feat: update lc problems

Author: yanglbme

solution/0000-0099/0015.3Sum/README_EN.md

@@ -15,7 +15,7 @@
 <strong>Input:</strong> nums = [-1,0,1,2,-1,-4]
 <strong>Output:</strong> [[-1,-1,2],[-1,0,1]]
 <strong>Explanation:</strong> 
-nums[0] + nums[1] + nums[1] = (-1) + 0 + 1 = 0.
+nums[0] + nums[1] + nums[2] = (-1) + 0 + 1 = 0.
 nums[1] + nums[2] + nums[4] = 0 + 1 + (-1) = 0.
 nums[0] + nums[3] + nums[4] = (-1) + 2 + (-1) = 0.
 The distinct triplets are [-1,0,1] and [-1,-1,2].

solution/0000-0099/0030.Substring with Concatenation of All Words/README_EN.md

@@ -37,10 +37,9 @@ The output order does not matter, returning [9,0] is fine too.
 
 <ul>
 	<li><code>1 &lt;= s.length &lt;= 10<sup>4</sup></code></li>
-	<li><code>s</code> consists of lower-case English letters.</li>
 	<li><code>1 &lt;= words.length &lt;= 5000</code></li>
 	<li><code>1 &lt;= words[i].length &lt;= 30</code></li>
-	<li><code>words[i]</code>&nbsp;consists of lower-case English letters.</li>
+	<li><code>s</code> and <code>words[i]</code> consist of lowercase English letters.</li>
 </ul>
 
 ## Solutions

solution/0200-0299/0215.Kth Largest Element in an Array/README.md

@@ -30,7 +30,7 @@
 <p><strong>提示： </strong></p>
 
 <ul>
-	<li><code>1 &lt;= k &lt;= nums.length &lt;= 10<sup>4</sup></code></li>
+	<li><code>1 &lt;= k &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
 	<li><code>-10<sup>4</sup>&nbsp;&lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
 </ul>
 

solution/0200-0299/0297.Serialize and Deserialize Binary Tree/README_EN.md

@@ -8,7 +8,7 @@
 
 <p>Design an algorithm to serialize and deserialize a binary tree. There is no restriction on how your serialization/deserialization algorithm should work. You just need to ensure that a binary tree can be serialized to a string and this string can be deserialized to the original tree structure.</p>
 
-<p><strong>Clarification:</strong> The input/output format is the same as <a href="/faq/#binary-tree" target="_blank">how LeetCode serializes a binary tree</a>. You do not necessarily need to follow this format, so please be creative and come up with different approaches yourself.</p>
+<p><strong>Clarification:</strong> The input/output format is the same as <a href="https://support.leetcode.com/hc/en-us/articles/360011883654-What-does-1-null-2-3-mean-in-binary-tree-representation-" target="_blank">how LeetCode serializes a binary tree</a>. You do not necessarily need to follow this format, so please be creative and come up with different approaches yourself.</p>
 
 <p>&nbsp;</p>
 <p><strong>Example 1:</strong></p>

solution/0300-0399/0336.Palindrome Pairs/README.md

@@ -32,6 +32,8 @@
 <strong>输出：</strong>[[0,1],[1,0]]
 </pre>
 
+
+
 <p><strong>提示：</strong></p>
 
 <ul>

solution/0500-0599/0554.Brick Wall/README.md

@@ -28,6 +28,8 @@
 <strong>输出：</strong>3
 </pre>
 
+
+
 <p><strong>提示：</strong></p>
 
 <ul>

solution/0500-0599/0592.Fraction Addition and Subtraction/Solution.py

@@ -10,7 +10,7 @@ def fractionAddition(self, expression: str) -> str:
             j = i
             while j < n and expression[j] not in '+-':
                 j += 1
-            s = expression[i: j]
+            s = expression[i:j]
             a, b = s.split('/')
             x += sign * int(a) * y // int(b)
             i = j

solution/0700-0799/0775.Global and Local Inversions/README.md

@@ -42,6 +42,8 @@
 <strong>解释：</strong>有 2 个全局倒置，和 1 个局部倒置。
 </pre>
 
+
+
 <p><strong>提示：</strong></p>
 
 <ul>

solution/0800-0899/0830.Positions of Large Groups/README.md

@@ -48,6 +48,8 @@
 <strong>输出：</strong>[]
 </pre>
 
+
+
 <p><strong>提示：</strong></p>
 
 <ul>

solution/1000-1099/1030.Matrix Cells in Distance Order/README.md

@@ -6,7 +6,7 @@
 
 <!-- 这里写题目描述 -->
 
-<p>给定四个整数 <code>row</code>&nbsp;,&nbsp; &nbsp;<code>cols</code> ,&nbsp; <code>rCenter</code> 和 <code>cCenter</code> 。有一个&nbsp;<code>rows x cols</code>&nbsp;的矩阵，你在单元格上的坐标是&nbsp;<code>(rCenter, cCenter)</code> 。</p>
+<p>给定四个整数 <code>rows</code>&nbsp;,&nbsp; &nbsp;<code>cols</code> ,&nbsp; <code>rCenter</code> 和 <code>cCenter</code> 。有一个&nbsp;<code>rows x cols</code>&nbsp;的矩阵，你在单元格上的坐标是&nbsp;<code>(rCenter, cCenter)</code> 。</p>
 
 <p>返回矩阵中的所有单元格的坐标，并按与<em>&nbsp;</em><code>(rCenter, cCenter)</code><em>&nbsp;</em>的 <strong>距离</strong> 从最小到最大的顺序排。你可以按 <strong>任何</strong> 满足此条件的顺序返回答案。</p>
 

solution/1100-1199/1146.Snapshot Array/README_EN.md

@@ -7,10 +7,10 @@
 <p>Implement a SnapshotArray that supports the following interface:</p>
 
 <ul>
-	<li><code>SnapshotArray(int length)</code> initializes an array-like data structure with the given length.&nbsp; <strong>Initially, each element equals 0</strong>.</li>
+	<li><code>SnapshotArray(int length)</code> initializes an array-like data structure with the given length. <strong>Initially, each element equals 0</strong>.</li>
 	<li><code>void set(index, val)</code> sets the element at the given <code>index</code> to be equal to <code>val</code>.</li>
-	<li><code>int snap()</code>&nbsp;takes a snapshot of the array and returns the <code>snap_id</code>: the total number of times we called <code>snap()</code> minus <code>1</code>.</li>
-	<li><code>int get(index, snap_id)</code>&nbsp;returns the value at the given <code>index</code>, at the time we took the snapshot with the given <code>snap_id</code></li>
+	<li><code>int snap()</code> takes a snapshot of the array and returns the <code>snap_id</code>: the total number of times we called <code>snap()</code> minus <code>1</code>.</li>
+	<li><code>int get(index, snap_id)</code> returns the value at the given <code>index</code>, at the time we took the snapshot with the given <code>snap_id</code></li>
 </ul>
 
 <p>&nbsp;</p>
@@ -31,11 +31,11 @@ snapshotArr.get(0,0);  // Get the value of array[0] with snap_id = 0, return 5</
 <p><strong>Constraints:</strong></p>
 
 <ul>
-	<li><code>1 &lt;= length&nbsp;&lt;= 50000</code></li>
-	<li>At most <code>50000</code>&nbsp;calls will be made to <code>set</code>, <code>snap</code>, and <code>get</code>.</li>
-	<li><code>0 &lt;= index&nbsp;&lt;&nbsp;length</code></li>
-	<li><code>0 &lt;=&nbsp;snap_id &lt;&nbsp;</code>(the total number of times we call <code>snap()</code>)</li>
-	<li><code>0 &lt;=&nbsp;val &lt;= 10^9</code></li>
+	<li><code>1 &lt;= length &lt;= 5 * 10<sup>4</sup></code></li>
+	<li><code>0 &lt;= index &lt; length</code></li>
+	<li><code>0 &lt;= val &lt;= 10<sup>9</sup></code></li>
+	<li><code>0 &lt;= snap_id &lt; </code>(the total number of times we call <code>snap()</code>)</li>
+	<li>At most <code>5 * 10<sup>4</sup></code> calls will be made to <code>set</code>, <code>snap</code>, and <code>get</code>.</li>
 </ul>
 
 ## Solutions

solution/1200-1299/1206.Design Skiplist/README.md

@@ -356,7 +356,7 @@ public:
         head = new Node(-1, maxLevel);
         level = 0;
     }
-    
+
     bool search(int target) {
         Node* curr = head;
         for (int i = level - 1; ~i; --i)
@@ -366,7 +366,7 @@ public:
         }
         return false;
     }
-    
+
     void add(int num) {
         Node* curr = head;
         int lv = randomLevel();
@@ -382,7 +382,7 @@ public:
             }
         }
     }
-    
+
     bool erase(int num) {
         Node* curr = head;
         bool ok = false;

solution/1200-1299/1206.Design Skiplist/README_EN.md

@@ -344,7 +344,7 @@ public:
         head = new Node(-1, maxLevel);
         level = 0;
     }
-    
+
     bool search(int target) {
         Node* curr = head;
         for (int i = level - 1; ~i; --i)
@@ -354,7 +354,7 @@ public:
         }
         return false;
     }
-    
+
     void add(int num) {
         Node* curr = head;
         int lv = randomLevel();
@@ -370,7 +370,7 @@ public:
             }
         }
     }
-    
+
     bool erase(int num) {
         Node* curr = head;
         bool ok = false;

solution/1200-1299/1227.Airplane Seat Assignment Probability/Solution.py

@@ -1,3 +1,3 @@
 class Solution:
     def nthPersonGetsNthSeat(self, n: int) -> float:
-        return 1 if n == 1 else .5
+        return 1 if n == 1 else 0.5

solution/1500-1599/1505.Minimum Possible Integer After at Most K Adjacent Swaps On Digits/README_EN.md

@@ -39,7 +39,7 @@
 <ul>
 	<li><code>1 &lt;= num.length &lt;= 3 * 10<sup>4</sup></code></li>
 	<li><code>num</code> consists of only <strong>digits</strong> and does not contain <strong>leading zeros</strong>.</li>
-	<li><code>1 &lt;= k &lt;= 10<sup>4</sup></code></li>
+	<li><code>1 &lt;= k &lt;= 10<sup>9</sup></code></li>
 </ul>
 
 ## Solutions

solution/2300-2399/2332.The Latest Time to Catch a Bus/README_EN.md

@@ -8,7 +8,14 @@
 
 <p>You are given an integer <code>capacity</code>, which represents the <strong>maximum</strong> number of passengers that can get on each bus.</p>
 
-<p>The passengers will get on the next available bus. You can get on a bus that will depart at <code>x</code> minutes if you arrive at <code>y</code> minutes where <code>y &lt;= x</code>, and the bus is not full. Passengers with the <strong>earliest</strong> arrival times get on the bus first.</p>
+<p>When a passenger arrives, they will wait in line for the next available bus. You can get on a bus that departs at <code>x</code> minutes if you arrive at <code>y</code> minutes where <code>y &lt;= x</code>, and the bus is not full. Passengers with the <strong>earliest</strong> arrival times get on the bus first.</p>
+
+<p>More formally when a bus arrives, either:</p>
+
+<ul>
+	<li>If <code>capacity</code> or fewer passengers are waiting for a bus, they will <strong>all</strong> get on the bus, or</li>
+	<li>The <code>capacity</code> passengers with the <strong>earliest</strong> arrival times will get on the bus.</li>
+</ul>
 
 <p>Return <em>the latest time you may arrive at the bus station to catch a bus</em>. You <strong>cannot</strong> arrive at the same time as another passenger.</p>
 
@@ -20,21 +27,21 @@
 <pre>
 <strong>Input:</strong> buses = [10,20], passengers = [2,17,18,19], capacity = 2
 <strong>Output:</strong> 16
-<strong>Explanation:</strong> 
-The 1<sup>st</sup> bus departs with the 1<sup>st</sup> passenger. 
-The 2<sup>nd</sup> bus departs with you and the 2<sup>nd</sup> passenger.
-Note that you must not arrive at the same time as the passengers, which is why you must arrive before the 2<sup>nd</sup><sup> </sup>passenger to catch the bus.</pre>
+<strong>Explanation:</strong> Suppose you arrive at time 16.
+At time 10, the first bus departs with the 0<sup>th</sup> passenger. 
+At time 20, the second bus departs with you and the 1<sup>st</sup> passenger.
+Note that you may not arrive at the same time as another passenger, which is why you must arrive before the 1<sup>st</sup> passenger to catch the bus.</pre>
 
 <p><strong>Example 2:</strong></p>
 
 <pre>
 <strong>Input:</strong> buses = [20,30,10], passengers = [19,13,26,4,25,11,21], capacity = 2
 <strong>Output:</strong> 20
-<strong>Explanation:</strong> 
-The 1<sup>st</sup> bus departs with the 4<sup>th</sup> passenger. 
-The 2<sup>nd</sup> bus departs with the 6<sup>th</sup>&nbsp;and 2<sup>nd</sup><sup> </sup>passengers.
-The 3<sup>rd</sup> bus departs with the 1<sup>s</sup><sup>t</sup> passenger and you.
-</pre>
+<strong>Explanation:</strong> Suppose you arrive at time 20.
+At time 10, the first bus departs with the 3<sup>rd</sup> passenger. 
+At time 20, the second bus departs with the 5<sup>th</sup> and 1<sup>st</sup> passengers.
+At time 30, the third bus departs with the 0<sup>th</sup> passenger and you.
+Notice if you had arrived any later, then the 6<sup>th</sup> passenger would have taken your seat on the third bus.</pre>
 
 <p>&nbsp;</p>
 <p><strong>Constraints:</strong></p>

solution/2300-2399/2352.Equal Row and Column Pairs/README.md

@@ -14,7 +14,7 @@
 
 <p><strong>示例 1：</strong></p>
 
-<p><img alt="" src="https://assets.leetcode.com/uploads/2022/06/01/ex1.jpg" style="width: 150px; height: 153px;" /></p>
+<p><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2300-2399/2352.Equal%20Row%20and%20Column%20Pairs/images/ex1.jpg" style="width: 150px; height: 153px;" /></p>
 
 <pre>
 <strong>输入：</strong>grid = [[3,2,1],[1,7,6],[2,7,7]]
@@ -25,7 +25,7 @@
 
 <p><strong>示例 2：</strong></p>
 
-<p><img alt="" src="https://assets.leetcode.com/uploads/2022/06/01/ex2.jpg" style="width: 200px; height: 209px;" /></p>
+<p><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2300-2399/2352.Equal%20Row%20and%20Column%20Pairs/images/ex2.jpg" style="width: 200px; height: 209px;" /></p>
 
 <pre>
 <strong>输入：</strong>grid = [[3,1,2,2],[1,4,4,5],[2,4,2,2],[2,4,2,2]]

solution/2300-2399/2352.Equal Row and Column Pairs/README_EN.md

@@ -10,7 +10,7 @@
 
 <p>&nbsp;</p>
 <p><strong>Example 1:</strong></p>
-<img alt="" src="https://assets.leetcode.com/uploads/2022/06/01/ex1.jpg" style="width: 150px; height: 153px;" />
+<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2300-2399/2352.Equal%20Row%20and%20Column%20Pairs/images/ex1.jpg" style="width: 150px; height: 153px;" />
 <pre>
 <strong>Input:</strong> grid = [[3,2,1],[1,7,6],[2,7,7]]
 <strong>Output:</strong> 1
@@ -19,7 +19,7 @@
 </pre>
 
 <p><strong>Example 2:</strong></p>
-<img alt="" src="https://assets.leetcode.com/uploads/2022/06/01/ex2.jpg" style="width: 200px; height: 209px;" />
+<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2300-2399/2352.Equal%20Row%20and%20Column%20Pairs/images/ex2.jpg" style="width: 200px; height: 209px;" />
 <pre>
 <strong>Input:</strong> grid = [[3,1,2,2],[1,4,4,5],[2,4,2,2],[2,4,2,2]]
 <strong>Output:</strong> 3

solution/2300-2399/2352.Equal Row and Column Pairs/images/ex1.jpg

@@ -17,14 +17,14 @@
 
 <ul>
 	<li><code>FoodRatings(String[] foods, String[] cuisines, int[] ratings)</code> 初始化系统。食物由 <code>foods</code>、<code>cuisines</code> 和 <code>ratings</code> 描述，长度均为 <code>n</code> 。
-	<ul>
-		<li><code>foods[i]</code> 是第 <code>i</code> 种食物的名字。</li>
-		<li><code>cuisines[i]</code> 是第 <code>i</code> 种食物的烹饪方式。</li>
-		<li><code>ratings[i]</code> 是第 <code>i</code> 种食物的最初评分。</li>
-	</ul>
-	</li>
-	<li><code>void changeRating(String food, int newRating)</code> 修改名字为 <code>food</code> 的食物的评分。</li>
-	<li><code>String highestRated(String cuisine)</code> 返回指定烹饪方式 <code>cuisine</code> 下评分最高的食物的名字。如果存在并列，返回 <strong>字典序较小</strong> 的名字。</li>
+    <ul>
+    	<li><code>foods[i]</code> 是第 <code>i</code> 种食物的名字。</li>
+    	<li><code>cuisines[i]</code> 是第 <code>i</code> 种食物的烹饪方式。</li>
+    	<li><code>ratings[i]</code> 是第 <code>i</code> 种食物的最初评分。</li>
+    </ul>
+    </li>
+    <li><code>void changeRating(String food, int newRating)</code> 修改名字为 <code>food</code> 的食物的评分。</li>
+    <li><code>String highestRated(String cuisine)</code> 返回指定烹饪方式 <code>cuisine</code> 下评分最高的食物的名字。如果存在并列，返回 <strong>字典序较小</strong> 的名字。</li>
 </ul>
 
 <p>注意，字符串 <code>x</code> 的字典序比字符串 <code>y</code> 更小的前提是：<code>x</code> 在字典中出现的位置在 <code>y</code> 之前，也就是说，要么 <code>x</code> 是 <code>y</code> 的前缀，或者在满足&nbsp;<code>x[i] != y[i]</code> 的第一个位置 <code>i</code> 处，<code>x[i]</code> 在字母表中出现的位置在 <code>y[i]</code> 之前。</p>

solution/2300-2399/2352.Equal Row and Column Pairs/images/ex2.jpg

@@ -15,14 +15,14 @@
 
 <ul>
 	<li><code>FoodRatings(String[] foods, String[] cuisines, int[] ratings)</code> Initializes the system. The food items are described by <code>foods</code>, <code>cuisines</code> and <code>ratings</code>, all of which have a length of <code>n</code>.
-	<ul>
-		<li><code>foods[i]</code> is the name of the <code>i<sup>th</sup></code> food,</li>
-		<li><code>cuisines[i]</code> is the type of cuisine of the <code>i<sup>th</sup></code> food, and</li>
-		<li><code>ratings[i]</code> is the initial rating of the <code>i<sup>th</sup></code> food.</li>
-	</ul>
-	</li>
-	<li><code>void changeRating(String food, int newRating)</code> Changes the rating of the food item with the name <code>food</code>.</li>
-	<li><code>String highestRated(String cuisine)</code> Returns the name of the food item that has the highest rating for the given type of <code>cuisine</code>. If there is a tie, return the item with the <strong>lexicographically smaller</strong> name.</li>
+    <ul>
+    	<li><code>foods[i]</code> is the name of the <code>i<sup>th</sup></code> food,</li>
+    	<li><code>cuisines[i]</code> is the type of cuisine of the <code>i<sup>th</sup></code> food, and</li>
+    	<li><code>ratings[i]</code> is the initial rating of the <code>i<sup>th</sup></code> food.</li>
+    </ul>
+    </li>
+    <li><code>void changeRating(String food, int newRating)</code> Changes the rating of the food item with the name <code>food</code>.</li>
+    <li><code>String highestRated(String cuisine)</code> Returns the name of the food item that has the highest rating for the given type of <code>cuisine</code>. If there is a tie, return the item with the <strong>lexicographically smaller</strong> name.</li>
 </ul>
 
 <p>Note that a string <code>x</code> is lexicographically smaller than string <code>y</code> if <code>x</code> comes before <code>y</code> in dictionary order, that is, either <code>x</code> is a prefix of <code>y</code>, or if <code>i</code> is the first position such that <code>x[i] != y[i]</code>, then <code>x[i]</code> comes before <code>y[i]</code> in alphabetic order.</p>

solution/2300-2399/2353.Design a Food Rating System/README.md

@@ -60,7 +60,6 @@ It can be proven that 13 is the maximum number of books you can take.
 	<li><code>0 &lt;= books[i] &lt;= 10<sup>5</sup></code></li>
 </ul>
 
-
 ## Solutions
 
 <!-- tabs:start -->