feat: update lc problems

Author: yanglbme

solution/0000-0099/0097.Interleaving String/README_EN.md

@@ -6,7 +6,7 @@
 
 <p>Given strings <code>s1</code>, <code>s2</code>, and <code>s3</code>, find whether <code>s3</code> is formed by an <strong>interleaving</strong> of <code>s1</code> and <code>s2</code>.</p>
 
-<p>An <strong>interleaving</strong> of two strings <code>s</code> and <code>t</code> is a configuration where they are divided into <strong>non-empty</strong> substrings such that:</p>
+<p>An <strong>interleaving</strong> of two strings <code>s</code> and <code>t</code> is a configuration where <code>s</code> and <code>t</code> are divided into <code>n</code> and <code>m</code> <strong>non-empty</strong> substrings respectively, such that:</p>
 
 <ul>
 	<li><code>s = s<sub>1</sub> + s<sub>2</sub> + ... + s<sub>n</sub></code></li>
@@ -23,13 +23,18 @@
 <pre>
 <strong>Input:</strong> s1 = &quot;aabcc&quot;, s2 = &quot;dbbca&quot;, s3 = &quot;aadbbcbcac&quot;
 <strong>Output:</strong> true
+<strong>Explanation:</strong> One way to obtain s3 is:
+Split s1 into s1 = &quot;aa&quot; + &quot;bc&quot; + &quot;c&quot;, and s2 into s2 = &quot;dbbc&quot; + &quot;a&quot;.
+Interleaving the two splits, we get &quot;aa&quot; + &quot;dbbc&quot; + &quot;bc&quot; + &quot;a&quot; + &quot;c&quot; = &quot;aadbbcbcac&quot;.
+Since s3 can be obtained by interleaving s1 and s2, we return true.
 </pre>
 
 <p><strong>Example 2:</strong></p>
 
 <pre>
 <strong>Input:</strong> s1 = &quot;aabcc&quot;, s2 = &quot;dbbca&quot;, s3 = &quot;aadbbbaccc&quot;
 <strong>Output:</strong> false
+<strong>Explanation:</strong> Notice how it is impossible to interleave s2 with any other string to obtain s3.
 </pre>
 
 <p><strong>Example 3:</strong></p>

solution/0100-0199/0130.Surrounded Regions/README_EN.md

@@ -14,7 +14,11 @@
 <pre>
 <strong>Input:</strong> board = [[&quot;X&quot;,&quot;X&quot;,&quot;X&quot;,&quot;X&quot;],[&quot;X&quot;,&quot;O&quot;,&quot;O&quot;,&quot;X&quot;],[&quot;X&quot;,&quot;X&quot;,&quot;O&quot;,&quot;X&quot;],[&quot;X&quot;,&quot;O&quot;,&quot;X&quot;,&quot;X&quot;]]
 <strong>Output:</strong> [[&quot;X&quot;,&quot;X&quot;,&quot;X&quot;,&quot;X&quot;],[&quot;X&quot;,&quot;X&quot;,&quot;X&quot;,&quot;X&quot;],[&quot;X&quot;,&quot;X&quot;,&quot;X&quot;,&quot;X&quot;],[&quot;X&quot;,&quot;O&quot;,&quot;X&quot;,&quot;X&quot;]]
-<strong>Explanation:</strong> Surrounded regions should not be on the border, which means that any &#39;O&#39; on the border of the board are not flipped to &#39;X&#39;. Any &#39;O&#39; that is not on the border and it is not connected to an &#39;O&#39; on the border will be flipped to &#39;X&#39;. Two cells are connected if they are adjacent cells connected horizontally or vertically.
+<strong>Explanation:</strong> Notice that an &#39;O&#39; should not be flipped if:
+- It is on the border, or
+- It is adjacent to an &#39;O&#39; that should not be flipped.
+The bottom &#39;O&#39; is on the border, so it is not flipped.
+The other three &#39;O&#39; form a surrounded region, so they are flipped.
 </pre>
 
 <p><strong>Example 2:</strong></p>

solution/0100-0199/0139.Word Break/README.md

@@ -90,7 +90,7 @@ class Trie:
     def __init__(self):
         self.children = [None] * 26
         self.is_end = False
-    
+
     def insert(self, w):
         node = self
         for c in w:
@@ -99,7 +99,7 @@ class Trie:
                 node.children[idx] = Trie()
             node = node.children[idx]
         node.is_end = True
-    
+
     def search(self, w):
         node = self
         for c in w:
@@ -114,7 +114,7 @@ class Solution:
         @cache
         def dfs(s):
             return not s or any(trie.search(s[:i]) and dfs(s[i:]) for i in range(1, len(s) + 1))
-        
+
         trie = Trie()
         for w in wordDict:
             trie.insert(w)
@@ -269,7 +269,7 @@ public:
 
     bool wordBreak(string s, vector<string>& wordDict) {
         for (auto w : wordDict) trie->insert(w);
-        return dfs(s);    
+        return dfs(s);
     }
 
     bool dfs(string s) {

solution/0100-0199/0139.Word Break/README_EN.md

@@ -72,7 +72,7 @@ class Trie:
     def __init__(self):
         self.children = [None] * 26
         self.is_end = False
-    
+
     def insert(self, w):
         node = self
         for c in w:
@@ -81,7 +81,7 @@ class Trie:
                 node.children[idx] = Trie()
             node = node.children[idx]
         node.is_end = True
-    
+
     def search(self, w):
         node = self
         for c in w:
@@ -96,7 +96,7 @@ class Solution:
         @cache
         def dfs(s):
             return not s or any(trie.search(s[:i]) and dfs(s[i:]) for i in range(1, len(s) + 1))
-        
+
         trie = Trie()
         for w in wordDict:
             trie.insert(w)
@@ -249,7 +249,7 @@ public:
 
     bool wordBreak(string s, vector<string>& wordDict) {
         for (auto w : wordDict) trie->insert(w);
-        return dfs(s);    
+        return dfs(s);
     }
 
     bool dfs(string s) {

solution/0200-0299/0212.Word Search II/README.md

@@ -208,7 +208,7 @@ public:
         if(node->w != "") res.insert(node->w);
         char c = board[i][j];
         board[i][j] = '0';
-        
+
         for (int k = 0; k < 4; ++k)
         {
             int x = i + dirs[k], y = j + dirs[k + 1];

solution/0200-0299/0212.Word Search II/README_EN.md

@@ -198,7 +198,7 @@ public:
         if(node->w != "") res.insert(node->w);
         char c = board[i][j];
         board[i][j] = '0';
-        
+
         for (int k = 0; k < 4; ++k)
         {
             int x = i + dirs[k], y = j + dirs[k + 1];

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

@@ -10,28 +10,28 @@
 
 <p>请注意，你需要找的是数组排序后的第 <code>k</code> 个最大的元素，而不是第 <code>k</code> 个不同的元素。</p>
 
-<p> </p>
+<p>&nbsp;</p>
 
 <p><strong>示例 1:</strong></p>
 
 <pre>
-<strong>输入:</strong> <code>[3,2,1,5,6,4] 和</code> k = 2
+<strong>输入:</strong> <code>[3,2,1,5,6,4],</code> k = 2
 <strong>输出:</strong> 5
 </pre>
 
-<p><strong>示例 2:</strong></p>
+<p><strong>示例&nbsp;2:</strong></p>
 
 <pre>
-<strong>输入:</strong> <code>[3,2,3,1,2,4,5,5,6] 和</code> k = 4
+<strong>输入:</strong> <code>[3,2,3,1,2,4,5,5,6], </code>k = 4
 <strong>输出:</strong> 4</pre>
 
-<p> </p>
+<p>&nbsp;</p>
 
 <p><strong>提示： </strong></p>
 
 <ul>
-	<li><code>1 <= k <= nums.length <= 10<sup>4</sup></code></li>
-	<li><code>-10<sup>4</sup> <= nums[i] <= 10<sup>4</sup></code></li>
+	<li><code>1 &lt;= k &lt;= nums.length &lt;= 10<sup>4</sup></code></li>
+	<li><code>-10<sup>4</sup>&nbsp;&lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
 </ul>
 
 ## 解法

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

@@ -8,6 +8,8 @@
 
 <p>Note that it is the <code>k<sup>th</sup></code> largest element in the sorted order, not the <code>k<sup>th</sup></code> distinct element.</p>
 
+<p>You must solve it in <code>O(n)</code> time complexity.</p>
+
 <p>&nbsp;</p>
 <p><strong>Example 1:</strong></p>
 <pre><strong>Input:</strong> nums = [3,2,1,5,6,4], k = 2
@@ -20,7 +22,7 @@
 <p><strong>Constraints:</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> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
 </ul>
 

solution/0200-0299/0235.Lowest Common Ancestor of a Binary Search Tree/README_EN.md

@@ -4,9 +4,9 @@
 
 ## Description
 
-<p>Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST.</p>
+<p>Given a binary search tree (BST), find the lowest common ancestor (LCA) node of two given nodes in the BST.</p>
 
-<p>According to the <a href="https://en.wikipedia.org/wiki/Lowest_common_ancestor" target="_blank">definition of LCA on Wikipedia</a>: &ldquo;The lowest common ancestor is defined between two nodes <code>p</code> and <code>q</code> as the lowest node in <code>T</code> that has both <code>p</code> and <code>q</code> as descendants (where we allow <b>a node to be a descendant of itself</b>).&rdquo;</p>
+<p>According to the <a href="https://en.wikipedia.org/wiki/Lowest_common_ancestor" target="_blank">definition of LCA on Wikipedia</a>: &ldquo;The lowest common ancestor is defined between two nodes <code>p</code> and <code>q</code> as the lowest node in <code>T</code> that has both <code>p</code> and <code>q</code> as descendants (where we allow <strong>a node to be a descendant of itself</strong>).&rdquo;</p>
 
 <p>&nbsp;</p>
 <p><strong>Example 1:</strong></p>

solution/0200-0299/0243.Shortest Word Distance/README_EN.md

@@ -25,7 +25,7 @@
 <p><strong>Constraints:</strong></p>
 
 <ul>
-	<li><code>1 &lt;= wordsDict.length &lt;= 3 * 10<sup>4</sup></code></li>
+	<li><code>2 &lt;= wordsDict.length &lt;= 3 * 10<sup>4</sup></code></li>
 	<li><code>1 &lt;= wordsDict[i].length &lt;= 10</code></li>
 	<li><code>wordsDict[i]</code> consists of lowercase English letters.</li>
 	<li><code>word1</code> and <code>word2</code> are in <code>wordsDict</code>.</li>