Minimize Malware Spread
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.
Example 1:
Input: graph = [[1,1,0],[1,1,0],[0,0,1]], initial = [0,1]
Output: 0
Example 2:
Input: graph = [[1,0,0],[0,1,0],[0,0,1]], initial = [0,2]
Output: 0
Example 3:
Input: graph = [[1,1,1],[1,1,1],[1,1,1]], initial = [1,2]Output: 1
Constraints:
n == graph.lengthn == graph[i].length2 <= n <= 300graph[i][j]is0or1.graph[i][j] == graph[j][i]graph[i][i] == 11 <= initial.length <= n0 <= initial[i] <= n - 1- All the integers in
initialare unique.
Solve it on LeetCode
Approaches
1. DFS for Component Identification
Intuition:
The problem is essentially about identifying connected components in a graph and determining the effect of removing a malware-infected node. We can use Depth First Search (DFS) to identify the size of each connected component and keep track of nodes that are initially infected.
- First, mark all connected nodes using DFS for each unvisited node, which helps identify components.
- For each initially infected node, simulate its removal and determine how many other nodes can be saved by using the largest saved component strategy.
Algorithm:
- Traverse the graph using DFS to identify components and their sizes.
- Record infected nodes within these components.
- Evaluate each infected node's removal impact. If multiple nodes result in the same number of saved nodes, choose the node with the smallest index.
Code:
- Time Complexity: O(N^2) due to the DFS which traverses all nodes and edges.
- Space Complexity: O(N) for storing visited states and component information.
2. Union Find (Disjoint Set)
Intuition:
A more efficient approach than DFS is using a Union Find (Disjoint Set) data structure to determine connected components. This method efficiently merges nodes and provides component size information.
- Use Union-Find to identify connected components.
- Count infected nodes within each component.
- By eliminating a node that is the only infected node within a component, maximize saved nodes.
Algorithm:
- Implement Union-Find to determine component connections.
- Count the infected nodes in each component.
- Identify the node whose removal saves the maximum nodes.
Code:
- Time Complexity: O(N^2) amortized due to the union-find operations needed for dense graph connections.
- Space Complexity: O(N) for storing parent and rank information in Union-Find.