feat: update lc problems

solution/0000-0099/0057.Insert Interval/README.md

@@ -8,13 +8,17 @@
 
 <!-- 这里写题目描述 -->
 
-<p>给你一个<strong> 无重叠的</strong><em> ，</em>按照区间起始端点排序的区间列表。</p>
+<p>给你一个<strong> 无重叠的</strong><em> ，</em>按照区间起始端点排序的区间列表 <code>intervals</code>，其中&nbsp;<code>intervals[i] = [start<sub>i</sub>, end<sub>i</sub>]</code>&nbsp;表示第&nbsp;<code>i</code>&nbsp;个区间的开始和结束，并且&nbsp;<code>intervals</code>&nbsp;按照&nbsp;<code>start<sub>i</sub></code>&nbsp;升序排列。同样给定一个区间&nbsp;<code>newInterval = [start, end]</code>&nbsp;表示另一个区间的开始和结束。</p>
 
-<p>在列表中插入一个新的区间，你需要确保列表中的区间仍然有序且不重叠（如果有必要的话，可以合并区间）。</p>
+<p>在&nbsp;<code>intervals</code> 中插入区间&nbsp;<code>newInterval</code>，使得&nbsp;<code>intervals</code>&nbsp;依然按照&nbsp;<code>start<sub>i</sub></code>&nbsp;升序排列，且区间之间不重叠（如果有必要的话，可以合并区间）。</p>
 
-<p> </p>
+<p>返回插入之后的&nbsp;<code>intervals</code>。</p>
 
-<p><strong>示例 1：</strong></p>
+<p><strong>注意</strong> 你不需要原地修改&nbsp;<code>intervals</code>。你可以创建一个新数组然后返回它。</p>
+
+<p>&nbsp;</p>
+
+<p><strong>示例&nbsp;1：</strong></p>
 
 <pre>
 <strong>输入：</strong>intervals = [[1,3],[6,9]], newInterval = [2,5]
@@ -26,40 +30,20 @@
 <pre>
 <strong>输入：</strong>intervals = [[1,2],[3,5],[6,7],[8,10],[12,16]], newInterval = [4,8]
 <strong>输出：</strong>[[1,2],[3,10],[12,16]]
-<strong>解释：</strong>这是因为新的区间 <code>[4,8]</code> 与 <code>[3,5],[6,7],[8,10]</code> 重叠。</pre>
-
-<p><strong>示例 3：</strong></p>
-
-<pre>
-<strong>输入：</strong>intervals = [], newInterval = [5,7]
-<strong>输出：</strong>[[5,7]]
-</pre>
-
-<p><strong>示例 4：</strong></p>
-
-<pre>
-<strong>输入：</strong>intervals = [[1,5]], newInterval = [2,3]
-<strong>输出：</strong>[[1,5]]
-</pre>
-
-<p><strong>示例 5：</strong></p>
-
-<pre>
-<strong>输入：</strong>intervals = [[1,5]], newInterval = [2,7]
-<strong>输出：</strong>[[1,7]]
+<strong>解释：</strong>这是因为新的区间 <code>[4,8]</code> 与 <code>[3,5],[6,7],[8,10]</code>&nbsp;重叠。
 </pre>
 
-<p> </p>
+<p>&nbsp;</p>
 
 <p><strong>提示：</strong></p>
 
 <ul>
-	<li><code>0 <= intervals.length <= 10<sup>4</sup></code></li>
+	<li><code>0 &lt;= intervals.length &lt;= 10<sup>4</sup></code></li>
 	<li><code>intervals[i].length == 2</code></li>
-	<li><code>0 <= intervals[i][0] <= intervals[i][1] <= 10<sup>5</sup></code></li>
-	<li><code>intervals</code> 根据 <code>intervals[i][0]</code> 按 <strong>升序</strong> 排列</li>
+	<li><code>0 &lt;=&nbsp;start<sub>i</sub> &lt;=&nbsp;end<sub>i</sub> &lt;= 10<sup>5</sup></code></li>
+	<li><code>intervals</code> 根据 <code>start<sub>i</sub></code> 按 <strong>升序</strong> 排列</li>
 	<li><code>newInterval.length == 2</code></li>
-	<li><code>0 <= newInterval[0] <= newInterval[1] <= 10<sup>5</sup></code></li>
+	<li><code>0 &lt;=&nbsp;start &lt;=&nbsp;end &lt;= 10<sup>5</sup></code></li>
 </ul>
 
 ## 解法

solution/0000-0099/0057.Insert Interval/README_EN.md

@@ -12,6 +12,8 @@
 
 <p>Return <code>intervals</code><em> after the insertion</em>.</p>
 
+<p><strong>Note</strong> that you don&#39;t need to modify <code>intervals</code> in-place. You can make a new array and return it.</p>
+
 <p>&nbsp;</p>
 <p><strong class="example">Example 1:</strong></p>
 

solution/0000-0099/0090.Subsets II/README.md

@@ -8,7 +8,7 @@
 
 <!-- 这里写题目描述 -->
 
-<p>给你一个整数数组 <code>nums</code> ，其中可能包含重复元素，请你返回该数组所有可能的&lt;span data-keyword="subset"&gt;子集&lt;/span&gt;（幂集）。</p>
+<p>给你一个整数数组 <code>nums</code> ，其中可能包含重复元素，请你返回该数组所有可能的 <span data-keyword="subset">子集</span>（幂集）。</p>
 
 <p>解集 <strong>不能</strong> 包含重复的子集。返回的解集中，子集可以按 <strong>任意顺序</strong> 排列。</p>
 

solution/0300-0399/0310.Minimum Height Trees/README.md

@@ -8,7 +8,7 @@
 
 <!-- 这里写题目描述 -->
 
-<p>树是一个无向图，其中任何两个顶点只通过一条路径连接。 换句话说，一个任何没有简单环路的连通图都是一棵树。</p>
+<p>树是一个无向图，其中任何两个顶点只通过一条路径连接。 换句话说，任何一个没有简单环路的连通图都是一棵树。</p>
 
 <p>给你一棵包含&nbsp;<code>n</code>&nbsp;个节点的树，标记为&nbsp;<code>0</code>&nbsp;到&nbsp;<code>n - 1</code> 。给定数字&nbsp;<code>n</code>&nbsp;和一个有 <code>n - 1</code> 条无向边的 <code>edges</code>&nbsp;列表（每一个边都是一对标签），其中 <code>edges[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> 表示树中节点 <code>a<sub>i</sub></code> 和 <code>b<sub>i</sub></code> 之间存在一条无向边。</p>
 

solution/0700-0799/0791.Custom Sort String/README_EN.md

@@ -13,7 +13,7 @@
 <p>Return <em>any permutation of </em><code>s</code><em> that satisfies this property</em>.</p>
 
 <p>&nbsp;</p>
-<p><strong class="example">Example 1: </strong></p>
+<p><strong class="example">Example 1:</strong></p>
 
 <div class="example-block" style="border-color: var(--border-tertiary); border-left-width: 2px; color: var(--text-secondary); font-size: .875rem; margin-bottom: 1rem; margin-top: 1rem; overflow: visible; padding-left: 1rem;">
 <p><strong>Input: </strong> <span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;"> order = &quot;cba&quot;, s = &quot;abcd&quot; </span></p>
@@ -25,7 +25,7 @@
 <p>Since <code>&quot;d&quot;</code> does not appear in <code>order</code>, it can be at any position in the returned string. <code>&quot;dcba&quot;</code>, <code>&quot;cdba&quot;</code>, <code>&quot;cbda&quot;</code> are also valid outputs.</p>
 </div>
 
-<p><strong class="example">Example 2: </strong></p>
+<p><strong class="example">Example 2:</strong></p>
 
 <div class="example-block" style="border-color: var(--border-tertiary); border-left-width: 2px; color: var(--text-secondary); font-size: .875rem; margin-bottom: 1rem; margin-top: 1rem; overflow: visible; padding-left: 1rem;">
 <p><strong>Input: </strong> <span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;"> order = &quot;bcafg&quot;, s = &quot;abcd&quot; </span></p>

solution/1000-1099/1018.Binary Prefix Divisible By 5/README.md

@@ -2,7 +2,7 @@
 
 [English Version](/solution/1000-1099/1018.Binary%20Prefix%20Divisible%20By%205/README_EN.md)
 
-<!-- tags:数组 -->
+<!-- tags:位运算,数组 -->
 
 ## 题目描述
 

solution/1000-1099/1018.Binary Prefix Divisible By 5/README_EN.md

@@ -2,7 +2,7 @@
 
 [中文文档](/solution/1000-1099/1018.Binary%20Prefix%20Divisible%20By%205/README.md)
 
-<!-- tags:Array -->
+<!-- tags:Bit Manipulation,Array -->
 
 ## Description
 

solution/1000-1099/1035.Uncrossed Lines/README.md

@@ -10,7 +10,7 @@
 
 <p>在两条独立的水平线上按给定的顺序写下 <code>nums1</code> 和 <code>nums2</code> 中的整数。</p>
 
-<p>现在，可以绘制一些连接两个数字 <code>nums1[i]</code>&nbsp;和 <code>nums2[j]</code>&nbsp;的直线，这些直线需要同时满足满足：</p>
+<p>现在，可以绘制一些连接两个数字 <code>nums1[i]</code>&nbsp;和 <code>nums2[j]</code>&nbsp;的直线，这些直线需要同时满足：</p>
 
 <ul>
 	<li>&nbsp;<code>nums1[i] == nums2[j]</code></li>

solution/1100-1199/1151.Minimum Swaps to Group All 1's Together/README.md

@@ -64,16 +64,15 @@
 
 我们先统计数组中 $1$ 的个数，记为 $k$。然后我们使用滑动窗口，窗口大小为 $k$，窗口右边界从左向右移动，统计窗口内 $1$ 的个数，记为 $t$。每次移动窗口时，都更新 $t$ 的值，最后窗口右边界移动到数组末尾时，窗口内 $1$ 的个数最多，记为 $mx$。最后答案为 $k - mx$。
 
-时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为数组长度。
+时间复杂度 $O(n)$，其中 $n$ 为数组长度。空间复杂度 $O(1)$。
 
 <!-- tabs:start -->
 
 ```python
 class Solution:
     def minSwaps(self, data: List[int]) -> int:
         k = data.count(1)
-        t = sum(data[:k])
-        mx = t
+        mx = t = sum(data[:k])
         for i in range(k, len(data)):
             t += data[i]
             t -= data[i - k]
@@ -159,6 +158,22 @@ function minSwaps(data: number[]): number {
 }
 ```
 
+```cs
+public class Solution {
+    public int MinSwaps(int[] data) {
+        int k = data.Count(x => x == 1);
+        int t = data.Take(k).Sum();
+        int mx = t;
+        for (int i = k; i < data.Length; ++i) {
+            t += data[i];
+            t -= data[i - k];
+            mx = Math.Max(mx, t);
+        }
+        return k - mx;
+    }
+}
+```
+
 <!-- tabs:end -->
 
 <!-- end -->

solution/1100-1199/1151.Minimum Swaps to Group All 1's Together/README_EN.md

@@ -59,8 +59,7 @@ The time complexity is $O(n)$, and the space complexity is $O(1)$. Here, $n$ is
 class Solution:
     def minSwaps(self, data: List[int]) -> int:
         k = data.count(1)
-        t = sum(data[:k])
-        mx = t
+        mx = t = sum(data[:k])
         for i in range(k, len(data)):
             t += data[i]
             t -= data[i - k]
@@ -146,6 +145,22 @@ function minSwaps(data: number[]): number {
 }
 ```
 
+```cs
+public class Solution {
+    public int MinSwaps(int[] data) {
+        int k = data.Count(x => x == 1);
+        int t = data.Take(k).Sum();
+        int mx = t;
+        for (int i = k; i < data.Length; ++i) {
+            t += data[i];
+            t -= data[i - k];
+            mx = Math.Max(mx, t);
+        }
+        return k - mx;
+    }
+}
+```
+
 <!-- tabs:end -->
 
 <!-- end -->

solution/1100-1199/1151.Minimum Swaps to Group All 1's Together/Solution.cs

@@ -0,0 +1,13 @@
+public class Solution {
+    public int MinSwaps(int[] data) {
+        int k = data.Count(x => x == 1);
+        int t = data.Take(k).Sum();
+        int mx = t;
+        for (int i = k; i < data.Length; ++i) {
+            t += data[i];
+            t -= data[i - k];
+            mx = Math.Max(mx, t);
+        }
+        return k - mx;
+    }
+}

solution/1100-1199/1151.Minimum Swaps to Group All 1's Together/Solution.py

@@ -1,8 +1,7 @@
 class Solution:
     def minSwaps(self, data: List[int]) -> int:
         k = data.count(1)
-        t = sum(data[:k])
-        mx = t
+        mx = t = sum(data[:k])
         for i in range(k, len(data)):
             t += data[i]
             t -= data[i - k]

solution/1300-1399/1361.Validate Binary Tree Nodes/README.md

@@ -20,33 +20,28 @@
 
 <p><strong>示例 1：</strong></p>
 
-<p><strong><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1300-1399/1361.Validate%20Binary%20Tree%20Nodes/images/1503_ex1.png" style="height: 287px; width: 195px;"></strong></p>
+<p><strong><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1300-1399/1361.Validate%20Binary%20Tree%20Nodes/images/1503_ex1.png" style="height: 287px; width: 195px;" /></strong></p>
 
-<pre><strong>输入：</strong>n = 4, leftChild = [1,-1,3,-1], rightChild = [2,-1,-1,-1]
+<pre>
+<strong>输入：</strong>n = 4, leftChild = [1,-1,3,-1], rightChild = [2,-1,-1,-1]
 <strong>输出：</strong>true
 </pre>
 
 <p><strong>示例 2：</strong></p>
 
-<p><strong><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1300-1399/1361.Validate%20Binary%20Tree%20Nodes/images/1503_ex2.png" style="height: 272px; width: 183px;"></strong></p>
+<p><strong><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1300-1399/1361.Validate%20Binary%20Tree%20Nodes/images/1503_ex2.png" style="height: 272px; width: 183px;" /></strong></p>
 
-<pre><strong>输入：</strong>n = 4, leftChild = [1,-1,3,-1], rightChild = [2,3,-1,-1]
+<pre>
+<strong>输入：</strong>n = 4, leftChild = [1,-1,3,-1], rightChild = [2,3,-1,-1]
 <strong>输出：</strong>false
 </pre>
 
 <p><strong>示例 3：</strong></p>
 
-<p><strong><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1300-1399/1361.Validate%20Binary%20Tree%20Nodes/images/1503_ex3.png" style="height: 174px; width: 82px;"></strong></p>
+<p><strong><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1300-1399/1361.Validate%20Binary%20Tree%20Nodes/images/1503_ex3.png" style="height: 174px; width: 82px;" /></strong></p>
 
-<pre><strong>输入：</strong>n = 2, leftChild = [1,0], rightChild = [-1,-1]
-<strong>输出：</strong>false
-</pre>
-
-<p><strong>示例 4：</strong></p>
-
-<p><strong><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1300-1399/1361.Validate%20Binary%20Tree%20Nodes/images/1503_ex4.png" style="height: 191px; width: 470px;"></strong></p>
-
-<pre><strong>输入：</strong>n = 6, leftChild = [1,-1,-1,4,-1,-1], rightChild = [2,-1,-1,5,-1,-1]
+<pre>
+<strong>输入：</strong>n = 2, leftChild = [1,0], rightChild = [-1,-1]
 <strong>输出：</strong>false
 </pre>
 
@@ -55,8 +50,8 @@
 <p><strong>提示：</strong></p>
 
 <ul>
-	<li><code>1 &lt;= n &lt;= 10^4</code></li>
-	<li><code>leftChild.length == rightChild.length == n</code></li>
+	<li><code>n == leftChild.length == rightChild.length</code></li>
+	<li><code>1 &lt;= n &lt;= 10<sup>4</sup></code></li>
 	<li><code>-1 &lt;= leftChild[i], rightChild[i] &lt;= n - 1</code></li>
 </ul>
 

solution/1300-1399/1385.Find the Distance Value Between Two Arrays/README_EN.md

@@ -8,7 +8,7 @@
 
 <p>Given two integer arrays <code>arr1</code> and <code>arr2</code>, and the integer <code>d</code>, <em>return the distance value between the two arrays</em>.</p>
 
-<p>The distance value is defined as the number of elements <code>arr1[i]</code> such that there is not any element <code>arr2[j]</code> where <code>|arr1[i]-arr2[j]| &lt;= d</code>.</p>
+<p>The distance value is defined as the number of elements <code>arr1[i]</code> such that there is not any element <code>arr2[j]</code> where <code>|arr1[i]-arr2[j]| &lt; d</code>.</p>
 
 <p>&nbsp;</p>
 <p><strong class="example">Example 1:</strong></p>