<p>You are given a network of <code>n</code> nodes represented as an <code>n x n</code> adjacency matrix <code>graph</code>, where the <code>i<sup>th</sup></code> node is directly connected to the <code>j<sup>th</sup></code> node if <code>graph[i][j] == 1</code>.</p>

<p>Some nodes <code>initial</code> 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.</p>

<p>Suppose <code>M(initial)</code> is the final number of nodes infected with malware in the entire network after the spread of malware stops. We will remove <strong>exactly one node</strong> from <code>initial</code>.</p>

<p>Return the node that, if removed, would minimize <code>M(initial)</code>. If multiple nodes could be removed to minimize <code>M(initial)</code>, return such a node with <strong>the smallest index</strong>.</p>

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

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> graph = [[1,1,0],[1,1,0],[0,0,1]], initial = [0,1]
<strong>Output:</strong> 0
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> graph = [[1,0,0],[0,1,0],[0,0,1]], initial = [0,2]
<strong>Output:</strong> 0
</pre><p><strong class="example">Example 3:</strong></p>
<pre><strong>Input:</strong> graph = [[1,1,1],[1,1,1],[1,1,1]], initial = [1,2]
<strong>Output:</strong> 1
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == graph.length</code></li>
	<li><code>n == graph[i].length</code></li>
	<li><code>2 &lt;= n &lt;= 300</code></li>
	<li><code>graph[i][j]</code> is <code>0</code> or <code>1</code>.</li>
	<li><code>graph[i][j] == graph[j][i]</code></li>
	<li><code>graph[i][i] == 1</code></li>
	<li><code>1 &lt;= initial.length &lt;= n</code></li>
	<li><code>0 &lt;= initial[i] &lt;= n - 1</code></li>
	<li>All the integers in <code>initial</code> are <strong>unique</strong>.</li>
</ul>

from collections import defaultdict

class Solution:
    def minMalwareSpread(self, graph: List[List[int]], initial: List[int]) -> int:
        n = len(graph)
        
        # Function to perform DFS (Depth-First Search) and mark connected components
        def dfs(node, component):
            if node in visited:
                return
            visited.add(node)
            components[node] = component
            for neighbor in range(n):
                if graph[node][neighbor] == 1:
                    dfs(neighbor, component)
        
        # Find all connected components
        visited = set()
        components = {}
        component_id = 0
        for i in range(n):
            if i not in visited:
                dfs(i, component_id)
                component_id += 1
        
        # Count initial infected nodes in each component
        infected_count = defaultdict(int)
        for node in initial:
            infected_count[components[node]] += 1
        
        # Calculate the impact of removing each initially infected node
        impact = defaultdict(int)
        for node in initial:
            component = components[node]
            if infected_count[component] == 1:
                impact[node] = sum(1 for i in range(n) if components[i] == component)
        
        # Find the node with maximum impact (or smallest index if tied)
        max_impact = max(impact.values(), default=0)
        candidates = [node for node, imp in impact.items() if imp == max_impact]
        
        return min(candidates) if candidates else min(initial)

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