feat: update lc problems

solution/0100-0199/0195.Tenth Line/README_EN.md

@@ -23,26 +23,41 @@ tags:
 <p>Assume that <code>file.txt</code> has the following content:</p>
 
 <pre>
+
 Line 1
+
 Line 2
+
 Line 3
+
 Line 4
+
 Line 5
+
 Line 6
+
 Line 7
+
 Line 8
+
 Line 9
+
 Line 10
+
 </pre>
 
 <p>Your script should output the tenth line, which is:</p>
 
 <pre>
+
 Line 10
+
 </pre>
 
 <div class="spoilers"><b>Note:</b><br />
+
 1. If the file contains less than 10 lines, what should you output?<br />
+
 2. There&#39;s at least three different solutions. Try to explore all possibilities.</div>
 
 <!-- description:end -->

solution/0600-0699/0636.Exclusive Time of Functions/README.md

@@ -73,7 +73,7 @@ tags:
 
 <ul>
 	<li><code>1 &lt;= n &lt;= 100</code></li>
-	<li><code>1 &lt;= logs.length &lt;= 500</code></li>
+	<li><code>2 &lt;= logs.length &lt;= 500</code></li>
 	<li><code>0 &lt;= function_id &lt; n</code></li>
 	<li><code>0 &lt;= timestamp &lt;= 10<sup>9</sup></code></li>
 	<li>两个开始事件不会在同一时间戳发生</li>

solution/0600-0699/0636.Exclusive Time of Functions/README_EN.md

@@ -73,7 +73,7 @@ So function 0 spends 2 + 4 + 1 = 7 units of total time executing, and function 1
 
 <ul>
 	<li><code>1 &lt;= n &lt;= 100</code></li>
-	<li><code>1 &lt;= logs.length &lt;= 500</code></li>
+	<li><code>2 &lt;= logs.length &lt;= 500</code></li>
 	<li><code>0 &lt;= function_id &lt; n</code></li>
 	<li><code>0 &lt;= timestamp &lt;= 10<sup>9</sup></code></li>
 	<li>No two start events will happen at the same timestamp.</li>

solution/0700-0799/0790.Domino and Tromino Tiling/README_EN.md

@@ -28,7 +28,7 @@ tags:
 <pre>
 <strong>Input:</strong> n = 3
 <strong>Output:</strong> 5
-<strong>Explanation:</strong> The five different ways are show above.
+<strong>Explanation:</strong> The five different ways are shown above.
 </pre>
 
 <p><strong class="example">Example 2:</strong></p>

solution/0700-0799/0799.Champagne Tower/README_EN.md

@@ -27,35 +27,51 @@ tags:
 <p>Now after pouring some non-negative integer cups of champagne, return how full the <code>j<sup>th</sup></code> glass in the <code>i<sup>th</sup></code> row is (both <code>i</code> and <code>j</code> are 0-indexed.)</p>
 
 <p>&nbsp;</p>
+
 <p><strong class="example">Example 1:</strong></p>
 
 <pre>
+
 <strong>Input:</strong> poured = 1, query_row = 1, query_glass = 1
+
 <strong>Output:</strong> 0.00000
+
 <strong>Explanation:</strong> We poured 1 cup of champange to the top glass of the tower (which is indexed as (0, 0)). There will be no excess liquid so all the glasses under the top glass will remain empty.
+
 </pre>
 
 <p><strong class="example">Example 2:</strong></p>
 
 <pre>
+
 <strong>Input:</strong> poured = 2, query_row = 1, query_glass = 1
+
 <strong>Output:</strong> 0.50000
+
 <strong>Explanation:</strong> We poured 2 cups of champange to the top glass of the tower (which is indexed as (0, 0)). There is one cup of excess liquid. The glass indexed as (1, 0) and the glass indexed as (1, 1) will share the excess liquid equally, and each will get half cup of champange.
+
 </pre>
 
 <p><strong class="example">Example 3:</strong></p>
 
 <pre>
+
 <strong>Input:</strong> poured = 100000009, query_row = 33, query_glass = 17
+
 <strong>Output:</strong> 1.00000
+
 </pre>
 
 <p>&nbsp;</p>
+
 <p><strong>Constraints:</strong></p>
 
 <ul>
-	<li><code>0 &lt;=&nbsp;poured &lt;= 10<sup>9</sup></code></li>
-	<li><code>0 &lt;= query_glass &lt;= query_row&nbsp;&lt; 100</code></li>
+
+    <li><code>0 &lt;=&nbsp;poured &lt;= 10<sup>9</sup></code></li>
+
+    <li><code>0 &lt;= query_glass &lt;= query_row&nbsp;&lt; 100</code></li>
+
 </ul>
 
 <!-- description:end -->

solution/0800-0899/0819.Most Common Word/README.md

@@ -23,6 +23,8 @@ tags:
 
 <p><code>paragraph</code> 中的单词 <strong>不区分大小写</strong> ，答案应以 <strong>小写 </strong>形式返回。</p>
 
+<p><b>注意</b>&nbsp;单词不包含标点符号。</p>
+
 <p>&nbsp;</p>
 
 <p><strong class="example">示例 1：</strong></p>

solution/1100-1199/1108.Defanging an IP Address/README_EN.md

@@ -23,18 +23,29 @@ tags:
 <p>A <em>defanged&nbsp;IP address</em>&nbsp;replaces every period <code>&quot;.&quot;</code> with <code>&quot;[.]&quot;</code>.</p>
 
 <p>&nbsp;</p>
+
 <p><strong class="example">Example 1:</strong></p>
+
 <pre><strong>Input:</strong> address = "1.1.1.1"
+
 <strong>Output:</strong> "1[.]1[.]1[.]1"
+
 </pre><p><strong class="example">Example 2:</strong></p>
+
 <pre><strong>Input:</strong> address = "255.100.50.0"
+
 <strong>Output:</strong> "255[.]100[.]50[.]0"
+
 </pre>
+
 <p>&nbsp;</p>
+
 <p><strong>Constraints:</strong></p>
 
 <ul>
-	<li>The given <code>address</code> is a valid IPv4 address.</li>
+
+    <li>The given <code>address</code> is a valid IPv4 address.</li>
+
 </ul>
 
 <!-- description:end -->

solution/1100-1199/1111.Maximum Nesting Depth of Two Valid Parentheses Strings/README_EN.md

@@ -22,17 +22,25 @@ tags:
 <p>A string is a <em>valid parentheses string</em>&nbsp;(denoted VPS) if and only if it consists of <code>&quot;(&quot;</code> and <code>&quot;)&quot;</code> characters only, and:</p>
 
 <ul>
-	<li>It is the empty string, or</li>
-	<li>It can be written as&nbsp;<code>AB</code>&nbsp;(<code>A</code>&nbsp;concatenated with&nbsp;<code>B</code>), where&nbsp;<code>A</code>&nbsp;and&nbsp;<code>B</code>&nbsp;are VPS&#39;s, or</li>
-	<li>It can be written as&nbsp;<code>(A)</code>, where&nbsp;<code>A</code>&nbsp;is a VPS.</li>
+
+    <li>It is the empty string, or</li>
+
+    <li>It can be written as&nbsp;<code>AB</code>&nbsp;(<code>A</code>&nbsp;concatenated with&nbsp;<code>B</code>), where&nbsp;<code>A</code>&nbsp;and&nbsp;<code>B</code>&nbsp;are VPS&#39;s, or</li>
+
+    <li>It can be written as&nbsp;<code>(A)</code>, where&nbsp;<code>A</code>&nbsp;is a VPS.</li>
+
 </ul>
 
 <p>We can&nbsp;similarly define the <em>nesting depth</em> <code>depth(S)</code> of any VPS <code>S</code> as follows:</p>
 
 <ul>
-	<li><code>depth(&quot;&quot;) = 0</code></li>
-	<li><code>depth(A + B) = max(depth(A), depth(B))</code>, where <code>A</code> and <code>B</code> are VPS&#39;s</li>
-	<li><code>depth(&quot;(&quot; + A + &quot;)&quot;) = 1 + depth(A)</code>, where <code>A</code> is a VPS.</li>
+
+    <li><code>depth(&quot;&quot;) = 0</code></li>
+
+    <li><code>depth(A + B) = max(depth(A), depth(B))</code>, where <code>A</code> and <code>B</code> are VPS&#39;s</li>
+
+    <li><code>depth(&quot;(&quot; + A + &quot;)&quot;) = 1 + depth(A)</code>, where <code>A</code> is a VPS.</li>
+
 </ul>
 
 <p>For example,&nbsp; <code>&quot;&quot;</code>,&nbsp;<code>&quot;()()&quot;</code>, and&nbsp;<code>&quot;()(()())&quot;</code>&nbsp;are VPS&#39;s (with nesting depths 0, 1, and 2), and <code>&quot;)(&quot;</code> and <code>&quot;(()&quot;</code> are not VPS&#39;s.</p>

solution/1100-1199/1138.Alphabet Board Path/README_EN.md

@@ -28,31 +28,49 @@ tags:
 <p>We may make the following moves:</p>
 
 <ul>
-	<li><code>&#39;U&#39;</code> moves our position up one row, if the position exists on the board;</li>
-	<li><code>&#39;D&#39;</code> moves our position down one row, if the position exists on the board;</li>
-	<li><code>&#39;L&#39;</code> moves our position left one column, if the position exists on the board;</li>
-	<li><code>&#39;R&#39;</code> moves our position right one column, if the position exists on the board;</li>
-	<li><code>&#39;!&#39;</code>&nbsp;adds the character <code>board[r][c]</code> at our current position <code>(r, c)</code>&nbsp;to the&nbsp;answer.</li>
+
+    <li><code>&#39;U&#39;</code> moves our position up one row, if the position exists on the board;</li>
+
+    <li><code>&#39;D&#39;</code> moves our position down one row, if the position exists on the board;</li>
+
+    <li><code>&#39;L&#39;</code> moves our position left one column, if the position exists on the board;</li>
+
+    <li><code>&#39;R&#39;</code> moves our position right one column, if the position exists on the board;</li>
+
+    <li><code>&#39;!&#39;</code>&nbsp;adds the character <code>board[r][c]</code> at our current position <code>(r, c)</code>&nbsp;to the&nbsp;answer.</li>
+
 </ul>
 
 <p>(Here, the only positions that exist on the board are positions with letters on them.)</p>
 
 <p>Return a sequence of moves that makes our answer equal to <code>target</code>&nbsp;in the minimum number of moves.&nbsp; You may return any path that does so.</p>
 
 <p>&nbsp;</p>
+
 <p><strong class="example">Example 1:</strong></p>
+
 <pre><strong>Input:</strong> target = "leet"
+
 <strong>Output:</strong> "DDR!UURRR!!DDD!"
+
 </pre><p><strong class="example">Example 2:</strong></p>
+
 <pre><strong>Input:</strong> target = "code"
+
 <strong>Output:</strong> "RR!DDRR!UUL!R!"
+
 </pre>
+
 <p>&nbsp;</p>
+
 <p><strong>Constraints:</strong></p>
 
 <ul>
-	<li><code>1 &lt;= target.length &lt;= 100</code></li>
-	<li><code>target</code> consists only of English lowercase letters.</li>
+
+    <li><code>1 &lt;= target.length &lt;= 100</code></li>
+
+    <li><code>target</code> consists only of English lowercase letters.</li>
+
 </ul>
 
 <!-- description:end -->

solution/1100-1199/1139.Largest 1-Bordered Square/README_EN.md

@@ -23,27 +23,39 @@ tags:
 <p>Given a 2D <code>grid</code> of <code>0</code>s and <code>1</code>s, return the number of elements in&nbsp;the largest <strong>square</strong>&nbsp;subgrid that has all <code>1</code>s on its <strong>border</strong>, or <code>0</code> if such a subgrid&nbsp;doesn&#39;t exist in the <code>grid</code>.</p>
 
 <p>&nbsp;</p>
+
 <p><strong class="example">Example 1:</strong></p>
 
 <pre>
+
 <strong>Input:</strong> grid = [[1,1,1],[1,0,1],[1,1,1]]
+
 <strong>Output:</strong> 9
+
 </pre>
 
 <p><strong class="example">Example 2:</strong></p>
 
 <pre>
+
 <strong>Input:</strong> grid = [[1,1,0,0]]
+
 <strong>Output:</strong> 1
+
 </pre>
 
 <p>&nbsp;</p>
+
 <p><strong>Constraints:</strong></p>
 
 <ul>
-	<li><code>1 &lt;= grid.length &lt;= 100</code></li>
-	<li><code>1 &lt;= grid[0].length &lt;= 100</code></li>
-	<li><code>grid[i][j]</code> is <code>0</code> or <code>1</code></li>
+
+    <li><code>1 &lt;= grid.length &lt;= 100</code></li>
+
+    <li><code>1 &lt;= grid[0].length &lt;= 100</code></li>
+
+    <li><code>grid[i][j]</code> is <code>0</code> or <code>1</code></li>
+
 </ul>
 
 <!-- description:end -->

solution/1100-1199/1160.Find Words That Can Be Formed by Characters/README.md

@@ -21,13 +21,11 @@ tags:
 
 <!-- description:start -->
 
-<p>给你一份『词汇表』（字符串数组）&nbsp;<code>words</code>&nbsp;和一张『字母表』（字符串）&nbsp;<code>chars</code>。</p>
+<p>给定一个字符串数组&nbsp;<code>words</code>&nbsp;和一个字符串 <code>chars</code>。</p>
 
-<p>假如你可以用&nbsp;<code>chars</code>&nbsp;中的『字母』（字符）拼写出 <code>words</code>&nbsp;中的某个『单词』（字符串），那么我们就认为你掌握了这个单词。</p>
+<p>如果字符串可以由 <code>chars</code> 中的字符组成（每个字符在 <strong>每个</strong>&nbsp;<code>words</code> 中只能使用一次），则认为它是好的。</p>
 
-<p>注意：每次拼写（指拼写词汇表中的一个单词）时，<code>chars</code> 中的每个字母都只能用一次。</p>
-
-<p>返回词汇表&nbsp;<code>words</code>&nbsp;中你掌握的所有单词的 <strong>长度之和</strong>。</p>
+<p>返回&nbsp;<code>words</code>&nbsp;中所有好的字符串的长度之和。</p>
 
 <p>&nbsp;</p>
 
@@ -56,7 +54,7 @@ tags:
 <ul>
 	<li><code>1 &lt;= words.length &lt;= 1000</code></li>
 	<li><code>1 &lt;= words[i].length, chars.length&nbsp;&lt;= 100</code></li>
-	<li>所有字符串中都仅包含小写英文字母</li>
+	<li><code>words[i]</code>&nbsp;和&nbsp;<code>chars</code> 中都仅包含小写英文字母</li>
 </ul>
 
 <!-- description:end -->

solution/1100-1199/1184.Distance Between Bus Stops/README_EN.md

@@ -25,13 +25,17 @@ tags:
 <p>Return the shortest distance between the given&nbsp;<code>start</code>&nbsp;and <code>destination</code>&nbsp;stops.</p>
 
 <p>&nbsp;</p>
+
 <p><strong class="example">Example 1:</strong></p>
 
 <p><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1100-1199/1184.Distance%20Between%20Bus%20Stops/images/untitled-diagram-1.jpg" style="width: 388px; height: 240px;" /></p>
 
 <pre>
+
 <strong>Input:</strong> distance = [1,2,3,4], start = 0, destination = 1
+
 <strong>Output:</strong> 1
+
 <strong>Explanation:</strong> Distance between 0 and 1 is 1 or 9, minimum is 1.</pre>
 
 <p>&nbsp;</p>
@@ -41,9 +45,13 @@ tags:
 <p><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1100-1199/1184.Distance%20Between%20Bus%20Stops/images/untitled-diagram-1-1.jpg" style="width: 388px; height: 240px;" /></p>
 
 <pre>
+
 <strong>Input:</strong> distance = [1,2,3,4], start = 0, destination = 2
+
 <strong>Output:</strong> 3
+
 <strong>Explanation:</strong> Distance between 0 and 2 is 3 or 7, minimum is 3.
+
 </pre>
 
 <p>&nbsp;</p>
@@ -53,19 +61,29 @@ tags:
 <p><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1100-1199/1184.Distance%20Between%20Bus%20Stops/images/untitled-diagram-1-2.jpg" style="width: 388px; height: 240px;" /></p>
 
 <pre>
+
 <strong>Input:</strong> distance = [1,2,3,4], start = 0, destination = 3
+
 <strong>Output:</strong> 4
+
 <strong>Explanation:</strong> Distance between 0 and 3 is 6 or 4, minimum is 4.
+
 </pre>
 
 <p>&nbsp;</p>
+
 <p><strong>Constraints:</strong></p>
 
 <ul>
-	<li><code>1 &lt;= n&nbsp;&lt;= 10^4</code></li>
-	<li><code>distance.length == n</code></li>
-	<li><code>0 &lt;= start, destination &lt; n</code></li>
-	<li><code>0 &lt;= distance[i] &lt;= 10^4</code></li>
+
+    <li><code>1 &lt;= n&nbsp;&lt;= 10^4</code></li>
+
+    <li><code>distance.length == n</code></li>
+
+    <li><code>0 &lt;= start, destination &lt; n</code></li>
+
+    <li><code>0 &lt;= distance[i] &lt;= 10^4</code></li>
+
 </ul>
 
 <!-- description:end -->