any order</strong>.</p>
 
 <p>&nbsp;</p>
 <p><strong>Example 1:</strong></p>
 
 <pre>
 <strong>Input:</strong> list1 = [&quot;Shogun&quot;,&quot;Tapioca Express&quot;,&quot;Burger King&quot;,&quot;KFC&quot;], list2 = [&quot;Piatti&quot;,&quot;The Grill at Torrey Pines&quot;,&quot;Hungry Hunter Steakhouse&quot;,&quot;Shogun&quot;]
 <strong>Output:</strong> [&quot;Shogun&quot;]
-<strong>Explanation:</strong> The only restaurant they both like is &quot;Shogun&quot;.
+<strong>Explanation:</strong> The only common string is &quot;Shogun&quot;.
 </pre>
 
 <p><strong>Example 2:</strong></p>
 
 <pre>
 <strong>Input:</strong> list1 = [&quot;Shogun&quot;,&quot;Tapioca Express&quot;,&quot;Burger King&quot;,&quot;KFC&quot;], list2 = [&quot;KFC&quot;,&quot;Shogun&quot;,&quot;Burger King&quot;]
 <strong>Output:</strong> [&quot;Shogun&quot;]
-<strong>Explanation:</strong> The restaurant they both like and have the least index sum is &quot;Shogun&quot; with index sum 1 (0+1).
+<strong>Explanation:</strong> The common string with the least index sum is &quot;Shogun&quot; with index sum = (0 + 1) = 1.
+</pre>
+
+<p><strong>Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> list1 = [&quot;happy&quot;,&quot;sad&quot;,&quot;good&quot;], list2 = [&quot;sad&quot;,&quot;happy&quot;,&quot;good&quot;]
+<strong>Output:</strong> [&quot;sad&quot;,&quot;happy&quot;]
+<strong>Explanation:</strong> There are three common strings:
+&quot;happy&quot; with index sum = (0 + 1) = 1.
+&quot;sad&quot; with index sum = (1 + 0) = 1.
+&quot;good&quot; with index sum = (2 + 2) = 4.
+The strings with the least index sum are &quot;sad&quot; and &quot;happy&quot;.
 </pre>
 
 <p>&nbsp;</p>
@@ -32,8 +48,8 @@
 	<li><code>1 &lt;= list1.length, list2.length &lt;= 1000</code></li>
 	<li><code>1 &lt;= list1[i].length, list2[i].length &lt;= 30</code></li>
 	<li><code>list1[i]</code> and <code>list2[i]</code> consist of spaces <code>&#39; &#39;</code> and English letters.</li>
-	<li>All the stings of <code>list1</code> are <strong>unique</strong>.</li>
-	<li>All the stings of <code>list2</code>&nbsp;are <strong>unique</strong>.</li>
+	<li>All the strings of <code>list1</code> are <strong>unique</strong>.</li>
+	<li>All the strings of <code>list2</code> are <strong>unique</strong>.</li>
 </ul>
 
 ## Solutions

solution/0600-0699/0662.Maximum Width of Binary Tree/README_EN.md

@@ -94,7 +94,7 @@ class Solution:
                 ans = max(ans, i - t[depth] + 1)
             dfs(root.left, depth + 1, i << 1)
             dfs(root.right, depth + 1, i << 1 | 1)
-        
+
         ans = 1
         t = []
         dfs(root, 0, 1)

solution/0600-0699/0662.Maximum Width of Binary Tree/images/maximum-width-of-binary-tree-v3.jpg

@@ -49,7 +49,7 @@ We made 2 refueling stops along the way, so we return 2.
 <ul>
 	<li><code>1 &lt;= target, startFuel &lt;= 10<sup>9</sup></code></li>
 	<li><code>0 &lt;= stations.length &lt;= 500</code></li>
-	<li><code>0 &lt;= position<sub>i</sub> &lt;= position<sub>i+1</sub> &lt; target</code></li>
+	<li><code>1 &lt;= position<sub>i</sub> &lt; position<sub>i+1</sub> &lt; target</code></li>
 	<li><code>1 &lt;= fuel<sub>i</sub> &lt; 10<sup>9</sup></code></li>
 </ul>
 

solution/0800-0899/0871.Minimum Number of Refueling Stops/README_EN.md

@@ -149,7 +149,7 @@ public:
 
     int longestZigZag(TreeNode* root) {
         dfs(root, 0, 0);
-        return ans;    
+        return ans;
     }
 
     void dfs(TreeNode* root, int l, int r) {

solution/1300-1399/1372.Longest ZigZag Path in a Binary Tree/README.md

@@ -136,7 +136,7 @@ public:
 
     int longestZigZag(TreeNode* root) {
         dfs(root, 0, 0);
-        return ans;    
+        return ans;
     }
 
     void dfs(TreeNode* root, int l, int r) {

solution/1300-1399/1372.Longest ZigZag Path in a Binary Tree/README_EN.md

@@ -65,7 +65,6 @@
 
 时间复杂度 $O(n\log n)$，其中 $n$ 是 `nums` 数组元素的个数。
 
-
 <!-- tabs:start -->
 
 ### **Python3**

solution/1400-1499/1424.Diagonal Traverse II/README.md

@@ -57,7 +57,7 @@ class Solution {
         for (int i = 0; i < nums.size(); ++i) {
             for (int j = 0; j < nums.get(i).size(); ++j) {
                 arr.add(new int[]{i + j, j, nums.get(i).get(j)});
-            } 
+            }
         }
         arr.sort((a, b) -> a[0] == b[0] ? a[1] - b[1] : a[0] - b[0]);
         int[] ans = new int[arr.size()];

solution/1400-1499/1424.Diagonal Traverse II/README_EN.md

@@ -12,39 +12,28 @@
 
 <p>&nbsp;</p>
 <p><strong>Example 1:</strong></p>
-
-<p><strong><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1400-1499/1494.Parallel%20Courses%20II/images/leetcode_parallel_courses_1.png" style="width: 300px; height: 164px;" /></strong></p>
-
+<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1400-1499/1494.Parallel%20Courses%20II/images/leetcode_parallel_courses_1.png" style="width: 269px; height: 147px;" />
 <pre>
-<strong>Input:</strong> n = 4, dependencies = [[2,1],[3,1],[1,4]], k = 2
-<strong>Output:</strong> 3 
+<strong>Input:</strong> n = 4, relations = [[2,1],[3,1],[1,4]], k = 2
+<strong>Output:</strong> 3
 <strong>Explanation:</strong> The figure above represents the given graph.
 In the first semester, you can take courses 2 and 3.
 In the second semester, you can take course 1.
 In the third semester, you can take course 4.
 </pre>
 
 <p><strong>Example 2:</strong></p>
-
-<p><strong><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1400-1499/1494.Parallel%20Courses%20II/images/leetcode_parallel_courses_2.png" style="width: 300px; height: 234px;" /></strong></p>
-
+<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1400-1499/1494.Parallel%20Courses%20II/images/leetcode_parallel_courses_2.png" style="width: 271px; height: 211px;" />
 <pre>
-<strong>Input:</strong> n = 5, dependencies = [[2,1],[3,1],[4,1],[1,5]], k = 2
-<strong>Output:</strong> 4 
+<strong>Input:</strong> n = 5, relations = [[2,1],[3,1],[4,1],[1,5]], k = 2
+<strong>Output:</strong> 4
 <strong>Explanation:</strong> The figure above represents the given graph.
-In the first semester, you can take courses 2 and 3 only since you cannot take more than two per semester.
+In the first semester, you can only take courses 2 and 3 since you cannot take more than two per semester.
 In the second semester, you can take course 4.
 In the third semester, you can take course 1.
 In the fourth semester, you can take course 5.
 </pre>
 
-<p><strong>Example 3:</strong></p>
-
-<pre>
-<strong>Input:</strong> n = 11, dependencies = [], k = 2
-<strong>Output:</strong> 6
-</pre>
-
 <p>&nbsp;</p>
 <p><strong>Constraints:</strong></p>
 

solution/1400-1499/1494.Parallel Courses II/README_EN.md

@@ -7,61 +7,44 @@
 <p>A split of an integer array is <strong>good</strong> if:</p>
 
 <ul>
-    <li>The array is split into three <strong>non-empty</strong> contiguous subarrays - named <code>left</code>, <code>mid</code>, <code>right</code> respectively from left to right.</li>
-    <li>The sum of the elements in <code>left</code> is less than or equal to the sum of the elements in <code>mid</code>, and the sum of the elements in <code>mid</code> is less than or equal to the sum of the elements in <code>right</code>.</li>
-
+	<li>The array is split into three <strong>non-empty</strong> contiguous subarrays - named <code>left</code>, <code>mid</code>, <code>right</code> respectively from left to right.</li>
+	<li>The sum of the elements in <code>left</code> is less than or equal to the sum of the elements in <code>mid</code>, and the sum of the elements in <code>mid</code> is less than or equal to the sum of the elements in <code>right</code>.</li>
 </ul>
 
 <p>Given <code>nums</code>, an array of <strong>non-negative</strong> integers, return <em>the number of <strong>good</strong> ways to split</em> <code>nums</code>. As the number may be too large, return it <strong>modulo</strong> <code>10<sup>9 </sup>+ 7</code>.</p>
 
 <p>&nbsp;</p>
-
 <p><strong>Example 1:</strong></p>
 
 <pre>
-
 <strong>Input:</strong> nums = [1,1,1]
-
 <strong>Output:</strong> 1
-
 <strong>Explanation:</strong> The only good way to split nums is [1] [1] [1].</pre>
 
 <p><strong>Example 2:</strong></p>
 
 <pre>
-
 <strong>Input:</strong> nums = [1,2,2,2,5,0]
-
 <strong>Output:</strong> 3
-
 <strong>Explanation:</strong> There are three good ways of splitting nums:
-
 [1] [2] [2,2,5,0]
-
 [1] [2,2] [2,5,0]
-
 [1,2] [2,2] [5,0]
-
 </pre>
 
 <p><strong>Example 3:</strong></p>
 
 <pre>
-
 <strong>Input:</strong> nums = [3,2,1]
-
 <strong>Output:</strong> 0
-
 <strong>Explanation:</strong> There is no good way to split nums.</pre>
 
 <p>&nbsp;</p>
-
 <p><strong>Constraints:</strong></p>
 
 <ul>
-    <li><code>3 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
-    <li><code>0 &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
-
+	<li><code>3 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>0 &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
 </ul>
 
 ## Solutions

solution/1700-1799/1712.Ways to Split Array Into Three Subarrays/README_EN.md

@@ -61,7 +61,6 @@ For ID 2, the previous value that is not null is from ID 1. We replace the null
 Note that the rows in the output are the same as in the input.
 </pre>
 
-
 ## 解法
 
 <!-- 这里可写通用的实现逻辑 -->

solution/2300-2399/2388.Change Null Values in a Table to the Previous Value/README.md

@@ -59,7 +59,6 @@ For ID 2, the previous value that is not null is from ID 1. We replace the null
 Note that the rows in the output are the same as in the input.
 </pre>
 
-
 ## Solutions
 
 <!-- tabs:start -->