Problem Description

In this LeetCode problem, we are given a network of n computers, indexed from 0 to n-1. The connections between these computers are represented as a list of pairs, where each pair [ai, bi] indicates a direct ethernet cable connection between the computers ai and bi. Even though direct connections exist only between certain pairs, every computer can be reached from any other computer through a series of connections, forming a network.

Our task is to make all computers directly or indirectly connected with the least number of operations. An operation consists of disconnecting an ethernet cable between two directly connected computers and reconnecting it to a pair of disconnected computers.

We seek to find the minimum number of such operations needed to connect the entire network. If it is impossible to connect all computers due to an insufficient number of cables, the function should return -1.

Flowchart Walkthrough

Let's analyze LeetCode 1319, "Number of Operations to Make Network Connected" using the Flowchart. I'll guide you through the decision-making process based on this problem's specifics:

Is it a graph?

• Yes: The problem explicitly describes a network of computers connected by cables, forming a graph where computers are nodes, and cables are edges.

Is it a tree?

• No: The graph can have cycles since extra cables could form loops.

Is the problem related to directed acyclic graphs (DAGs)?

• No: The problem deals with an undirected graph; cables do not have a direction and form a plain network.

Is the problem related to shortest paths?

• No: The task is to find the minimum number of operations to connect all computers, not the shortest path between them.

Does the problem involve connectivity?

• Yes: The goal is to ensure that all computers are connected, forming one entire network (one connected component).

Is the graph weighted?

• No: The graph is unweighted; each cable/connection has the same import without any specific weight(characteristics).

Conclusion: Based on the flowchart, using a graph-based method focusing on connectivity for an unweighted graph, DFS is appropriate. DFS can efficiently explore each connected component to determine how many additional connections (operations) are needed to connect all components into a single network.

Intuition

The solution revolves around the concept of the Union-Find algorithm, a common data structure used for disjoint-set operations. The core idea is to establish a way to check quickly if two elements are in the same subset and to unify two subsets into one.

Here's an intuitive breakdown of the approach:

1. Initialization: We start by assuming each computer forms its own single-computer network. This is represented by an array where each index represents a computer and contains its "parent." Initially, all computers are their own parents, forming n independent sets.

2. Counting Redundant Connections: As we iterate through the list of connections, we use the Union-Find with path compression to determine if two computers are already connected. If they are in the same subset, this connection is redundant. We count redundant connections because they represent extra cables that we can use to connect other more significant parts of the network.

3. Unifying Sets: For each non-redundant connection, we unify the sets that the two computers belong to. This action reduces the number of independent sets by one since we are connecting two previously separate networks.

4. Determine Minimum Operations: After all direct connections are processed, the number of independent sets remaining (n) minus one (since n-1 connections are required to connect n computers in a single network) gives us the number of operations needed to connect the entire network.

5. Check for Possibility: If we have more independent networks than redundant connections (extra cables) remaining, it's impossible to connect all computers, and we return -1.

6. Result: If it's possible to connect all computers, we return the number of operations needed (n-1), which corresponds to the number of extra connections required to connect all disjoint networks.

By using Union-Find, we efficiently manage the merging of networks and avoid unnecessary complexity, leading us to the optimal number of operations required to connect the entire network.

Learn more about Depth-First Search, Breadth-First Search, Union Find and Graph patterns.

Solution Approach

The solution for connecting all computers with a minimum amount of operations applies the Union-Find algorithm, which includes methods to find the set to which an element belongs, and to unite two sets. Here's a step-by-step walkthrough of how the implementation works:

1. Creating a find function: This function is recursive and is used to find the root parent of a computer. If the computer is its own parent, we return its value; otherwise, we recursively find the parent, applying path compression along the way by setting p[x] to the root parent. Path compression flattens the structure, reducing the time complexity of subsequent find operations.

   def find(x):
       if p[x] != x:
           p[x] = find(p[x])
       return p[x]

2. Initializing the Parent List: A list p of range n is initialized, signifying that each computer is initially its own parent, forming n separate sets.

   p = list(range(n))

3. Iterating Over Connections: For each connection [a, b] in connections, the algorithm checks whether a and b belong to the same set by comparing the root parent found by find(a) and find(b).

   for a, b in connections:
       if find(a) == find(b):
           cnt += 1
       else:
           p[find(a)] = find(b)
           n -= 1

   • If they are in the same set, this connection is redundant (extra), and a counter cnt is incremented.
   • If they are not, we unite the sets by making the root parent of a the parent of the root parent of b, effectively merging the two sets. We also decrement the count of separate networks (n) by 1.

4. Calculating Minimum Operations: Once all pairs are processed, the algorithm checks if there are enough spare (redundant) connections to connect the remaining separate networks. This is done by comparing n - 1 (the minimum number of extra connections required) with cnt (the number of spare connections). If n - 1 is greater than cnt, it's impossible to connect all networks, and the function returns -1. Otherwise, it returns n - 1 as the number of operations required.

   return -1 if n - 1 > cnt else n - 1

The underlying logic of Union-Find with path compression ensures that the time complexity of the find operation is almost constant (amortized O(alpha(n))), where alpha is the inverse Ackermann function, which grows extremely slowly and is less than 5 for all practical values of n. This property makes our solution highly efficient for the given problem.

Example Walkthrough

Let's assume we have a network with n = 4 computers and the following list of ethernet cable connections (connections): [[0, 1], [1, 2], [1, 3], [2, 3]]. We want to find the minimum number of operations to connect the network.

1. Creating the Parent List: We initialize a list p = [0, 1, 2, 3], which signifies that each computer is initially its own parent.

2. Iterating Over Connections: We process each pair in connections and track redundant connections with a counter cnt = 0.

   • For [0, 1], find(0) returns 0 and find(1) returns 1. Since they have different parents, we connect them by setting p[0] as the parent of p[1] (or vice versa). Now p = [0, 0, 2, 3], and n = 4 - 1 = 3.

   • For [1, 2], find(1) returns 0 and find(2) returns 2. They are not in the same set, so we connect them, updating p[2] to 0 and n becomes 2. Now p = [0, 0, 0, 3].

   • For [1, 3], find(1) gives 0 and find(3) gives 3. They belong to different sets, so we merge them, setting p[3] to 0 and n becomes 1. Now p = [0, 0, 0, 0].

   • Finally, for [2, 3], find(2) and find(3) both return 0, indicating they are in the same set. This means the connection is redundant, we increment cnt to 1.

3. Calculating Minimum Operations: We have n - 1 = 1 - 1 = 0 minimum extra connections required to connect all computers and cnt = 1 redundant connections.

   Since we already have one redundant connection and we need zero connections to unify the network (all computers have been connected through previous steps), we can perform 0 operations to connect the network.

Therefore, our function would return 0 for this example.

The example illustrates that by using Union-Find, we were able to merge disjoint sets efficiently, track redundant connections, and detect that no further operations were needed to fully connect the network.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the given code mainly depends on the number of connections and the union-find operations performed on them. In the best-case scenario, where all connections are already in their own set, the find operation takes constant time, thus the time complexity would be O(C), where C is the number of connections. However, in the average and worst case, where the find function may traverse up to n nodes (with path compression, it would be less), the time complexity of each find operation can be approximated to O(log*n), which is the inverse Ackermann function, very slowly growing and often considered almost constant. Therefore, for C connections, the average time complexity is O(C * log*n).

The final check for whether the number of extra connections is sufficient to connect all components has a constant time complexity of O(1).

So, the overall time complexity of the algorithm is O(C * log*n).

Space Complexity

The space complexity is determined by the space needed to store the parent array p, which contains n elements. Hence, the space complexity is O(n) for the disjoint set data structure.

Additionally, no other auxiliary space that scales with the problem size is used, so the total space complexity remains O(n).

Learn more about how to find time and space complexity quickly using problem constraints.