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Minimize Malware Spread

Difficulty: Hard

Problem Description

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. 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. Note that if a node was removed from the initial list of infected nodes, it might still be infected later due to the malware spread.

Key Insights

  • The problem can be viewed as a graph traversal problem where we need to find the impact of removing each initially infected node on the total spread of malware.
  • We can use either Depth-First Search (DFS) or Breadth-First Search (BFS) to explore the connected components of the graph.
  • The goal is to identify which initially infected node's removal leads to the least number of final infections, considering the structure of the graph.

Space and Time Complexity

Time Complexity: O(n^2) - The graph traversal requires examining the adjacency matrix, which has a size of n^2. Space Complexity: O(n) - Space is needed for storing visited nodes and the components.

Solution

To solve this problem, we can use a graph traversal algorithm (DFS or BFS) to identify the connected components of the graph. We will simulate the removal of each initially infected node one by one, and for each simulation, we will calculate how many nodes would eventually get infected. The node removal that results in the minimum spread of malware will be our answer. If there are ties, we will return the node with the smallest index.

Code Solutions

def minMalwareSpread(graph, initial):
    def dfs(node, visited):
        stack = [node]
        count = 0
        while stack:
            curr = stack.pop()
            if curr in visited:
                continue
            visited.add(curr)
            count += 1
            for neighbor in range(len(graph)):
                if graph[curr][neighbor] == 1 and neighbor not in visited:
                    stack.append(neighbor)
        return count
    
    initial_set = set(initial)
    visited = set()
    component_count = {}
    
    # Count the size of components influenced by each node
    for node in range(len(graph)):
        if node not in visited:
            size = dfs(node, visited)
            for n in visited:
                component_count[n] = size
    
    # Calculate the impact of removing each initially infected node
    impact = {}
    for node in initial:
        infected_nodes = set()
        for n in visited:
            if n in initial_set and n != node:
                infected_nodes.add(n)
        total_infected = 0
        for inf_node in infected_nodes:
            total_infected += component_count[inf_node]
        impact[node] = total_infected
    
    # Find the minimum impacted node
    min_node = min(initial, key=lambda x: (impact[x], x))
    return min_node
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