Introduction

In this blog post, we will explore how to solve the LeetCode problem "547" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

There are n cities. Some of them are connected, while some are not. If city a is connected directly with city b, and city b is connected directly with city c, then city a is connected indirectly with city c. A province is a group of directly or indirectly connected cities and no other cities outside of the group. You are given an n x n matrix isConnected where isConnected[i][j] = 1 if the ith city and the jth city are directly connected, and isConnected[i][j] = 0 otherwise. Return the total number of provinces. Example 1: Input: isConnected = [[1,1,0],[1,1,0],[0,0,1]] Output: 2 Example 2: Input: isConnected = [[1,0,0],[0,1,0],[0,0,1]] Output: 3 Constraints: 1 <= n <= 200 n == isConnected.length n == isConnected[i].length isConnected[i][j] is 1 or 0. isConnected[i][i] == 1 isConnected[i][j] == isConnected[j][i]

Explanation

Here's the solution to the province counting problem, focusing on efficiency and clarity:

• Core Idea: Treat the cities as nodes in a graph and the connections as edges. Use Depth-First Search (DFS) or Breadth-First Search (BFS) to explore connected components. Each connected component represents a province.

• Algorithm: Iterate through each city. If a city hasn't been visited yet, it marks the start of a new province. Perform DFS/BFS from that city to mark all connected cities as visited.

• Optimization: Using an efficient visited array to prevent redundant traversals and using DFS, which is well-suited for exploring connected components.

• Complexity: O(n^2) time complexity (due to traversing the adjacency matrix) and O(n) space complexity (for the visited array and recursion stack in DFS, though the stack can technically reach O(n^2) in worst-case complete graph scenario, it's practically O(n) in most typical province connection structures)

Code

    def findCircleNum(isConnected):
    """
    Finds the number of provinces in a given connectivity matrix.

    Args:
        isConnected: A list of lists representing the connectivity matrix.

    Returns:
        The number of provinces.
    """

    n = len(isConnected)
    visited = [False] * n
    province_count = 0

    def dfs(city):
        """
        Performs Depth-First Search to mark all connected cities as visited.

        Args:
            city: The index of the current city being visited.
        """
        visited[city] = True
        for neighbor in range(n):
            if isConnected[city][neighbor] == 1 and not visited[neighbor]:
                dfs(neighbor)

    for city in range(n):
        if not visited[city]:
            dfs(city)
            province_count += 1

    return province_count