What is the time complexity of Minimize Malware Spread II?

The optimal solution has a time complexity of O(n^2*m).

What is the space complexity of Minimize Malware Spread II?

The space complexity is O(n+m).

What data structures are used to solve Minimize Malware Spread II?

This problem uses Array, Hash Table, Depth-First Search, Breadth-First Search, Union Find, Graph. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Minimize Malware Spread II?

Insight 1: Removing a node from 'initial' can prevent the spread of malware from that node, potentially reducing the total number of infected nodes. Insight 2: Breadth-First Search (BFS) or Depth-First Search (DFS) can be used to simulate the malware spread effectively within the graph.

Which companies ask LeetCode 928. Minimize Malware Spread II in interviews?

This problem is frequently asked at Dropbox.

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Minimize Malware Spread II - LeetCode 928 Solution

Minimize Malware Spread II - Complete Solution Guide

Minimize Malware Spread II is LeetCode problem 928, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given a network of n nodes represented as an n x n adjacency matrix graph , where the i th node is directly connected to the j th 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 i

Detailed Explanation

The problem presents a network of nodes represented by an adjacency matrix, where `graph[i][j] = 1` indicates a direct connection between nodes `i` and `j`. Some nodes are initially infected with malware. The malware spreads through connected nodes until no further nodes can be infected. The goal is to find the single node from the `initial` set of infected nodes that, when removed, minimizes the total number of infected nodes after the spread stops. If multiple such nodes exist, the node with the smallest index should be returned.

Solution Approach

The provided solution iterates through each node in the 'initial' array, considering each as a candidate to be removed. For each candidate, it simulates the spread of malware from the remaining infected nodes using Breadth-First Search (BFS). It then counts the number of infected nodes after the spread. The candidate node that results in the minimum number of infected nodes is selected as the node to remove.

Step-by-Step Algorithm

Step 1: Sort the 'initial' array to ensure that if multiple nodes result in the same minimum infected count, the smallest index will be chosen.
Step 2: Initialize 'min_infected_count' to infinity and 'result_node' to the first element of the sorted 'initial' array. These variables will store the minimum infected count found so far and the corresponding node to remove.
Step 3: Iterate through each node 'node_to_remove' in the 'initial' array.
Step 4: For each 'node_to_remove', create a set 'initial_sources' containing the remaining initial infected nodes (excluding 'node_to_remove').
Step 5: Perform a BFS starting from the nodes in 'initial_sources'. During the BFS, ensure that 'node_to_remove' is not visited or considered as a potential source of infection.
Step 6: Keep track of the infected nodes in a 'infected' set. Each time a node is infected during the BFS, add it to this set.
Step 7: After the BFS completes, count the number of nodes in the 'infected' set. This is the 'current_infected_count'.
Step 8: Compare 'current_infected_count' with 'min_infected_count'. If 'current_infected_count' is smaller, update 'min_infected_count' with 'current_infected_count' and update 'result_node' with 'node_to_remove'.
Step 9: After iterating through all nodes in 'initial', return 'result_node'.

Key Insights

Insight 1: Removing a node from 'initial' can prevent the spread of malware from that node, potentially reducing the total number of infected nodes.
Insight 2: Breadth-First Search (BFS) or Depth-First Search (DFS) can be used to simulate the malware spread effectively within the graph.
Insight 3: Sorting the 'initial' array allows for consistent selection of the smallest-indexed node when multiple nodes lead to the same minimum infection count.

Complexity Analysis

Time Complexity: O(n^2*m)

Space Complexity: O(n+m)

Topics

This problem involves: Array, Hash Table, Depth-First Search, Breadth-First Search, Union Find, Graph.

Companies

Asked at: Dropbox.

Minimize Malware Spread II | LeetCode 928

Hard

Array Hash Table Depth-First Search Breadth-First Search Union Find Graph

Time: O(n^2*m)Space: O(n+m)

Solution Code

Available in:

import java.util.*;

class Solution {
    public int minMalwareSpread(int[][] graph, int[] initial) {
        int n = graph.length;
        Arrays.sort(initial);

        int min_infected_count = Integer.MAX_VALUE;
        int result_node = initial[0];

        for (int node_to_remove : initial) {
            Set<Integer> initial_sources = new HashSet<>();
            for (int node : initial) {
                initial_sources.add(node);
            }
            initial_sources.remove(node_to_remove);

            Queue<Integer> q = new LinkedList<>(initial_sources);
            Set<Integer> infected = new HashSet<>(initial_sources);

            while (!q.isEmpty()) {
                int u = q.poll();

                for (int v = 0; v < n; v++) {
                    if (graph[u][v] == 1 && v != node_to_remove && !infected.contains(v)) {
                        infected.add(v);
                        q.offer(v);
                    }
                }
            }

            int current_infected_count = infected.size();

            if (current_infected_count < min_infected_count) {
                min_infected_count = current_infected_count;
                result_node = node_to_remove;
            }

        }

        return result_node;
    }
}

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Problem Info

Description

Problem 928: Minimize Malware Spread II - Detailed explanation and solution approaches.

Topics

Array Hash Table Depth-First Search Breadth-First Search Union Find Graph

Companies

Dropbox

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:928

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

9 steps

1
Step 1: Sort the 'initial' array to ensure that if multiple nodes result in the same minimum infected count, the smallest index will be chosen.
2
Step 2: Initialize 'min_infected_count' to infinity and 'result_node' to the first element of the sorted 'initial' array. These variables will store the minimum infected count found so far and the corresponding node to remove.
3
Step 3: Iterate through each node 'node_to_remove' in the 'initial' array.
4
Step 4: For each 'node_to_remove', create a set 'initial_sources' containing the remaining initial infected nodes (excluding 'node_to_remove').
5
Step 5: Perform a BFS starting from the nodes in 'initial_sources'. During the BFS, ensure that 'node_to_remove' is not visited or considered as a potential source of infection.
6
Step 6: Keep track of the infected nodes in a 'infected' set. Each time a node is infected during the BFS, add it to this set.
7
Step 7: After the BFS completes, count the number of nodes in the 'infected' set. This is the 'current_infected_count'.
8
Step 8: Compare 'current_infected_count' with 'min_infected_count'. If 'current_infected_count' is smaller, update 'min_infected_count' with 'current_infected_count' and update 'result_node' with 'node_to_remove'.
9
Step 9: After iterating through all nodes in 'initial', return 'result_node'.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n^2*m)

💾

Space Complexity

O(n+m)

📝

Why?

The outer loop iterates through the 'initial' array (length 'm'). The BFS traversal visits each node and edge at most once, taking O(n+e) time, where 'n' is the number of nodes and 'e' is the number of edges. In a dense graph e can go up to n^2 making BFS as O(n^2). For each element in initial we are doing this. The space is determined by the queue in BFS and the infected nodes sets, which can hold up to n nodes.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Union Find

Union Find (Disjoint Set) can be used to identify connected components in the graph. By analyzing how removing each initial node affects the size and number of infected components, we can determine the optimal node to remove.

Time: O(n^2 + m*alpha(n)), where n is number of nodes, m is the initial size and alpha is the very slowly growing inverse Ackermann function.Space: O(n)

Trade-offs: Union Find might be more efficient for larger graphs where connected components are more sparse. The constant factors in the implementation can also impact performance compared to BFS.

✨

Key Insights

1Insight 1: Removing a node from 'initial' can prevent the spread of malware from that node, potentially reducing the total number of infected nodes.
2Insight 2: Breadth-First Search (BFS) or Depth-First Search (DFS) can be used to simulate the malware spread effectively within the graph.
3Insight 3: Sorting the 'initial' array allows for consistent selection of the smallest-indexed node when multiple nodes lead to the same minimum infection count.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Graph

Nodes connected by edges. Like a social network or city map!

Edge Cases

Edge case 1: If 'initial' contains only one node, removing it will result in no initial infections. The solution correctly handles this as it iterates through the single element.
Edge case 2: If the graph is disconnected, the BFS will only explore the connected component containing the initial infected nodes. This is handled correctly as the BFS naturally limits the spread to reachable nodes.
Edge case 3: If removing one node leaves the initial infections unchanged or reduced, then solution ensures that first of such node index is picked. This is ensured by sorting intitial and updating result only when current count is strictly less than the minimum.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Failing to properly remove the node and its connections during the BFS, leading to an incorrect simulation of the spread.

✗Common mistake 2: Not considering the case where multiple nodes in 'initial' may result in the same minimum infected count and failing to return the node with the smallest index.

✗Common mistake 3: Incorrectly calculating the number of infected nodes after the BFS, for example, counting the removed node as infected.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Identifying and isolating a compromised server in a network to minimize the spread of a virus or malware. The 'initial' set represents the initially infected servers, and the goal is to remove the server that will minimize the impact of the infection.

🏦Real-world scenario 2: Contact tracing during an epidemic. The 'graph' represents contacts between individuals, the 'initial' set represents the initially infected individuals, and the goal is to isolate an individual who, if isolated, would minimize the spread of the disease.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: June 29, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 14

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Minimize Malware Spread II"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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