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Copy file name to clipboardExpand all lines: solution/1300-1399/1319.Number of Operations to Make Network Connected/README_EN.md
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### Solution 1
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### Solution 1: Union-Find
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We can use a union-find data structure to maintain the connectivity between computers. Traverse all connections, and for each connection $(a, b),ドル 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,ドル and decrement the number of connected components.
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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.
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The time complexity is $O(m \times \log n),ドル and the space complexity is $O(n)$. Here, $n$ and $m$ are the number of computers and the number of connections, respectively.
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Collapse file: solution/1300-1399/1320.Minimum Distance to Type a Word Using Two Fingers/README.md
Copy file name to clipboardExpand all lines: solution/1300-1399/1320.Minimum Distance to Type a Word Using Two Fingers/README_EN.md
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<pre>
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<strong>Input:</strong> word = "CAKE"
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<strong>Output:</strong> 3
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<strong>Explanation:</strong> Using two fingers, one optimal way to type "CAKE" is:
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Finger 1 on letter 'C' -> cost = 0
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Finger 1 on letter 'A' -> cost = Distance from letter 'C' to letter 'A' = 2
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Finger 2 on letter 'K' -> cost = 0
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Finger 2 on letter 'E' -> cost = Distance from letter 'K' to letter 'E' = 1
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<strong>Explanation:</strong> Using two fingers, one optimal way to type "CAKE" is:
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Finger 1 on letter 'C' -> cost = 0
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Finger 1 on letter 'A' -> cost = Distance from letter 'C' to letter 'A' = 2
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Finger 2 on letter 'K' -> cost = 0
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Finger 2 on letter 'E' -> cost = Distance from letter 'K' to letter 'E' = 1
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Total distance = 3
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</pre>
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### Solution 1
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### Solution 1: Dynamic Programming
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We define $f[i][j][k]$ to represent the minimum distance after typing $\textit{word}[i],ドル 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, $f[i][j][k] = \infty$.
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We implement a function $\textit{dist}(a, b)$ to represent the distance between positions $a$ and $b,ドル i.e., $\textit{dist}(a, b) = |\frac{a}{6} - \frac{b}{6}| + |a \bmod 6 - b \bmod 6|$.
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Next, we consider typing $\textit{word}[0],ドル i.e., the case with only one letter. There are two choices:
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- Finger 1 is at the position of $\textit{word}[0],ドル and finger 2 is at any position. In this case, $f[0][\textit{word}[0]][k] = 0,ドル where $k \in [0,..25]$.
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- Finger 2 is at the position of $\textit{word}[0],ドル and finger 1 is at any position. In this case, $f[0][k][\textit{word}[0]] = 0,ドル where $k \in [0,..25]$.
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Then we consider typing $\textit{word}[1,..n-1]$. Let the positions of the previous letter and the current letter be $a$ and $b,ドル respectively. Next, we discuss the following cases:
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If the current finger 1 is at position $b,ドル 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,ドル i.e., $j = a,ドル then we enumerate the position $k$ of finger 1 in the previous position. In this case, $f[i][b][j] = \min(f[i][b][j], f[i-1][k][a] + \textit{dist}(k, b))$.
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Similarly, if the current finger 2 is at position $b,ドル 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,ドル i.e., $j = a,ドル then we enumerate the position $k$ of finger 2 in the previous position. In this case, $f[i][j][b] = \min(f[i][j][b], f[i-1][a][k] + \textit{dist}(k, b))$.
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Finally, we enumerate the positions of finger 1 and finger 2 at the last position and take the minimum value as the answer.
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The time complexity is $O(n \times |\Sigma|^2),ドル and the space complexity is $O(n \times |\Sigma|^2)$. Here, $n$ is the length of the string $\textit{word},ドル and $|\Sigma|$ is the size of the alphabet, which is 26ドル$ in this problem.
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