928. Minimize Malware Spread II LeetCode Solution

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Problem Statement
Complexity Analysis
Minimize Malware Spread II solution in C++
Minimize Malware Spread II solution in Java
Minimize Malware Spread II solution in Python
Additional Resources

928. Minimize Malware Spread II LeetCode Solution image

Problem Statement of Minimize Malware Spread II

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.

Complexity Analysis

Time Complexity: O(n^2 \cdot |\texttt{initial}|)
Space Complexity: O(n)

928. Minimize Malware Spread II LeetCode Solution in C++

class Solution {
 public:
  int minMalwareSpread(vector<vector<int>>& graph, vector<int>& initial) {
    int ans = 0;
    int minCount = graph.size();

    ranges::sort(initial);

    for (const int i : initial) {
      const int count = bfs(graph, i, initial);
      if (count < minCount) {
        minCount = count;
        ans = i;
      }
    }

    return ans;
  }

 private:
  int bfs(const vector<vector<int>>& graph, int removed, vector<int>& initial) {
    queue<int> q;
    vector<bool> seen(graph.size());
    seen[removed] = true;

    int count = 0;

    for (const int i : initial)
      if (i != removed) {
        q.push(i);
        seen[i] = true;
      }

    while (!q.empty()) {
      const int u = q.front();
      q.pop();
      ++count;
      for (int i = 0; i < graph.size(); ++i) {
        if (seen[i])
          continue;
        if (i != u && graph[i][u]) {
          q.push(i);
          seen[i] = true;
        }
      }
    }

    return count;
  }
};
/* code provided by PROGIEZ */

928. Minimize Malware Spread II LeetCode Solution in Java

class Solution {
  public int minMalwareSpread(int[][] graph, int[] initial) {
    int ans = 0;
    int minCount = graph.length;

    Arrays.sort(initial);

    for (final int i : initial) {
      final int count = bfs(graph, i, initial);
      if (count < minCount) {
        minCount = count;
        ans = i;
      }
    }

    return ans;
  }

  private int bfs(int[][] graph, int removed, int[] initial) {
    Queue<Integer> q = new ArrayDeque<>();
    boolean[] seen = new boolean[graph.length];
    seen[removed] = true;

    int count = 0;

    for (final int i : initial)
      if (i != removed) {
        q.offer(i);
        seen[i] = true;
      }

    while (!q.isEmpty()) {
      final int u = q.poll();
      ++count;
      for (int i = 0; i < graph.length; ++i) {
        if (seen[i])
          continue;
        if (i != u && graph[i][u] == 1) {
          q.offer(i);
          seen[i] = true;
        }
      }
    }

    return count;
  }
}
// code provided by PROGIEZ