Introduction

In this blog post, we will explore how to solve the LeetCode problem "928" 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

You are given a network of n nodes represented as an n x n adjacency matrix graph, where the ith node is directly connected to the jth node if graph[i][j] == 1. Some nodes initial are initially infected by malware. Whenever two nodes are directly connected, and at least one of those two nodes is infected by malware, both nodes will be infected by malware. This spread of malware will continue until no more nodes can be infected in this manner. Suppose M(initial) is the final number of nodes infected with malware in the entire network after the spread of malware stops. We will remove exactly one node from initial, completely removing it and any connections from this node to any other node. Return the node that, if removed, would minimize M(initial). If multiple nodes could be removed to minimize M(initial), return such a node with the smallest index. Example 1: Input: graph = [[1,1,0],[1,1,0],[0,0,1]], initial = [0,1] Output: 0 Example 2: Input: graph = [[1,1,0],[1,1,1],[0,1,1]], initial = [0,1] Output: 1 Example 3: Input: graph = [[1,1,0,0],[1,1,1,0],[0,1,1,1],[0,0,1,1]], initial = [0,1] Output: 1 Constraints: n == graph.length n == graph[i].length 2 <= n <= 300 graph[i][j] is 0 or 1. graph[i][j] == graph[j][i] graph[i][i] == 1 1 <= initial.length < n 0 <= initial[i] <= n - 1 All the integers in initial are unique.

Explanation

Here's the breakdown of the solution:

• Identify Connected Components: Use Depth-First Search (DFS) or Breadth-First Search (BFS) to find connected components in the graph. The initial malware nodes might belong to different connected components.
• Analyze Impact of Removing Each Initial Node: For each node in the initial list, temporarily remove it from the initial list and determine how many nodes will be infected after the malware spread. Compare the resulting infected node counts to find the node that minimizes the final infected count.

• Handle Ties: If multiple nodes minimize the infected count, choose the node with the smallest index.

• Runtime Complexity: O(n * (n + m)), where n is the number of nodes and m is the number of edges in the graph. This is because for each initial node, we potentially perform a DFS/BFS on the graph. Storage Complexity: O(n) due to the visited array used in DFS/BFS.

Code

    def minMalwareSpread(graph, initial):
    """
    Finds the node to remove from the initial set to minimize the final malware spread.

    Args:
        graph: An adjacency matrix representing the network.
        initial: A list of initially infected nodes.

    Returns:
        The node to remove that minimizes the malware spread.
    """

    n = len(graph)
    initial = sorted(initial)  # Sort initial for tie-breaking

    def dfs(node, infected, visited, current_component):
        visited[node] = True
        current_component.add(node)
        for neighbor in range(n):
            if graph[node][neighbor] == 1 and not visited[neighbor] and neighbor not in infected:
                dfs(neighbor, infected, visited, current_component)

    def get_infected_count(infected_nodes):
        visited = [False] * n
        infected_count = 0

        for node in range(n):
            if node in infected_nodes and not visited[node]:
                dfs(node, infected_nodes, visited, set())  # Start DFS from infected node

        infected_count = sum(visited)
        return infected_count


    best_node = -1
    min_infected = float('inf')

    for node_to_remove in initial:
        temp_initial = set(initial)
        temp_initial.remove(node_to_remove)

        infected_count = get_infected_count(temp_initial)

        if infected_count < min_infected:
            min_infected = infected_count
            best_node = node_to_remove
        elif infected_count == min_infected and node_to_remove < best_node:
            best_node = node_to_remove

    return best_node if best_node != -1 else min(initial)  # Handle empty initial case