feat: update lc problems

Author: yanglbme

solution/0100-0199/0179.Largest Number/README.md

@@ -128,7 +128,7 @@ public class Comparer: IComparer<string>
     {
         return Compare(left, right, 0, 0);
     }
-    
+
     private int Compare(string left, string right, int lBegin, int rBegin)
     {
         var len = Math.Min(left.Length - lBegin, right.Length - rBegin);
@@ -139,7 +139,7 @@ public class Comparer: IComparer<string>
                 return left[lBegin + i] < right[rBegin + i] ? -1 : 1;
             }
         }
-        
+
         if (left.Length - lBegin == right.Length - rBegin)
         {
             return 0;
@@ -159,7 +159,7 @@ public class Solution {
     public string LargestNumber(int[] nums) {
         var sb = new StringBuilder();
         var strs = nums.Select(n => n.ToString(CultureInfo.InvariantCulture)).OrderByDescending(s => s, new Comparer());
-        
+
         var nonZeroOccurred = false;
         foreach (var str in strs)
         {

solution/0100-0199/0179.Largest Number/README_EN.md

@@ -115,7 +115,7 @@ public class Comparer: IComparer<string>
     {
         return Compare(left, right, 0, 0);
     }
-    
+
     private int Compare(string left, string right, int lBegin, int rBegin)
     {
         var len = Math.Min(left.Length - lBegin, right.Length - rBegin);
@@ -126,7 +126,7 @@ public class Comparer: IComparer<string>
                 return left[lBegin + i] < right[rBegin + i] ? -1 : 1;
             }
         }
-        
+
         if (left.Length - lBegin == right.Length - rBegin)
         {
             return 0;
@@ -146,7 +146,7 @@ public class Solution {
     public string LargestNumber(int[] nums) {
         var sb = new StringBuilder();
         var strs = nums.Select(n => n.ToString(CultureInfo.InvariantCulture)).OrderByDescending(s => s, new Comparer());
-        
+
         var nonZeroOccurred = false;
         foreach (var str in strs)
         {

solution/0300-0399/0315.Count of Smaller Numbers After Self/README_EN.md

@@ -4,7 +4,7 @@
 
 ## Description
 
-<p>You are given an integer array <code>nums</code> and you have to return a new <code>counts</code> array. The <code>counts</code> array has the property where <code>counts[i]</code> is the number of smaller elements to the right of <code>nums[i]</code>.</p>
+<p>Given an integer array <code>nums</code>, return<em> an integer array </em><code>counts</code><em> where </em><code>counts[i]</code><em> is the number of smaller elements to the right of </em><code>nums[i]</code>.</p>
 
 <p>&nbsp;</p>
 <p><strong>Example 1:</strong></p>

solution/0300-0399/0347.Top K Frequent Elements/README_EN.md

@@ -19,6 +19,7 @@
 
 <ul>
 	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>-10<sup>4</sup> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
 	<li><code>k</code> is in the range <code>[1, the number of unique elements in the array]</code>.</li>
 	<li>It is <strong>guaranteed</strong> that the answer is <strong>unique</strong>.</li>
 </ul>

solution/0500-0599/0593.Valid Square/README.md

@@ -8,7 +8,7 @@
 
 <p>给定2D空间中四个点的坐标&nbsp;<code>p1</code>,&nbsp;<code>p2</code>,&nbsp;<code>p3</code>&nbsp;和&nbsp;<code>p4</code>，如果这四个点构成一个正方形，则返回 <code>true</code> 。</p>
 
-<p>点的坐标&nbsp;<code>p<sub>i</sub></code> 表示为 <code>[xi, yi]</code> 。输入 <strong>不是</strong> 按任何顺序给出的。</p>
+<p>点的坐标&nbsp;<code>p<sub>i</sub></code> 表示为 <code>[xi, yi]</code> 。 <code>输入没有任何顺序</code> 。</p>
 
 <p>一个 <strong>有效的正方形</strong> 有四条等边和四个等角(90度角)。</p>
 

solution/0600-0699/0622.Design Circular Queue/README.md

@@ -195,37 +195,37 @@ public:
         q = vector<int>(k);
         front = size = 0;
     }
-    
+
     bool enQueue(int value) {
         if (isFull()) return false;
         int idx = (front + size) % capacity;
         q[idx] = value;
         ++size;
         return true;
     }
-    
+
     bool deQueue() {
         if (isEmpty()) return false;
         front = (front + 1) % capacity;
         --size;
         return true;
     }
-    
+
     int Front() {
         if (isEmpty()) return -1;
         return q[front];
     }
-    
+
     int Rear() {
         if (isEmpty()) return -1;
         int idx = (front + size - 1) % capacity;
         return q[idx];
     }
-    
+
     bool isEmpty() {
         return size == 0;
     }
-    
+
     bool isFull() {
         return size == capacity;
     }

solution/0600-0699/0622.Design Circular Queue/README_EN.md

@@ -194,37 +194,37 @@ public:
         q = vector<int>(k);
         front = size = 0;
     }
-    
+
     bool enQueue(int value) {
         if (isFull()) return false;
         int idx = (front + size) % capacity;
         q[idx] = value;
         ++size;
         return true;
     }
-    
+
     bool deQueue() {
         if (isEmpty()) return false;
         front = (front + 1) % capacity;
         --size;
         return true;
     }
-    
+
     int Front() {
         if (isEmpty()) return -1;
         return q[front];
     }
-    
+
     int Rear() {
         if (isEmpty()) return -1;
         int idx = (front + size - 1) % capacity;
         return q[idx];
     }
-    
+
     bool isEmpty() {
         return size == 0;
     }
-    
+
     bool isFull() {
         return size == capacity;
     }

solution/0600-0699/0638.Shopping Offers/README_EN.md

@@ -39,11 +39,9 @@ You cannot add more items, though only $9 for 2A ,2B and 1C.
 <p><strong>Constraints:</strong></p>
 
 <ul>
-	<li><code>n == price.length</code></li>
-	<li><code>n == needs.length</code></li>
+	<li><code>n == price.length == needs.length</code></li>
 	<li><code>1 &lt;= n &lt;= 6</code></li>
-	<li><code>0 &lt;= price[i] &lt;= 10</code></li>
-	<li><code>0 &lt;= needs[i] &lt;= 10</code></li>
+	<li><code>0 &lt;= price[i], needs[i] &lt;= 10</code></li>
 	<li><code>1 &lt;= special.length &lt;= 100</code></li>
 	<li><code>special[i].length == n + 1</code></li>
 	<li><code>0 &lt;= special[i][j] &lt;= 50</code></li>

solution/0800-0899/0802.Find Eventual Safe States/README_EN.md

@@ -322,7 +322,7 @@ var eventualSafeNodes = function (graph) {
  * @param {number[][]} graph
  * @return {number[]}
  */
- var eventualSafeNodes = function(graph) {
+var eventualSafeNodes = function (graph) {
     const n = graph.length;
     const color = new Array(n).fill(0);
     function dfs(i) {

solution/0800-0899/0802.Find Eventual Safe States/Solution.js

@@ -2,7 +2,7 @@
  * @param {number[][]} graph
  * @return {number[]}
  */
- var eventualSafeNodes = function(graph) {
+var eventualSafeNodes = function (graph) {
     const n = graph.length;
     const color = new Array(n).fill(0);
     function dfs(i) {
@@ -25,4 +25,4 @@
         }
     }
     return ans;
-};
+};

solution/1100-1199/1161.Maximum Level Sum of a Binary Tree/README.md

@@ -52,8 +52,6 @@ BFS 层次遍历，求每一层的节点和，找出节点和最大的层，若
 
 **方法二：DFS**
 
-
-
 <!-- tabs:start -->
 
 ### **Python3**

solution/1200-1299/1209.Remove All Adjacent Duplicates in String II/README_EN.md

@@ -8,7 +8,7 @@
 
 <p>We repeatedly make <code>k</code> <strong>duplicate removals</strong> on <code>s</code> until we no longer can.</p>
 
-<p>Return the final string after all such duplicate removals have been made. It is guaranteed that the answer is unique.</p>
+<p>Return <em>the final string after all such duplicate removals have been made</em>. It is guaranteed that the answer is <strong>unique</strong>.</p>
 
 <p>&nbsp;</p>
 <p><strong>Example 1:</strong></p>
@@ -41,7 +41,7 @@ Finally delete &quot;ddd&quot;, get &quot;aa&quot;</pre>
 <ul>
 	<li><code>1 &lt;= s.length &lt;= 10<sup>5</sup></code></li>
 	<li><code>2 &lt;= k &lt;= 10<sup>4</sup></code></li>
-	<li><code>s</code> only contains lower case English letters.</li>
+	<li><code>s</code> only contains lowercase English letters.</li>
 </ul>
 
 ## Solutions

solution/1500-1599/1579.Remove Max Number of Edges to Keep Graph Fully Traversable/README_EN.md

@@ -4,17 +4,17 @@
 
 ## Description
 
-<p>Alice and Bob have an undirected graph of&nbsp;<code>n</code>&nbsp;nodes&nbsp;and 3 types of edges:</p>
+<p>Alice and Bob have an undirected graph of <code>n</code> nodes and three types of edges:</p>
 
 <ul>
 	<li>Type 1: Can be traversed by Alice only.</li>
 	<li>Type 2: Can be traversed by Bob only.</li>
-	<li>Type 3: Can by traversed by both Alice and Bob.</li>
+	<li>Type 3: Can be traversed by both Alice and Bob.</li>
 </ul>
 
-<p>Given an array&nbsp;<code>edges</code>&nbsp;where&nbsp;<code>edges[i] = [type<sub>i</sub>, u<sub>i</sub>, v<sub>i</sub>]</code>&nbsp;represents a bidirectional edge of type&nbsp;<code>type<sub>i</sub></code>&nbsp;between nodes&nbsp;<code>u<sub>i</sub></code>&nbsp;and&nbsp;<code>v<sub>i</sub></code>, find the maximum number of edges you can remove so that after removing the edges, the graph can still be fully traversed by both Alice and Bob. The graph is fully traversed by Alice and Bob if starting from any node, they can reach all other nodes.</p>
+<p>Given an array <code>edges</code> where <code>edges[i] = [type<sub>i</sub>, u<sub>i</sub>, v<sub>i</sub>]</code> represents a bidirectional edge of type <code>type<sub>i</sub></code> between nodes <code>u<sub>i</sub></code> and <code>v<sub>i</sub></code>, find the maximum number of edges you can remove so that after removing the edges, the graph can still be fully traversed by both Alice and Bob. The graph is fully traversed by Alice and Bob if starting from any node, they can reach all other nodes.</p>
 
-<p>Return <em>the maximum number of edges you can remove, or return</em> <code>-1</code> <em>if it&#39;s impossible for the graph to be fully traversed by Alice and Bob.</em></p>
+<p>Return <em>the maximum number of edges you can remove, or return</em> <code>-1</code> <em>if Alice and Bob cannot fully traverse the graph.</em></p>
 
 <p>&nbsp;</p>
 <p><strong>Example 1:</strong></p>
@@ -52,12 +52,12 @@
 <p><strong>Constraints:</strong></p>
 
 <ul>
-	<li><code>1 &lt;= n &lt;= 10^5</code></li>
-	<li><code>1 &lt;= edges.length &lt;= min(10^5, 3 * n * (n-1) / 2)</code></li>
+	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
+	<li><code>1 &lt;= edges.length &lt;= min(10<sup>5</sup>, 3 * n * (n - 1) / 2)</code></li>
 	<li><code>edges[i].length == 3</code></li>
-	<li><code>1 &lt;= edges[i][0] &lt;= 3</code></li>
-	<li><code>1 &lt;= edges[i][1] &lt; edges[i][2] &lt;= n</code></li>
-	<li>All tuples&nbsp;<code>(type<sub>i</sub>, u<sub>i</sub>, v<sub>i</sub>)</code>&nbsp;are distinct.</li>
+	<li><code>1 &lt;= type<sub>i</sub> &lt;= 3</code></li>
+	<li><code>1 &lt;= u<sub>i</sub> &lt; v<sub>i</sub> &lt;= n</code></li>
+	<li>All tuples <code>(type<sub>i</sub>, u<sub>i</sub>, v<sub>i</sub>)</code> are distinct.</li>
 </ul>
 
 ## Solutions

solution/1600-1699/1672.Richest Customer Wealth/README.md

@@ -159,7 +159,7 @@ class Solution {
             }
         }
         return max
-    }   
+    }
 }
 ```
 

solution/1600-1699/1672.Richest Customer Wealth/README_EN.md

@@ -152,7 +152,7 @@ class Solution {
             }
         }
         return max
-    }   
+    }
 }
 ```
 

solution/2200-2299/2261.K Divisible Elements Subarrays/README_EN.md

@@ -49,6 +49,11 @@ Since all subarrays are distinct, the total number of subarrays satisfying all t
 	<li><code>1 &lt;= k &lt;= nums.length</code></li>
 </ul>
 
+<p>&nbsp;</p>
+<p><strong>Follow up:</strong></p>
+
+<p>Can you solve this problem in O(n<sup>2</sup>) time complexity?</p>
+
 ## Solutions
 
 <!-- tabs:start -->

solution/2300-2399/2356.Number of Unique Subjects Taught by Each Teacher/README.md

@@ -63,7 +63,6 @@ Teacher 2:
   - They teach subject 4 in department 1.
 </pre>
 
-
 ## 解法
 
 <!-- 这里可写通用的实现逻辑 -->

solution/2300-2399/2356.Number of Unique Subjects Taught by Each Teacher/README_EN.md

@@ -61,7 +61,6 @@ Teacher 2:
   - They teach subject 4 in department 1.
 </pre>
 
-
 ## Solutions
 
 <!-- tabs:start -->

solution/2300-2399/2357.Make Array Zero by Subtracting Equal Amounts/README.md

@@ -44,7 +44,6 @@
 	<li><code>0 &lt;= nums[i] &lt;= 100</code></li>
 </ul>
 
-
 ## 解法
 
 <!-- 这里可写通用的实现逻辑 -->

solution/2300-2399/2357.Make Array Zero by Subtracting Equal Amounts/README_EN.md

@@ -41,7 +41,6 @@ In the third operation, choose x = 2. Now, nums = [0,0,0,0,0].
 	<li><code>0 &lt;= nums[i] &lt;= 100</code></li>
 </ul>
 
-
 ## Solutions
 
 <!-- tabs:start -->

solution/2300-2399/2358.Maximum Number of Groups Entering a Competition/README.md

@@ -44,7 +44,6 @@
 	<li><code>1 &lt;= grades[i] &lt;= 10<sup>5</sup></code></li>
 </ul>
 
-
 ## 解法
 
 <!-- 这里可写通用的实现逻辑 -->

solution/2300-2399/2358.Maximum Number of Groups Entering a Competition/README_EN.md

@@ -42,7 +42,6 @@ It can be shown that it is not possible to form more than 3 groups.
 	<li><code>1 &lt;= grades[i] &lt;= 10<sup>5</sup></code></li>
 </ul>
 
-
 ## Solutions
 
 <!-- tabs:start -->

solution/2300-2399/2359.Find Closest Node to Given Two Nodes/README.md

@@ -50,7 +50,6 @@
 	<li><code>0 &lt;= node1, node2 &lt; n</code></li>
 </ul>
 
-
 ## 解法
 
 <!-- 这里可写通用的实现逻辑 -->

solution/2300-2399/2359.Find Closest Node to Given Two Nodes/README_EN.md

@@ -44,7 +44,6 @@ The maximum of those two distances is 2. It can be proven that we cannot get a n
 	<li><code>0 &lt;= node1, node2 &lt; n</code></li>
 </ul>
 
-
 ## Solutions
 
 <!-- tabs:start -->

solution/2300-2399/2360.Longest Cycle in a Graph/README.md

@@ -48,7 +48,6 @@
 	<li><code>edges[i] != i</code></li>
 </ul>
 
-
 ## 解法
 
 <!-- 这里可写通用的实现逻辑 -->

solution/2300-2399/2360.Longest Cycle in a Graph/README_EN.md

@@ -40,7 +40,6 @@ The length of this cycle is 3, so 3 is returned.
 	<li><code>edges[i] != i</code></li>
 </ul>
 
-
 ## Solutions
 
 <!-- tabs:start -->