doocs · yanglbme · Jan 17, 2025 · Jan 17, 2025
diff --git a/solution/0400-0499/0480.Sliding Window Median/README_EN.md b/solution/0400-0499/0480.Sliding Window Median/README_EN.md
@@ -36,7 +36,7 @@ tags:
 <pre>
 <strong>Input:</strong> nums = [1,3,-1,-3,5,3,6,7], k = 3
 <strong>Output:</strong> [1.00000,-1.00000,-1.00000,3.00000,5.00000,6.00000]
-<strong>Explanation:</strong>
+<strong>Explanation:</strong> 
 Window position                Median
 ---------------                -----
 [<strong>1  3  -1</strong>] -3  5  3  6  7        1

diff --git a/solution/0500-0599/0588.Design In-Memory File System/README.md b/solution/0500-0599/0588.Design In-Memory File System/README.md
@@ -77,6 +77,7 @@ fileSystem.readContentFromFile("/a/b/c/d"); // 返回 "hello"</pre>
 	<li><code>path</code>&nbsp;和&nbsp;<code>filePath</code>&nbsp;都是绝对路径，除非是根目录&nbsp;<code>‘/’</code>&nbsp;自身，其他路径都是以&nbsp;<code>‘/’</code>&nbsp;开头且 <strong>不</strong> 以&nbsp;<code>‘/’</code>&nbsp;结束。</li>
 	<li>你可以假定所有操作的参数都是有效的，即用户不会获取不存在文件的内容，或者获取不存在文件夹和文件的列表。</li>
 	<li>你可以假定所有文件夹名字和文件名字都只包含小写字母，且同一文件夹下不会有相同名字的文件夹或文件。</li>
+	<li>你可以假定&nbsp;<code>addContentToFile</code> 中的文件的父目录都存在。</li>
 	<li><code>1 &lt;= content.length &lt;= 50</code></li>
 	<li><code>ls</code>,&nbsp;<code>mkdir</code>,&nbsp;<code>addContentToFile</code>, and&nbsp;<code>readContentFromFile</code>&nbsp;最多被调用&nbsp;<code>300</code>&nbsp;次</li>
 </ul>

diff --git a/solution/0700-0799/0731.My Calendar II/README_EN.md b/solution/0700-0799/0731.My Calendar II/README_EN.md
@@ -46,9 +46,9 @@ tags:
 
 <strong>Explanation</strong>
 MyCalendarTwo myCalendarTwo = new MyCalendarTwo();
-myCalendarTwo.book(10, 20); // return True, The event can be booked.
-myCalendarTwo.book(50, 60); // return True, The event can be booked.
-myCalendarTwo.book(10, 40); // return True, The event can be double booked.
+myCalendarTwo.book(10, 20); // return True, The event can be booked. 
+myCalendarTwo.book(50, 60); // return True, The event can be booked. 
+myCalendarTwo.book(10, 40); // return True, The event can be double booked. 
 myCalendarTwo.book(5, 15);  // return False, The event cannot be booked, because it would result in a triple booking.
 myCalendarTwo.book(5, 10); // return True, The event can be booked, as it does not use time 10 which is already double booked.
 myCalendarTwo.book(25, 55); // return True, The event can be booked, as the time in [25, 40) will be double booked with the third event, the time [40, 50) will be single booked, and the time [50, 55) will be double booked with the second event.

diff --git a/solution/0900-0999/0916.Word Subsets/README.md b/solution/0900-0999/0916.Word Subsets/README.md
@@ -28,47 +28,36 @@ tags:
 
 <p>如果对 <code>words2</code> 中的每一个单词&nbsp;<code>b</code>，<code>b</code> 都是 <code>a</code> 的子集，那么我们称&nbsp;<code>words1</code> 中的单词 <code>a</code> 是<em> </em><strong>通用单词</strong><em> </em>。</p>
 
-<p>以数组形式返回&nbsp;<code>words1</code> 中所有的通用单词。你可以按 <strong>任意顺序</strong> 返回答案。</p>
+<p>以数组形式返回&nbsp;<code>words1</code> 中所有的 <strong>通用</strong> 单词。你可以按 <strong>任意顺序</strong> 返回答案。</p>
 
 <p>&nbsp;</p>
 
 <ol>
 </ol>
 
-<p><strong>示例 1：</strong></p>
+<p><strong class="example">示例 1：</strong></p>
 
-<pre>
-<strong>输入：</strong>words1 = ["amazon","apple","facebook","google","leetcode"], words2 = ["e","o"]
-<strong>输出：</strong>["facebook","google","leetcode"]
-</pre>
+<div class="example-block">
+<p><span class="example-io"><b>输入：</b>words1 = ["amazon","apple","facebook","google","leetcode"], words2 = ["e","o"]</span></p>
 
-<p><strong>示例 2：</strong></p>
+<p><span class="example-io"><b>输出：</b>["facebook","google","leetcode"]</span></p>
+</div>
 
-<pre>
-<strong>输入：</strong>words1 = ["amazon","apple","facebook","google","leetcode"], words2 = ["l","e"]
-<strong>输出：</strong>["apple","google","leetcode"]
-</pre>
+<p><strong class="example">示例 2：</strong></p>
 
-<p><strong>示例 3：</strong></p>
+<div class="example-block">
+<p><span class="example-io"><b>输入：</b></span><span class="example-io">words1 = ["amazon","apple","facebook","google","leetcode"], words2 = ["lc","eo"]</span></p>
 
-<pre>
-<strong>输入：</strong>words1 = ["amazon","apple","facebook","google","leetcode"], words2 = ["e","oo"]
-<strong>输出：</strong>["facebook","google"]
-</pre>
+<p><span class="example-io"><b>输出：</b></span><span class="example-io">["leetcode"]</span></p>
+</div>
 
-<p><strong>示例 4：</strong></p>
+<p><strong class="example">示例 3：</strong></p>
 
-<pre>
-<strong>输入：</strong>words1 = ["amazon","apple","facebook","google","leetcode"], words2 = ["lo","eo"]
-<strong>输出：</strong>["google","leetcode"]
-</pre>
+<div class="example-block">
+<p><span class="example-io"><b>输入：</b></span><span class="example-io">words1 = ["acaac","cccbb","aacbb","caacc","bcbbb"], words2 = ["c","cc","b"]</span></p>
 
-<p><strong>示例 5：</strong></p>
-
-<pre>
-<strong>输入：</strong>words1 = ["amazon","apple","facebook","google","leetcode"], words2 = ["ec","oc","ceo"]
-<strong>输出：</strong>["facebook","leetcode"]
-</pre>
+<p><span class="example-io"><b>输出：</b></span><span class="example-io">["cccbb"]</span></p>
+</div>
 
 <p>&nbsp;</p>
 

diff --git a/solution/0900-0999/0975.Odd Even Jump/README_EN.md b/solution/0900-0999/0975.Odd Even Jump/README_EN.md
@@ -40,7 +40,7 @@ tags:
 <pre>
 <strong>Input:</strong> arr = [10,13,12,14,15]
 <strong>Output:</strong> 2
-<strong>Explanation:</strong>
+<strong>Explanation:</strong> 
 From starting index i = 0, we can make our 1st jump to i = 2 (since arr[2] is the smallest among arr[1], arr[2], arr[3], arr[4] that is greater or equal to arr[0]), then we cannot jump any more.
 From starting index i = 1 and i = 2, we can make our 1st jump to i = 3, then we cannot jump any more.
 From starting index i = 3, we can make our 1st jump to i = 4, so we have reached the end.
@@ -54,7 +54,7 @@ jumps.
 <pre>
 <strong>Input:</strong> arr = [2,3,1,1,4]
 <strong>Output:</strong> 3
-<strong>Explanation:</strong>
+<strong>Explanation:</strong> 
 From starting index i = 0, we make jumps to i = 1, i = 2, i = 3:
 During our 1st jump (odd-numbered), we first jump to i = 1 because arr[1] is the smallest value in [arr[1], arr[2], arr[3], arr[4]] that is greater than or equal to arr[0].
 During our 2nd jump (even-numbered), we jump from i = 1 to i = 2 because arr[2] is the largest value in [arr[2], arr[3], arr[4]] that is less than or equal to arr[1]. arr[3] is also the largest value, but 2 is a smaller index, so we can only jump to i = 2 and not i = 3

diff --git a/solution/1100-1199/1172.Dinner Plate Stacks/README.md b/solution/1100-1199/1172.Dinner Plate Stacks/README.md
@@ -65,14 +65,14 @@ D.popAtStack(0);   // 返回 20。栈的现状为：       4 21
                                             ﹈ ﹈ ﹈
 D.popAtStack(2);   // 返回 21。栈的现状为：       4
              &nbsp;                              1 &nbsp;3 &nbsp;5
-                                            ﹈ ﹈ ﹈
+                                            ﹈ ﹈ ﹈ 
 D.pop()            // 返回 5。栈的现状为：        4
-             &nbsp;                              1 &nbsp;3
-                                            ﹈ ﹈
-D.pop()            // 返回 4。栈的现状为：    1  3
-                                           ﹈ ﹈
-D.pop()            // 返回 3。栈的现状为：    1
-                                           ﹈
+             &nbsp;                              1 &nbsp;3 
+                                            ﹈ ﹈  
+D.pop()            // 返回 4。栈的现状为：    1  3 
+                                           ﹈ ﹈   
+D.pop()            // 返回 3。栈的现状为：    1 
+                                           ﹈   
 D.pop()            // 返回 1。现在没有栈。
 D.pop()            // 返回 -1。仍然没有栈。
 </pre>

diff --git a/solution/1100-1199/1172.Dinner Plate Stacks/README_EN.md b/solution/1100-1199/1172.Dinner Plate Stacks/README_EN.md
@@ -42,7 +42,7 @@ tags:
 <strong>Output</strong>
 [null, null, null, null, null, null, 2, null, null, 20, 21, 5, 4, 3, 1, -1]
 
-<strong>Explanation:</strong>
+<strong>Explanation:</strong> 
 DinnerPlates D = DinnerPlates(2);  // Initialize with capacity = 2
 D.push(1);
 D.push(2);
@@ -65,14 +65,14 @@ D.popAtStack(0);   // Returns 20.  The stacks are now:     4 21
                                                         ﹈ ﹈ ﹈
 D.popAtStack(2);   // Returns 21.  The stacks are now:     4
                                                         1  3  5
-                                                        ﹈ ﹈ ﹈
+                                                        ﹈ ﹈ ﹈ 
 D.pop()            // Returns 5.  The stacks are now:      4
-                                                        1  3
-                                                        ﹈ ﹈
-D.pop()            // Returns 4.  The stacks are now:   1  3
-                                                        ﹈ ﹈
-D.pop()            // Returns 3.  The stacks are now:   1
-                                                        ﹈
+                                                        1  3 
+                                                        ﹈ ﹈  
+D.pop()            // Returns 4.  The stacks are now:   1  3 
+                                                        ﹈ ﹈   
+D.pop()            // Returns 3.  The stacks are now:   1 
+                                                        ﹈   
 D.pop()            // Returns 1.  There are no stacks.
 D.pop()            // Returns -1.  There are still no stacks.
 </pre>

diff --git a/solution/1300-1399/1348.Tweet Counts Per Frequency/README.md b/solution/1300-1399/1348.Tweet Counts Per Frequency/README.md
@@ -68,7 +68,7 @@ tweetCounts.recordTweet("tweet3", 0);
 tweetCounts.recordTweet("tweet3", 60);
 tweetCounts.recordTweet("tweet3", 10);                             //&nbsp;"tweet3"&nbsp;发布推文的时间分别是&nbsp;0,&nbsp;10&nbsp;和&nbsp;60 。
 tweetCounts.getTweetCountsPerFrequency("minute", "tweet3", 0, 59); //&nbsp;返回&nbsp;[2]。统计频率是每分钟（60 秒），因此只有一个有效时间间隔 [0,60&gt;&nbsp;-&nbsp;&gt;&nbsp;2&nbsp;条推文。
-tweetCounts.getTweetCountsPerFrequency("minute", "tweet3", 0, 60); //&nbsp;返回&nbsp;[2,1]。统计频率是每分钟（60 秒），因此有两个有效时间间隔&nbsp;<strong>1)</strong>&nbsp;[0,60&gt;&nbsp;-&nbsp;&gt;&nbsp;2&nbsp;条推文，和&nbsp;<strong>2)</strong>&nbsp;[60,61&gt;&nbsp;-&nbsp;&gt;&nbsp;1&nbsp;条推文。
+tweetCounts.getTweetCountsPerFrequency("minute", "tweet3", 0, 60); //&nbsp;返回&nbsp;[2,1]。统计频率是每分钟（60 秒），因此有两个有效时间间隔&nbsp;<strong>1)</strong>&nbsp;[0,60&gt;&nbsp;-&nbsp;&gt;&nbsp;2&nbsp;条推文，和&nbsp;<strong>2)</strong>&nbsp;[60,61&gt;&nbsp;-&nbsp;&gt;&nbsp;1&nbsp;条推文。 
 tweetCounts.recordTweet("tweet3", 120);                            // "tweet3"&nbsp;发布推文的时间分别是 0, 10, 60 和 120 。
 tweetCounts.getTweetCountsPerFrequency("hour", "tweet3", 0, 210);  //&nbsp;返回&nbsp;[4]。统计频率是每小时（3600 秒），因此只有一个有效时间间隔 [0,211&gt;&nbsp;-&nbsp;&gt;&nbsp;4&nbsp;条推文。
 </pre>

diff --git a/solution/1400-1499/1400.Construct K Palindrome Strings/README.md b/solution/1400-1499/1400.Construct K Palindrome Strings/README.md
@@ -21,7 +21,7 @@ tags:
 
 <!-- description:start -->
 
-<p>给你一个字符串 <code>s</code>&nbsp;和一个整数 <code>k</code>&nbsp;。请你用 <code>s</code>&nbsp;字符串中 <strong>所有字符</strong>&nbsp;构造 <code>k</code>&nbsp;个非空 <strong>回文串</strong>&nbsp;。</p>
+<p>给你一个字符串 <code>s</code>&nbsp;和一个整数 <code>k</code>&nbsp;。请你用 <code>s</code>&nbsp;字符串中 <strong>所有字符</strong>&nbsp;构造 <code>k</code>&nbsp;个非空 <span data-keyword="palindrome-string">回文串</span>&nbsp;。</p>
 
 <p>如果你可以用&nbsp;<code>s</code>&nbsp;中所有字符构造&nbsp;<code>k</code>&nbsp;个回文字符串，那么请你返回 <strong>True</strong>&nbsp;，否则返回&nbsp;<strong>False</strong>&nbsp;。</p>
 
@@ -30,52 +30,36 @@ tags:
 <p><strong>示例 1：</strong></p>
 
 <pre>
-<strong>输入：</strong>s = &quot;annabelle&quot;, k = 2
+<strong>输入：</strong>s = "annabelle", k = 2
 <strong>输出：</strong>true
 <strong>解释：</strong>可以用 s 中所有字符构造 2 个回文字符串。
-一些可行的构造方案包括：&quot;anna&quot; + &quot;elble&quot;，&quot;anbna&quot; + &quot;elle&quot;，&quot;anellena&quot; + &quot;b&quot;
+一些可行的构造方案包括："anna" + "elble"，"anbna" + "elle"，"anellena" + "b"
 </pre>
 
 <p><strong>示例 2：</strong></p>
 
 <pre>
-<strong>输入：</strong>s = &quot;leetcode&quot;, k = 3
+<strong>输入：</strong>s = "leetcode", k = 3
 <strong>输出：</strong>false
 <strong>解释：</strong>无法用 s 中所有字符构造 3 个回文串。
 </pre>
 
 <p><strong>示例 3：</strong></p>
 
 <pre>
-<strong>输入：</strong>s = &quot;true&quot;, k = 4
+<strong>输入：</strong>s = "true", k = 4
 <strong>输出：</strong>true
 <strong>解释：</strong>唯一可行的方案是让 s 中每个字符单独构成一个字符串。
 </pre>
 
-<p><strong>示例 4：</strong></p>
-
-<pre>
-<strong>输入：</strong>s = &quot;yzyzyzyzyzyzyzy&quot;, k = 2
-<strong>输出：</strong>true
-<strong>解释：</strong>你只需要将所有的 z 放在一个字符串中，所有的 y 放在另一个字符串中。那么两个字符串都是回文串。
-</pre>
-
-<p><strong>示例 5：</strong></p>
-
-<pre>
-<strong>输入：</strong>s = &quot;cr&quot;, k = 7
-<strong>输出：</strong>false
-<strong>解释：</strong>我们没有足够的字符去构造 7 个回文串。
-</pre>
-
 <p>&nbsp;</p>
 
 <p><strong>提示：</strong></p>
 
 <ul>
-	<li><code>1 &lt;= s.length &lt;= 10^5</code></li>
+	<li><code>1 &lt;= s.length &lt;= 10<sup>5</sup></code></li>
 	<li><code>s</code>&nbsp;中所有字符都是小写英文字母。</li>
-	<li><code>1 &lt;= k &lt;= 10^5</code></li>
+	<li><code>1 &lt;= k &lt;= 10<sup>5</sup></code></li>
 </ul>
 
 <!-- description:end -->

diff --git a/...st Continuous Subarray With Absolute Diff Less Than or Equal to Limit/README.md b/...st Continuous Subarray With Absolute Diff Less Than or Equal to Limit/README.md
@@ -32,25 +32,25 @@ tags:
 <p><strong>示例 1：</strong></p>
 
 <pre><strong>输入：</strong>nums = [8,2,4,7], limit = 4
-<strong>输出：</strong>2
+<strong>输出：</strong>2 
 <strong>解释：</strong>所有子数组如下：
 [8] 最大绝对差 |8-8| = 0 &lt;= 4.
-[8,2] 最大绝对差 |8-2| = 6 &gt; 4.
+[8,2] 最大绝对差 |8-2| = 6 &gt; 4. 
 [8,2,4] 最大绝对差 |8-2| = 6 &gt; 4.
 [8,2,4,7] 最大绝对差 |8-2| = 6 &gt; 4.
 [2] 最大绝对差 |2-2| = 0 &lt;= 4.
 [2,4] 最大绝对差 |2-4| = 2 &lt;= 4.
 [2,4,7] 最大绝对差 |2-7| = 5 &gt; 4.
 [4] 最大绝对差 |4-4| = 0 &lt;= 4.
 [4,7] 最大绝对差 |4-7| = 3 &lt;= 4.
-[7] 最大绝对差 |7-7| = 0 &lt;= 4.
+[7] 最大绝对差 |7-7| = 0 &lt;= 4. 
 因此，满足题意的最长子数组的长度为 2 。
 </pre>
 
 <p><strong>示例 2：</strong></p>
 
 <pre><strong>输入：</strong>nums = [10,1,2,4,7,2], limit = 5
-<strong>输出：</strong>4
+<strong>输出：</strong>4 
 <strong>解释：</strong>满足题意的最长子数组是 [2,4,7,2]，其最大绝对差 |2-7| = 5 &lt;= 5 。
 </pre>
 

diff --git a/...Continuous Subarray With Absolute Diff Less Than or Equal to Limit/README_EN.md b/...Continuous Subarray With Absolute Diff Less Than or Equal to Limit/README_EN.md
@@ -30,26 +30,26 @@ tags:
 
 <pre>
 <strong>Input:</strong> nums = [8,2,4,7], limit = 4
-<strong>Output:</strong> 2
-<strong>Explanation:</strong> All subarrays are:
+<strong>Output:</strong> 2 
+<strong>Explanation:</strong> All subarrays are: 
 [8] with maximum absolute diff |8-8| = 0 &lt;= 4.
-[8,2] with maximum absolute diff |8-2| = 6 &gt; 4.
+[8,2] with maximum absolute diff |8-2| = 6 &gt; 4. 
 [8,2,4] with maximum absolute diff |8-2| = 6 &gt; 4.
 [8,2,4,7] with maximum absolute diff |8-2| = 6 &gt; 4.
 [2] with maximum absolute diff |2-2| = 0 &lt;= 4.
 [2,4] with maximum absolute diff |2-4| = 2 &lt;= 4.
 [2,4,7] with maximum absolute diff |2-7| = 5 &gt; 4.
 [4] with maximum absolute diff |4-4| = 0 &lt;= 4.
 [4,7] with maximum absolute diff |4-7| = 3 &lt;= 4.
-[7] with maximum absolute diff |7-7| = 0 &lt;= 4.
+[7] with maximum absolute diff |7-7| = 0 &lt;= 4. 
 Therefore, the size of the longest subarray is 2.
 </pre>
 
 <p><strong class="example">Example 2:</strong></p>
 
 <pre>
 <strong>Input:</strong> nums = [10,1,2,4,7,2], limit = 5
-<strong>Output:</strong> 4
+<strong>Output:</strong> 4 
 <strong>Explanation:</strong> The subarray [2,4,7,2] is the longest since the maximum absolute diff is |2-7| = 5 &lt;= 5.
 </pre>
 

diff --git a/...tion/1600-1699/1606.Find Servers That Handled Most Number of Requests/README.md b/...tion/1600-1699/1606.Find Servers That Handled Most Number of Requests/README.md
@@ -41,8 +41,8 @@ tags:
 <p><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1600-1699/1606.Find%20Servers%20That%20Handled%20Most%20Number%20of%20Requests/images/load-1.png" style="height: 221px; width: 389px;" /></p>
 
 <pre>
-<strong>输入：</strong>k = 3, arrival = [1,2,3,4,5], load = [5,2,3,3,3]
-<strong>输出：</strong>[1]
+<strong>输入：</strong>k = 3, arrival = [1,2,3,4,5], load = [5,2,3,3,3] 
+<strong>输出：</strong>[1] 
 <strong>解释：</strong>
 所有服务器一开始都是空闲的。
 前 3 个请求分别由前 3 台服务器依次处理。

diff --git a/...n/1600-1699/1606.Find Servers That Handled Most Number of Requests/README_EN.md b/...n/1600-1699/1606.Find Servers That Handled Most Number of Requests/README_EN.md
@@ -38,9 +38,9 @@ tags:
 <p><strong class="example">Example 1:</strong></p>
 <img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1600-1699/1606.Find%20Servers%20That%20Handled%20Most%20Number%20of%20Requests/images/load-1.png" style="width: 389px; height: 221px;" />
 <pre>
-<strong>Input:</strong> k = 3, arrival = [1,2,3,4,5], load = [5,2,3,3,3]
-<strong>Output:</strong> [1]
-<strong>Explanation:</strong>
+<strong>Input:</strong> k = 3, arrival = [1,2,3,4,5], load = [5,2,3,3,3] 
+<strong>Output:</strong> [1] 
+<strong>Explanation:</strong> 
 All of the servers start out available.
 The first 3 requests are handled by the first 3 servers in order.
 Request 3 comes in. Server 0 is busy, so it&#39;s assigned to the next available server, which is 1.
@@ -53,7 +53,7 @@ Servers 0 and 2 handled one request each, while server 1 handled two requests. H
 <pre>
 <strong>Input:</strong> k = 3, arrival = [1,2,3,4], load = [1,2,1,2]
 <strong>Output:</strong> [0]
-<strong>Explanation:</strong>
+<strong>Explanation:</strong> 
 The first 3 requests are handled by first 3 servers.
 Request 3 comes in. It is handled by server 0 since the server is available.
 Server 0 handled two requests, while servers 1 and 2 handled one request each. Hence server 0 is the busiest server.

diff --git a/solution/1800-1899/1825.Finding MK Average/README.md b/solution/1800-1899/1825.Finding MK Average/README.md
@@ -52,7 +52,7 @@ tags:
 [null, null, null, -1, null, 3, null, null, null, 5]
 
 <strong>解释：</strong>
-MKAverage obj = new MKAverage(3, 1);
+MKAverage obj = new MKAverage(3, 1); 
 obj.addElement(3);        // 当前元素为 [3]
 obj.addElement(1);        // 当前元素为 [3,1]
 obj.calculateMKAverage(); // 返回 -1 ，因为 m = 3 ，但数据流中只有 2 个元素

diff --git a/solution/1800-1899/1825.Finding MK Average/README_EN.md b/solution/1800-1899/1825.Finding MK Average/README_EN.md
@@ -51,7 +51,7 @@ tags:
 [null, null, null, -1, null, 3, null, null, null, 5]
 
 <strong>Explanation</strong>
-<code>MKAverage obj = new MKAverage(3, 1);
+<code>MKAverage obj = new MKAverage(3, 1); 
 obj.addElement(3);        // current elements are [3]
 obj.addElement(1);        // current elements are [3,1]
 obj.calculateMKAverage(); // return -1, because m = 3 and only 2 elements exist.

diff --git a/solution/2300-2399/2349.Design a Number Container System/README_EN.md b/solution/2300-2399/2349.Design a Number Container System/README_EN.md
@@ -54,7 +54,7 @@ nc.change(1, 10); // Your container at index 1 will be filled with number 10.
 nc.change(3, 10); // Your container at index 3 will be filled with number 10.
 nc.change(5, 10); // Your container at index 5 will be filled with number 10.
 nc.find(10); // Number 10 is at the indices 1, 2, 3, and 5. Since the smallest index that is filled with 10 is 1, we return 1.
-nc.change(1, 20); // Your container at index 1 will be filled with number 20. Note that index 1 was filled with 10 and then replaced with 20.
+nc.change(1, 20); // Your container at index 1 will be filled with number 20. Note that index 1 was filled with 10 and then replaced with 20. 
 nc.find(10); // Number 10 is at the indices 2, 3, and 5. The smallest index that is filled with 10 is 2. Therefore, we return 2.
 </pre>