deploy: 94e5644

Author: acbin

en/lc/3369/index.html

@@ -78046,9 +78046,9 @@
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-  <a href="#solution-1" class="md-nav__link">
+  <a href="#solution-1-bit-manipulation" class="md-nav__link">
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-      Solution 1
+      Solution 1: Bit Manipulation
     </span>
   </a>
 
@@ -80866,7 +80866,9 @@ <h2 id="description">Description</h2>
 <h2 id="solutions">Solutions</h2>
 <!-- solution:start -->
 
-<h3 id="solution-1">Solution 1</h3>
+<h3 id="solution-1-bit-manipulation">Solution 1: Bit Manipulation</h3>
+<p>We start with $x = 1$ and continuously left shift $x$ until $x - 1 \geq n$. At this point, $x - 1$ is the answer we are looking for.</p>
+<p>The time complexity is $O(\log n)$, and the space complexity is $O(1)$.</p>
 <div class="tabbed-set tabbed-alternate" data-tabs="1:5"><input checked="checked" id="__tabbed_1_1" name="__tabbed_1" type="radio" /><input id="__tabbed_1_2" name="__tabbed_1" type="radio" /><input id="__tabbed_1_3" name="__tabbed_1" type="radio" /><input id="__tabbed_1_4" name="__tabbed_1" type="radio" /><input id="__tabbed_1_5" name="__tabbed_1" type="radio" /><div class="tabbed-labels"><label for="__tabbed_1_1">Python3</label><label for="__tabbed_1_2">Java</label><label for="__tabbed_1_3">C++</label><label for="__tabbed_1_4">Go</label><label for="__tabbed_1_5">TypeScript</label></div>
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en/lc/3370/index.html

@@ -78067,9 +78067,9 @@
       <ul class="md-nav__list">
 
           <li class="md-nav__item">
-  <a href="#solution-1" class="md-nav__link">
+  <a href="#solution-1-hash-table-enumeration" class="md-nav__link">
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-      Solution 1
+      Solution 1: Hash Table + Enumeration
     </span>
   </a>
 
@@ -80882,7 +80882,11 @@ <h2 id="description">Description</h2>
 <h2 id="solutions">Solutions</h2>
 <!-- solution:start -->
 
-<h3 id="solution-1">Solution 1</h3>
+<h3 id="solution-1-hash-table-enumeration">Solution 1: Hash Table + Enumeration</h3>
+<p>We use a hash table $\textit{cnt}$ to record the frequency of each element in the array $\textit{nums}$.</p>
+<p>Next, we enumerate each element $x$ in the array $\textit{nums}$ as a possible outlier. For each $x$, we calculate the sum $t$ of all elements in the array $\textit{nums}$ except $x$. If $t$ is not even, or half of $t$ is not in $\textit{cnt}$, then $x$ does not meet the condition, and we skip this $x$. Otherwise, if $x$ is not equal to half of $t$, or $x$ appears more than once in $\textit{cnt}$, then $x$ is a possible outlier, and we update the answer.</p>
+<p>After enumerating all elements, we return the answer.</p>
+<p>The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $\textit{nums}$.</p>
 <div class="tabbed-set tabbed-alternate" data-tabs="1:5"><input checked="checked" id="__tabbed_1_1" name="__tabbed_1" type="radio" /><input id="__tabbed_1_2" name="__tabbed_1" type="radio" /><input id="__tabbed_1_3" name="__tabbed_1" type="radio" /><input id="__tabbed_1_4" name="__tabbed_1" type="radio" /><input id="__tabbed_1_5" name="__tabbed_1" type="radio" /><div class="tabbed-labels"><label for="__tabbed_1_1">Python3</label><label for="__tabbed_1_2">Java</label><label for="__tabbed_1_3">C++</label><label for="__tabbed_1_4">Go</label><label for="__tabbed_1_5">TypeScript</label></div>
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en/lc/3371/index.html

@@ -78088,9 +78088,9 @@
       <ul class="md-nav__list">
 
           <li class="md-nav__item">
-  <a href="#solution-1" class="md-nav__link">
+  <a href="#solution-1-enumeration-dfs" class="md-nav__link">
     <span class="md-ellipsis">
-      Solution 1
+      Solution 1: Enumeration + DFS
     </span>
   </a>
 
@@ -80879,7 +80879,14 @@ <h2 id="description">Description</h2>
 <h2 id="solutions">Solutions</h2>
 <!-- solution:start -->
 
-<h3 id="solution-1">Solution 1</h3>
+<h3 id="solution-1-enumeration-dfs">Solution 1: Enumeration + DFS</h3>
+<p>According to the problem description, to maximize the number of target nodes for node $i$, we must connect node $i$ to one of the nodes $j$ in the second tree. Therefore, the number of target nodes for node $i$ can be divided into two parts:</p>
+<ul>
+<li>In the first tree, the number of nodes reachable from node $i$ within a depth of $k$.</li>
+<li>In the second tree, the maximum number of nodes reachable from any node $j$ within a depth of $k - 1$.</li>
+</ul>
+<p>Thus, we can first calculate the number of nodes reachable within a depth of $k - 1$ for each node in the second tree. Then, we enumerate each node $i$ in the first tree, calculate the sum of the two parts mentioned above, and take the maximum value.</p>
+<p>The time complexity is $O(n^2 + m^2)$, and the space complexity is $O(n + m)$. Here, $n$ and $m$ are the number of nodes in the two trees, respectively.</p>
 <div class="tabbed-set tabbed-alternate" data-tabs="1:6"><input checked="checked" id="__tabbed_1_1" name="__tabbed_1" type="radio" /><input id="__tabbed_1_2" name="__tabbed_1" type="radio" /><input id="__tabbed_1_3" name="__tabbed_1" type="radio" /><input id="__tabbed_1_4" name="__tabbed_1" type="radio" /><input id="__tabbed_1_5" name="__tabbed_1" type="radio" /><input id="__tabbed_1_6" name="__tabbed_1" type="radio" /><div class="tabbed-labels"><label for="__tabbed_1_1">Python3</label><label for="__tabbed_1_2">Java</label><label for="__tabbed_1_3">C++</label><label for="__tabbed_1_4">Go</label><label for="__tabbed_1_5">TypeScript</label><label for="__tabbed_1_6">C#</label></div>
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en/lc/3372/index.html

@@ -78109,9 +78109,9 @@
       <ul class="md-nav__list">
 
           <li class="md-nav__item">
-  <a href="#solution-1" class="md-nav__link">
+  <a href="#solution-1-dfs" class="md-nav__link">
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-      Solution 1
+      Solution 1: DFS
     </span>
   </a>
 
@@ -80878,7 +80878,14 @@ <h2 id="description">Description</h2>
 <h2 id="solutions">Solutions</h2>
 <!-- solution:start -->
 
-<h3 id="solution-1">Solution 1</h3>
+<h3 id="solution-1-dfs">Solution 1: DFS</h3>
+<p>The number of target nodes for node $i$ can be divided into two parts:</p>
+<ul>
+<li>The number of nodes in the first tree with the same depth parity as node $i$.</li>
+<li>The maximum number of nodes in the second tree with the same depth parity.</li>
+</ul>
+<p>First, we use Depth-First Search (DFS) to calculate the number of nodes in the second tree with the same depth parity, denoted as $\textit{cnt2}$. Then, we calculate the number of nodes in the first tree with the same depth parity as node $i$, denoted as $\textit{cnt1}$. Therefore, the number of target nodes for node $i$ is $\max(\textit{cnt2}) + \textit{cnt1}$.</p>
+<p>The time complexity is $O(n + m)$, and the space complexity is $O(n + m)$. Here, $n$ and $m$ are the number of nodes in the first and second trees, respectively.</p>
 <div class="tabbed-set tabbed-alternate" data-tabs="1:6"><input checked="checked" id="__tabbed_1_1" name="__tabbed_1" type="radio" /><input id="__tabbed_1_2" name="__tabbed_1" type="radio" /><input id="__tabbed_1_3" name="__tabbed_1" type="radio" /><input id="__tabbed_1_4" name="__tabbed_1" type="radio" /><input id="__tabbed_1_5" name="__tabbed_1" type="radio" /><input id="__tabbed_1_6" name="__tabbed_1" type="radio" /><div class="tabbed-labels"><label for="__tabbed_1_1">Python3</label><label for="__tabbed_1_2">Java</label><label for="__tabbed_1_3">C++</label><label for="__tabbed_1_4">Go</label><label for="__tabbed_1_5">TypeScript</label><label for="__tabbed_1_6">C#</label></div>
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