feat: update lc problems

Author: yanglbme

solution/0000-0099/0005.Longest Palindromic Substring/README_EN.md

@@ -6,6 +6,8 @@
 
 <p>Given a string <code>s</code>, return <em>the longest palindromic substring</em> in <code>s</code>.</p>
 
+<p>A string is called a palindrome string if the reverse of that string is the same as the original string.</p>
+
 <p>&nbsp;</p>
 <p><strong>Example 1:</strong></p>
 

solution/0600-0699/0672.Bulb Switcher II/README.md

@@ -52,7 +52,7 @@
 <strong>解释：</strong>状态可以是：
 - 按压开关 1 ，[关, 关, 关]
 - 按压开关 2 ，[关, 开, 关]
-- 按压开关 3 ，[开, 关, 开]
+- 按压开关 3 ，[开, 开, 开]
 - 按压开关 4 ，[关, 开, 开]
 </pre>
 

solution/0600-0699/0682.Baseball Game/README_EN.md

@@ -4,18 +4,36 @@
 
 ## Description
 
-<p>You are keeping score for a baseball game with strange rules. The game consists of several rounds, where the scores of past rounds may affect future rounds&#39; scores.</p>
+<p>You are keeping the scores for a baseball game with strange rules. At the beginning of the game, you start with an empty record.</p>
 
-<p>At the beginning of the game, you start with an empty record. You are given a list of strings <code>ops</code>, where <code>ops[i]</code> is the <code>i<sup>th</sup></code> operation you must apply to the record and is one of the following:</p>
+<p>You are given a list of strings <code>operations</code>, where <code>operations[i]</code> is the <code>i<sup>th</sup></code> operation you must apply to the record and is one of the following:</p>
 
-<ol>
-	<li>An integer <code>x</code> - Record a new score of <code>x</code>.</li>
-	<li><code>&quot;+&quot;</code> - Record a new score that is the sum of the previous two scores. It is guaranteed there will always be two previous scores.</li>
-	<li><code>&quot;D&quot;</code> - Record a new score that is double the previous score. It is guaranteed there will always be a previous score.</li>
-	<li><code>&quot;C&quot;</code> - Invalidate the previous score, removing it from the record. It is guaranteed there will always be a previous score.</li>
-</ol>
+<ul>
+	<li>An integer <code>x</code>.
+    <ul>
+    	<li>Record a new score of <code>x</code>.</li>
+    </ul>
+    </li>
+    <li><code>&#39;+&#39;</code>.
+    <ul>
+    	<li>Record a new score that is the sum of the previous two scores.</li>
+    </ul>
+    </li>
+    <li><code>&#39;D&#39;</code>.
+    <ul>
+    	<li>Record a new score that is the double of the previous score.</li>
+    </ul>
+    </li>
+    <li><code>&#39;C&#39;</code>.
+    <ul>
+    	<li>Invalidate the previous score, removing it from the record.</li>
+    </ul>
+    </li>
+</ul>
+
+<p>Return <em>the sum of all the scores on the record after applying all the operations</em>.</p>
 
-<p>Return <em>the sum of all the scores on the record</em>. The test cases are generated so that the answer fits in a 32-bit integer.</p>
+<p>The test cases are generated such that the answer and all intermediate calculations fit in a <strong>32-bit</strong> integer and that all operations are valid.</p>
 
 <p>&nbsp;</p>
 <p><strong>Example 1:</strong></p>
@@ -64,8 +82,8 @@ Since the record is empty, the total sum is 0.
 <p><strong>Constraints:</strong></p>
 
 <ul>
-	<li><code>1 &lt;= ops.length &lt;= 1000</code></li>
-	<li><code>ops[i]</code> is <code>&quot;C&quot;</code>, <code>&quot;D&quot;</code>, <code>&quot;+&quot;</code>, or a string representing an integer in the range <code>[-3 * 10<sup>4</sup>, 3 * 10<sup>4</sup>]</code>.</li>
+	<li><code>1 &lt;= operations.length &lt;= 1000</code></li>
+	<li><code>operations[i]</code> is <code>&quot;C&quot;</code>, <code>&quot;D&quot;</code>, <code>&quot;+&quot;</code>, or a string representing an integer in the range <code>[-3 * 10<sup>4</sup>, 3 * 10<sup>4</sup>]</code>.</li>
 	<li>For operation <code>&quot;+&quot;</code>, there will always be at least two previous scores on the record.</li>
 	<li>For operations <code>&quot;C&quot;</code> and <code>&quot;D&quot;</code>, there will always be at least one previous score on the record.</li>
 </ul>

solution/0800-0899/0850.Rectangle Area II/README.md

@@ -226,7 +226,7 @@ class Solution {
             m.put(v, i);
             nums[i++] = v;
         }
-        
+
         SegmentTree tree = new SegmentTree(nums);
         long ans = 0;
         for (i = 0; i < segs.length; ++i) {

solution/0800-0899/0850.Rectangle Area II/README_EN.md

@@ -197,7 +197,7 @@ class Solution {
             m.put(v, i);
             nums[i++] = v;
         }
-        
+
         SegmentTree tree = new SegmentTree(nums);
         long ans = 0;
         for (i = 0; i < segs.length; ++i) {

solution/0800-0899/0850.Rectangle Area II/Solution.py

@@ -31,13 +31,11 @@ def modify(self, u, l, r, k):
 
     def pushup(self, u):
         if self.tr[u].cnt:
-            self.tr[u].length = self.nums[self.tr[u].r + 1] - \
-                self.nums[self.tr[u].l]
+            self.tr[u].length = self.nums[self.tr[u].r + 1] - self.nums[self.tr[u].l]
         elif self.tr[u].l == self.tr[u].r:
             self.tr[u].length = 0
         else:
-            self.tr[u].length = self.tr[u << 1].length + \
-                self.tr[u << 1 | 1].length
+            self.tr[u].length = self.tr[u << 1].length + self.tr[u << 1 | 1].length
 
     @property
     def length(self):

solution/0800-0899/0854.K-Similar Strings/README.md

@@ -6,9 +6,9 @@
 
 <!-- 这里写题目描述 -->
 
-<p>字符串 <code>s1</code> 和 <code>s2</code> 是 <strong><code>k</code> 相似</strong> 的(对于某些非负整数 <code>k</code> )，如果我们可以交换 <code>s1</code> 中两个字母的位置正好 <code>k</code> 次，使结果字符串等于 <code>s2</code> 。</p>
+<p>对于某些非负整数 <code>k</code> ，如果交换 <code>s1</code> 中两个字母的位置恰好 <code>k</code> 次，能够使结果字符串等于 <code>s2</code> ，则认为字符串 <code>s1</code> 和 <code>s2</code> 的<strong> 相似度为 </strong><code>k</code><strong> </strong><strong>。</strong></p>
 
-<p>给定两个字谜游戏 <code>s1</code> 和 <code>s2</code> ，返回 <code>s1</code> 和 <code>s2</code> 与 <strong><code>k</code> 相似&nbsp;</strong>的最小 <code>k</code> 。</p>
+<p>给你两个字母异位词 <code>s1</code> 和 <code>s2</code> ，返回 <code>s1</code> 和 <code>s2</code> 的相似度 <code>k</code><strong> </strong>的最小值。</p>
 
 <p>&nbsp;</p>
 
@@ -34,7 +34,7 @@
 	<li><code>1 &lt;= s1.length &lt;= 20</code></li>
 	<li><code>s2.length == s1.length</code></li>
 	<li><code>s1</code>&nbsp;和&nbsp;<code>s2</code>&nbsp;&nbsp;只包含集合&nbsp;<code>{'a', 'b', 'c', 'd', 'e', 'f'}</code>&nbsp;中的小写字母</li>
-	<li><code>s2</code> 是 <code>s1</code> 的一个字谜</li>
+	<li><code>s2</code> 是 <code>s1</code> 的一个字母异位词</li>
 </ul>
 
 ## 解法

solution/1100-1199/1106.Parsing A Boolean Expression/README_EN.md

@@ -4,47 +4,56 @@
 
 ## Description
 
-<p>Return the result of evaluating a given boolean <code>expression</code>, represented as a string.</p>
-
-<p>An expression can either be:</p>
+<p>A <strong>boolean expression</strong> is an expression that evaluates to either <code>true</code> or <code>false</code>. It can be in one of the following shapes:</p>
 
 <ul>
-	<li><code>&quot;t&quot;</code>, evaluating to <code>True</code>;</li>
-	<li><code>&quot;f&quot;</code>, evaluating to <code>False</code>;</li>
-	<li><code>&quot;!(expr)&quot;</code>, evaluating to the logical NOT of the inner expression <code>expr</code>;</li>
-	<li><code>&quot;&amp;(expr1,expr2,...)&quot;</code>, evaluating to the logical AND of 2 or more inner expressions <code>expr1, expr2, ...</code>;</li>
-	<li><code>&quot;|(expr1,expr2,...)&quot;</code>, evaluating to the logical OR of 2 or more inner expressions <code>expr1, expr2, ...</code></li>
+	<li><code>&#39;t&#39;</code> that evaluates to <code>true</code>.</li>
+	<li><code>&#39;f&#39;</code> that evaluates to <code>false</code>.</li>
+	<li><code>&#39;!(subExpr)&#39;</code> that evaluates to <strong>the logical NOT</strong> of the inner expression <code>subExpr</code>.</li>
+	<li><code>&#39;&amp;(subExpr<sub>1</sub>, subExpr<sub>2</sub>, ..., subExpr<sub>n</sub>)&#39;</code> that evaluates to <strong>the logical AND</strong> of the inner expressions <code>subExpr<sub>1</sub>, subExpr<sub>2</sub>, ..., subExpr<sub>n</sub></code> where <code>n &gt;= 1</code>.</li>
+	<li><code>&#39;|(subExpr<sub>1</sub>, subExpr<sub>2</sub>, ..., subExpr<sub>n</sub>)&#39;</code> that evaluates to <strong>the logical OR</strong> of the inner expressions <code>subExpr<sub>1</sub>, subExpr<sub>2</sub>, ..., subExpr<sub>n</sub></code> where <code>n &gt;= 1</code>.</li>
 </ul>
 
+<p>Given a string <code>expression</code> that represents a <strong>boolean expression</strong>, return <em>the evaluation of that expression</em>.</p>
+
+<p>It is <strong>guaranteed</strong> that the given expression is valid and follows the given rules.</p>
+
 <p>&nbsp;</p>
-<p><strong>Example 1:</strong></p>
+<p><strong class="example">Example 1:</strong></p>
 
 <pre>
-<strong>Input:</strong> expression = &quot;!(f)&quot;
-<strong>Output:</strong> true
+<strong>Input:</strong> expression = &quot;&amp;(|(f))&quot;
+<strong>Output:</strong> false
+<strong>Explanation:</strong> 
+First, evaluate |(f) --&gt; f. The expression is now &quot;&amp;(f)&quot;.
+Then, evaluate &amp;(f) --&gt; f. The expression is now &quot;f&quot;.
+Finally, return false.
 </pre>
 
-<p><strong>Example 2:</strong></p>
+<p><strong class="example">Example 2:</strong></p>
 
 <pre>
-<strong>Input:</strong> expression = &quot;|(f,t)&quot;
+<strong>Input:</strong> expression = &quot;|(f,f,f,t)&quot;
 <strong>Output:</strong> true
+<strong>Explanation:</strong> The evaluation of (false OR false OR false OR true) is true.
 </pre>
 
-<p><strong>Example 3:</strong></p>
+<p><strong class="example">Example 3:</strong></p>
 
 <pre>
-<strong>Input:</strong> expression = &quot;&amp;(t,f)&quot;
-<strong>Output:</strong> false
+<strong>Input:</strong> expression = &quot;!(&amp;(f,t))&quot;
+<strong>Output:</strong> true
+<strong>Explanation:</strong> 
+First, evaluate &amp;(f,t) --&gt; (false AND true) --&gt; false --&gt; f. The expression is now &quot;!(f)&quot;.
+Then, evaluate !(f) --&gt; NOT false --&gt; true. We return true.
 </pre>
 
 <p>&nbsp;</p>
 <p><strong>Constraints:</strong></p>
 
 <ul>
 	<li><code>1 &lt;= expression.length &lt;= 2 * 10<sup>4</sup></code></li>
-	<li><code>expression[i]</code> consists of characters in <code>{&#39;(&#39;, &#39;)&#39;, &#39;&amp;&#39;, &#39;|&#39;, &#39;!&#39;, &#39;t&#39;, &#39;f&#39;, &#39;,&#39;}</code>.</li>
-	<li><code>expression</code> is a valid expression representing a boolean, as given in the description.</li>
+	<li>expression[i] is one following characters: <code>&#39;(&#39;</code>, <code>&#39;)&#39;</code>, <code>&#39;&amp;&#39;</code>, <code>&#39;|&#39;</code>, <code>&#39;!&#39;</code>, <code>&#39;t&#39;</code>, <code>&#39;f&#39;</code>, and <code>&#39;,&#39;</code>.</li>
 </ul>
 
 ## Solutions

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

@@ -6,7 +6,7 @@
 
 <!-- 这里写题目描述 -->
 
-<p>给你一个整数&nbsp;<code>n</code>&nbsp;表示某所大学里课程的数目，编号为&nbsp;<code>1</code>&nbsp;到&nbsp;<code>n</code>&nbsp;，数组&nbsp;<code>dependencies</code>&nbsp;中，&nbsp;<code>dependencies[i] = [x<sub>i</sub>, y<sub>i</sub>]</code>&nbsp; 表示一个先修课的关系，也就是课程&nbsp;<code>x<sub>i</sub></code>&nbsp;必须在课程&nbsp;<code>y<sub>i</sub></code><sub>&nbsp;</sub>之前上。同时你还有一个整数&nbsp;<code>k</code>&nbsp;。</p>
+<p>给你一个整数&nbsp;<code>n</code>&nbsp;表示某所大学里课程的数目，编号为&nbsp;<code>1</code>&nbsp;到&nbsp;<code>n</code>&nbsp;，数组&nbsp;<code>relations</code>&nbsp;中，&nbsp;<code>relations[i] = [x<sub>i</sub>, y<sub>i</sub>]</code>&nbsp; 表示一个先修课的关系，也就是课程&nbsp;<code>x<sub>i</sub></code>&nbsp;必须在课程&nbsp;<code>y<sub>i</sub></code><sub>&nbsp;</sub>之前上。同时你还有一个整数&nbsp;<code>k</code>&nbsp;。</p>
 
 <p>在一个学期中，你 <strong>最多</strong>&nbsp;可以同时上 <code>k</code>&nbsp;门课，前提是这些课的先修课在之前的学期里已经上过了。</p>
 
@@ -16,25 +16,28 @@
 
 <p><strong>示例 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="height: 164px; width: 300px;"></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="height: 164px; width: 300px;" /></strong></p>
 
-<pre><strong>输入：</strong>n = 4, dependencies = [[2,1],[3,1],[1,4]], k = 2
+<pre>
+<strong>输入：</strong>n = 4, relations = [[2,1],[3,1],[1,4]], k = 2
 <strong>输出：</strong>3 
 <strong>解释：</strong>上图展示了题目输入的图。在第一个学期中，我们可以上课程 2 和课程 3 。然后第二个学期上课程 1 ，第三个学期上课程 4 。
 </pre>
 
 <p><strong>示例 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="height: 234px; width: 300px;"></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="height: 234px; width: 300px;" /></strong></p>
 
-<pre><strong>输入：</strong>n = 5, dependencies = [[2,1],[3,1],[4,1],[1,5]], k = 2
+<pre>
+<strong>输入：</strong>n = 5, relations = [[2,1],[3,1],[4,1],[1,5]], k = 2
 <strong>输出：</strong>4 
 <strong>解释：</strong>上图展示了题目输入的图。一个最优方案是：第一学期上课程 2 和 3，第二学期上课程 4 ，第三学期上课程 1 ，第四学期上课程 5 。
 </pre>
 
 <p><strong>示例 3：</strong></p>
 
-<pre><strong>输入：</strong>n = 11, dependencies = [], k = 2
+<pre>
+<strong>输入：</strong>n = 11, relations = [], k = 2
 <strong>输出：</strong>6
 </pre>
 
@@ -45,11 +48,11 @@
 <ul>
 	<li><code>1 &lt;= n &lt;= 15</code></li>
 	<li><code>1 &lt;= k &lt;= n</code></li>
-	<li><code>0 &lt;=&nbsp;dependencies.length &lt;= n * (n-1) / 2</code></li>
-	<li><code>dependencies[i].length == 2</code></li>
+	<li><code>0 &lt;=&nbsp;relations.length &lt;= n * (n-1) / 2</code></li>
+	<li><code>relations[i].length == 2</code></li>
 	<li><code>1 &lt;= x<sub>i</sub>, y<sub>i</sub>&nbsp;&lt;= n</code></li>
 	<li><code>x<sub>i</sub> != y<sub>i</sub></code></li>
-	<li>所有先修关系都是不同的，也就是说&nbsp;<code>dependencies[i] != dependencies[j]</code>&nbsp;。</li>
+	<li>所有先修关系都是不同的，也就是说&nbsp;<code>relations[i] != relations[j]</code>&nbsp;。</li>
 	<li>题目输入的图是个有向无环图。</li>
 </ul>