Problem Description

In this problem, we are given a total of n cities and a matrix called isConnected which is an n x n matrix. The element isConnected[i][j] will be 1 if there is a direct connection between city i and city j, and 0 if there is no direct connection. A set of cities is considered a province if all cities within the set are directly or indirectly connected to each other, and no city outside of the set is connected to any city within the set.

Our task is to determine how many provinces there are given the isConnected matrix.

Flowchart Walkthrough

To analyze Leetcode 547. Number of Provinces, let's go through the algorithm using the Flowchart. Here's a step-by-step walkthrough:

Is it a graph?

• Yes: The problem represents cities as nodes and if two cities are directly connected, there is an edge between them.

Is it a tree?

• No: The network can form multiple connected components and cycles, so it's not inherently a tree structure.

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

• No: The relationships are symmetric (if city A is connected to B, then B is connected to A), indicating undirected connectivity.

Is the problem related to shortest paths?

• No: The task is to count the number of connected components, not to find the shortest path between nodes.

Does the problem involve connectivity?

• Yes: The core of the problem is to determine how many connected groups of cities (provinces) there are.

Is the graph weighted?

• No: The connectivity between cities is simply binary (connected or not connected) with no weights associated with the connections.

Does the problem have small constraints?

• This varies based on the version of the problem and the test cases, but for DFS and BFS, typical constraints are manageable, so we don’t really need backtracking optimizations here.

Conclusion: The flowchart suggests using depth-first search (DFS) to efficiently explore and mark all cities within the same connected component (province) starting from any unvisited city, thereby effectively counting the number of such components, which equate to the number of provinces.

Intuition

To find the solution, we conceptualize the cities and the connections between them as a graph, where each city is a node and each direct connection is an edge. Now, the problem translates to finding the number of connected components in the graph. Each connected component will represent one province.

To do this, we use Depth-First Search (DFS). Here's the intuition behind using DFS:

1. We start with the first city and perform a DFS to mark all cities that are connected directly or indirectly to it. These cities form one province.
2. Once the DFS is completed, we look for the next city that hasn't been visited yet and perform a DFS from that city to find another province.
3. We repeat this process until all cities have been visited.

Each time we initiate a DFS from a new unvisited city, we know that we've found a new province, so we increment our province count. The DFS ensures that we navigate through all the cities within a province before moving on to the next one.

By doing the above steps using a vis (visited) list to keep track of which cities have been visited, we can effectively determine and count all the provinces.

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

Solution Approach

The solution uses a Depth-First Search (DFS) algorithm to explore the graph formed by the cities and connections. It utilizes an array vis to keep track of visited nodes (cities) to ensure we don't count the same province multiple times. Below is a step-by-step walk-through of the implementation:

1. Define a recursive function dfs(i: int) that will perform a depth-first search starting from city i.
2. Inside the dfs function, mark the current city i as visited by setting vis[i] to True.
3. Iterate over all cities using j (which correspond to the columns of isConnected[i]).
4. For each city j, check if j has not been visited (not vis[j]) and is directly connected to i (isConnected[i][j] == 1).
5. If that's the case, call dfs(j) to visit all cities connected to j, marking the entire connected component as visited.

The solution then follows these steps using the vis list:

6. Initialize the vis list to be of the same length as the number of cities (n), with all elements set to False, indicating that no cities have been visited yet.
7. Initialize a counter ans to 0, which will hold the number of provinces found.
8. Iterate through all cities i from 0 to n - 1.
9. For each city i, check if it has not been visited yet (not vis[i]).
10. If it hasn't, it means we've encountered a new province. Call dfs(i) to mark all cities within this new province as visited.
11. Increment the ans counter by 1 as we have found a new province.
12. Continue the loop until all cities have been visited and all provinces have been counted.

At the end of the loop, ans will contain the total number of provinces, which the function returns. This completes the solution implementation.

Example Walkthrough

Let's walk through a small example to illustrate the solution approach. Consider there are 4 cities, and the isConnected matrix is as follows:

isConnected = [
    [1, 1, 0, 0],
    [1, 1, 0, 0],
    [0, 0, 1, 1],
    [0, 0, 1, 1]
]

Here, cities 0 and 1 are connected, as well as cities 2 and 3, forming two distinct provinces.

We initialize our vis list as [False, False, False, False] and set our province counter ans to 0.

Now let's perform the steps of the algorithm:

1. We start with city 0 and run dfs(0).

   • In dfs(0), city 0 is marked visited: vis = [True, False, False, False].
   • We find that city 0 is connected to city 1, dfs(1) is called.
     • In dfs(1), city 1 is marked visited: vis = [True, True, False, False].
     • There are no unvisited cities connected to city 1, so dfs(1) ends.

2. Since all cities connected to city 0 are now visited, dfs(0) ends. We've found our first province, so we increment ans to 1.

3. Next, we move to city 1, but since it's already visited, we proceed to city 2 and run dfs(2).

   • In dfs(2), city 2 is marked visited: vis = [True, True, True, False].
   • We find that city 2 is connected to city 3, dfs(3) is called.
     • In dfs(3), city 3 is marked visited: vis = [True, True, True, True].
     • There are no unvisited cities connected to city 3, so dfs(3) ends.

4. At this point, all cities connected to city 2 are visited, ending dfs(2). We've found another province, incrementing ans to 2.

5. Finally, we move to city 3 and see it's already visited.

Now, we've visited all cities, and there are no unvisited cities to start a new dfs from. Thus, we conclude there are 2 provinces in total, which is the value of ans.

The full algorithm will perform similarly on a larger scale, incrementing the province count each time it initiates a DFS on an unvisited city, and continuing until all cities are visited. The final result is the total number of provinces.

Solution Implementation

Time and Space Complexity

The given code snippet represents the Depth-First Search (DFS) approach for finding the number of connected components (which can be referred to as 'circles') in an undirected graph represented by an adjacency matrix isConnected.

Time Complexity

The time complexity of the algorithm is O(N^2), where N is the number of vertices (or cities) in the graph. This complexity arises because the algorithm involves visiting every vertex once and, for each vertex, iterating through all possible adjacent vertices to explore the edges. In the worst-case scenario, this results in checking every entry in the isConnected matrix once, which has N^2 entries.

The dfs function explores all connected vertices through recursive calls. Since each edge and vertex will only be visited once in the DFS traversal, the total number of operations performed will be related to the total number of vertices and edges. However, because the graph is represented by an N x N adjacency matrix and it must be fully inspected to discover all connections, the time is bounded by the size of the matrix (N^2).

Space Complexity

The space complexity of the algorithm is O(N), which comes from the following:

1. The recursion stack for DFS, which in the worst case, could store up to N frames if the graph is implemented as a linked structure such as a list of nodes (a path with all nodes connected end-to-end).
2. The vis visited array, which is a boolean array of length N used to track whether each vertex has been visited or not to prevent cycles during the DFS.

Given that N recursion calls could happen in a singly connected component spanning all the vertices, the space taken by the call stack should be considered in the final space complexity, merging it with the space taken by the array to O(N).

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