Introduction

In this blog post, we will explore how to solve the LeetCode problem "924" 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. 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.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 approach and the Python code:

• High-Level Approach:

  • Identify Connected Components: Use Depth-First Search (DFS) to find connected components in the graph. Each component represents a group of nodes that will become infected together if any one of them is initially infected.
  • Analyze Initial Nodes in Components: For each connected component, count how many nodes from the initial list are present in that component.
  • Determine Optimal Removal: If a connected component contains only one node from the initial list, removing that node will prevent the entire component from becoming infected. We prioritize removing the node that leads to the largest reduction in the number of infected nodes (component size).

• Complexity:

  • Runtime Complexity: O(n^2), where n is the number of nodes in the graph (due to DFS).
  • Storage Complexity: O(n), for visited sets and component tracking.

Code

    def minMalwareSpread(graph, initial):
    n = len(graph)
    initial_set = set(initial)
    visited = set()
    components = []

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

    for node in range(n):
        if node not in visited:
            component = set()
            dfs(node, component)
            components.append(component)

    counts = {}
    for component in components:
        initial_in_component = []
        for node in component:
            if node in initial_set:
                initial_in_component.append(node)
        if len(initial_in_component) == 1:
            node = initial_in_component[0]
            counts[node] = counts.get(node, 0) + len(component)

    best_node = -1
    max_saved = -1
    for node in initial:
        saved = counts.get(node, 0)
        if saved > max_saved:
            max_saved = saved
            best_node = node
        elif saved == max_saved and node < best_node:
            best_node = node

    if best_node == -1:
        return min(initial)
    else:
        return best_node