#1319

Medium

auto_awesomeGraph traversal with depth-first search

LeetCode Problem Workspace

Number of Operations to Make Network Connected

Determine the minimum number of operations required to connect all computers in a network with a limited number of cables.

depth first search breadth first search union find graph

Problem Statement

You are given n computers connected by ethernet cables, where each connection is represented as a pair [ai, bi] meaning a cable between computers ai and bi. The goal is to find the minimum number of times you need to move cables between computers to ensure all computers are connected. If it is not possible to connect all computers, return -1.

You need to decide if there are enough cables to make the network fully connected. If the number of connections is less than n - 1, it is impossible to connect all computers. The problem tests your ability to use graph traversal techniques like DFS or BFS to manage connectivity between nodes.

Examples

Example 1

Input: n = 4, connections = [[0,1],[0,2],[1,2]]

Output: 1

Remove cable between computer 1 and 2 and place between computers 1 and 3.

Example 2

Input: n = 6, connections = [[0,1],[0,2],[0,3],[1,2],[1,3]]

Output: 2

Example details omitted.

Example 3

Input: n = 6, connections = [[0,1],[0,2],[0,3],[1,2]]

Output: -1

There are not enough cables.

Constraints

1 <= n <= 105
1 <= connections.length <= min(n * (n - 1) / 2, 105)
connections[i].length == 2
0 <= ai, bi < n
ai != bi
There are no repeated connections.
No two computers are connected by more than one cable.

Solution Approach

DFS / BFS Approach

Start by using DFS or BFS to explore all connected components in the graph. Count the number of disconnected components. If there are multiple components, you'll need extra cables to connect them, which equals the number of components minus one.

Union Find Approach

Union Find can be used to track connected components. For every pair of computers that are directly connected, perform a union operation. Count the number of distinct components remaining and subtract one to get the number of cables needed.

Edge Case Handling

Handle edge cases like insufficient cables where connections.length < n - 1, which immediately returns -1. Ensure to check for cases where there is no way to connect the network.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the chosen approach. For DFS/BFS, the complexity is O(n + m), where n is the number of computers and m is the number of connections. For Union Find, it is O(α(n)) per operation, where α(n) is the inverse Ackermann function, which is almost constant in practice. The space complexity is O(n) for both approaches due to the need to store the graph or the union-find structure.

What Interviewers Usually Probe

Check if the candidate recognizes the importance of connectivity before counting extra cables.
Observe whether they correctly identify the minimum number of cables needed based on graph traversal techniques.
Look for handling of edge cases such as insufficient connections or unreachable components.

Common Pitfalls or Variants

Common pitfalls

Misunderstanding the minimum cable count required when multiple disconnected components exist.
Failing to detect when it's impossible to connect all computers, like when the number of cables is less than n - 1.
Confusing the number of operations required with the number of components, leading to an incorrect solution.

Follow-up variants

Change the graph structure to a directed graph and adjust the problem accordingly.
Consider the case where the computers are initially connected in a cycle and add extra disconnections to the problem.
Increase the graph's size to test for scalability and performance in larger networks.

FAQ

How do you approach solving the "Number of Operations to Make Network Connected" problem?

Start by identifying the number of connected components in the graph using DFS, BFS, or Union Find. If the components exceed n - 1, calculate the extra cables needed.

What graph traversal techniques are useful in this problem?

DFS, BFS, and Union Find are all effective for counting connected components and determining the minimum number of operations needed.

Why do we need at least `n - 1` connections in this problem?

To connect all computers, the network must form a spanning tree, which requires at least n - 1 edges. If there are fewer, it is impossible to connect all computers.

Can this problem be solved with a greedy algorithm?

This problem is not well-suited to a greedy approach, as it requires graph traversal to ensure connectivity between components. Greedy algorithms would not efficiently handle disconnected components.

How does GhostInterview help with this problem?

GhostInterview guides you through the solution process, helps identify common pitfalls, and ensures you understand graph traversal techniques and their real-world applications.

terminal

Solution

Solution 1: Union-Find

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.

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class Solution:
    def makeConnected(self, n: int, connections: List[List[int]]) -> int:
        def find(x: int) -> int:
            if p[x] != x:
                p[x] = find(p[x])
            return p[x]

        cnt = 0
        p = list(range(n))
        for a, b in connections:
            pa, pb = find(a), find(b)
            if pa == pb:
                cnt += 1
            else:
                p[pa] = pb
                n -= 1
        return -1 if n - 1 > cnt else n - 1

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