deploy: 8cbbd41

Author: yanglbme

en/lc/1319/index.html

@@ -34963,9 +34963,9 @@
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-  <a href="#solution-1" class="md-nav__link">
+  <a href="#solution-1-union-find" class="md-nav__link">
     <span class="md-ellipsis">
-      Solution 1
+      Solution 1: Union-Find
     </span>
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@@ -80733,7 +80733,10 @@ <h2 id="description">Description</h2>
 <h2 id="solutions">Solutions</h2>
 <!-- solution:start -->
 
-<h3 id="solution-1">Solution 1</h3>
+<h3 id="solution-1-union-find">Solution 1: Union-Find</h3>
+<p>We can use a union-find data structure to maintain the connectivity between computers. Traverse all connections, and for each connection $(a, b)$, if $a$ and $b$ are already connected, then this connection is redundant, and we increment the count of redundant connections. Otherwise, we connect $a$ and $b$, and decrement the number of connected components.</p>
+<p>Finally, if the number of connected components minus one is greater than the number of redundant connections, it means we cannot connect all computers, so we return -1. Otherwise, we return the number of connected components minus one.</p>
+<p>The time complexity is $O(m \times \log n)$, and the space complexity is $O(n)$. Here, $n$ and $m$ are the number of computers and the number of connections, respectively.</p>
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en/lc/1320/index.html

@@ -34984,9 +34984,9 @@
       <ul class="md-nav__list">
 
           <li class="md-nav__item">
-  <a href="#solution-1" class="md-nav__link">
+  <a href="#solution-1-dynamic-programming" class="md-nav__link">
     <span class="md-ellipsis">
-      Solution 1
+      Solution 1: Dynamic Programming
     </span>
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@@ -80692,11 +80692,11 @@ <h2 id="description">Description</h2>
 <pre>
 <strong>Input:</strong> word = &quot;CAKE&quot;
 <strong>Output:</strong> 3
-<strong>Explanation:</strong> Using two fingers, one optimal way to type &quot;CAKE&quot; is: 
-Finger 1 on letter &#39;C&#39; -&gt; cost = 0 
-Finger 1 on letter &#39;A&#39; -&gt; cost = Distance from letter &#39;C&#39; to letter &#39;A&#39; = 2 
-Finger 2 on letter &#39;K&#39; -&gt; cost = 0 
-Finger 2 on letter &#39;E&#39; -&gt; cost = Distance from letter &#39;K&#39; to letter &#39;E&#39; = 1 
+<strong>Explanation:</strong> Using two fingers, one optimal way to type &quot;CAKE&quot; is:
+Finger 1 on letter &#39;C&#39; -&gt; cost = 0
+Finger 1 on letter &#39;A&#39; -&gt; cost = Distance from letter &#39;C&#39; to letter &#39;A&#39; = 2
+Finger 2 on letter &#39;K&#39; -&gt; cost = 0
+Finger 2 on letter &#39;E&#39; -&gt; cost = Distance from letter &#39;K&#39; to letter &#39;E&#39; = 1
 Total distance = 3
 </pre>
 
@@ -80727,7 +80727,19 @@ <h2 id="description">Description</h2>
 <h2 id="solutions">Solutions</h2>
 <!-- solution:start -->
 
-<h3 id="solution-1">Solution 1</h3>
+<h3 id="solution-1-dynamic-programming">Solution 1: Dynamic Programming</h3>
+<p>We define $f[i][j][k]$ to represent the minimum distance after typing $\textit{word}[i]$, with finger 1 at position $j$ and finger 2 at position $k$. Here, positions $j$ and $k$ represent the numbers corresponding to the letters, ranging from $[0,..25]$. Initially, $f[i][j][k] = \infty$.</p>
+<p>We implement a function $\textit{dist}(a, b)$ to represent the distance between positions $a$ and $b$, i.e., $\textit{dist}(a, b) = |\frac{a}{6} - \frac{b}{6}| + |a \bmod 6 - b \bmod 6|$.</p>
+<p>Next, we consider typing $\textit{word}[0]$, i.e., the case with only one letter. There are two choices:</p>
+<ul>
+<li>Finger 1 is at the position of $\textit{word}[0]$, and finger 2 is at any position. In this case, $f[0][\textit{word}[0]][k] = 0$, where $k \in [0,..25]$.</li>
+<li>Finger 2 is at the position of $\textit{word}[0]$, and finger 1 is at any position. In this case, $f[0][k][\textit{word}[0]] = 0$, where $k \in [0,..25]$.</li>
+</ul>
+<p>Then we consider typing $\textit{word}[1,..n-1]$. Let the positions of the previous letter and the current letter be $a$ and $b$, respectively. Next, we discuss the following cases:</p>
+<p>If the current finger 1 is at position $b$, we enumerate the position $j$ of finger 2. If the previous position $a$ was also the position of finger 1, then $f[i][b][j] = \min(f[i][b][j], f[i-1][a][j] + \textit{dist}(a, b))$. If the position of finger 2 is the same as the previous position $a$, i.e., $j = a$, then we enumerate the position $k$ of finger 1 in the previous position. In this case, $f[i][b][j] = \min(f[i][b][j], f[i-1][k][a] + \textit{dist}(k, b))$.</p>
+<p>Similarly, if the current finger 2 is at position $b$, we enumerate the position $j$ of finger 1. If the previous position $a$ was also the position of finger 2, then $f[i][j][b] = \min(f[i][j][b], f[i-1][j][a] + \textit{dist}(a, b))$. If the position of finger 1 is the same as the previous position $a$, i.e., $j = a$, then we enumerate the position $k$ of finger 2 in the previous position. In this case, $f[i][j][b] = \min(f[i][j][b], f[i-1][a][k] + \textit{dist}(k, b))$.</p>
+<p>Finally, we enumerate the positions of finger 1 and finger 2 at the last position and take the minimum value as the answer.</p>
+<p>The time complexity is $O(n \times |\Sigma|^2)$, and the space complexity is $O(n \times |\Sigma|^2)$. Here, $n$ is the length of the string $\textit{word}$, and $|\Sigma|$ is the size of the alphabet, which is $26$ in this problem.</p>
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