Introduction

In this blog post, we will explore how to solve the LeetCode problem "1319" 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 computers numbered from 0 to n - 1 connected by ethernet cables connections forming a network where connections[i] = [ai, bi] represents a connection between computers ai and bi. Any computer can reach any other computer directly or indirectly through the network. You are given an initial computer network connections. You can extract certain cables between two directly connected computers, and place them between any pair of disconnected computers to make them directly connected. Return the minimum number of times you need to do this in order to make all the computers connected. If it is not possible, return -1. Example 1: Input: n = 4, connections = [[0,1],[0,2],[1,2]] Output: 1 Explanation: 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 3: Input: n = 6, connections = [[0,1],[0,2],[0,3],[1,2]] Output: -1 Explanation: 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.

Explanation

Here's the breakdown of the solution:

• Connectivity Check: The core idea is to determine the number of connected components in the network. If the number of cables (connections.length) is less than n - 1, it's impossible to connect all computers.
• Connected Components: We use Depth-First Search (DFS) to find the connected components. Each DFS traversal explores a single component.

• Redundant Cables: The minimum number of moves required is the number of connected components minus 1. This is because we need to connect all the separated components together.

• Runtime Complexity: O(n + m) where n is the number of computers and m is the number of connections. Storage Complexity: O(n + m)

Code

    def makeConnected(n: int, connections: list[list[int]]) -> int:
    """
    Calculates the minimum number of moves to connect all computers in a network.

    Args:
        n: The number of computers.
        connections: A list of connections between computers.

    Returns:
        The minimum number of moves needed to connect all computers, or -1 if it's impossible.
    """

    if len(connections) < n - 1:
        return -1

    adj = [[] for _ in range(n)]
    for u, v in connections:
        adj[u].append(v)
        adj[v].append(u)

    visited = [False] * n
    num_components = 0

    def dfs(node: int):
        visited[node] = True
        for neighbor in adj[node]:
            if not visited[neighbor]:
                dfs(neighbor)

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

    return num_components - 1