<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>

class Solution:
    def minMalwareSpread(self, graph: List[List[int]], initial: List[int]) -> int:
        # Helper function to perform DFS and return all nodes in the same component
        def dfs(node, visited, graph):
            stack = [node]
            component = []
            while stack:
                u = stack.pop()
                if u not in visited:
                    visited.add(u)
                    component.append(u)
                    for v in range(len(graph)):
                        if graph[u][v] == 1 and v not in visited:
                            stack.append(v)
            return component
        
        n = len(graph)
        visited = set()
        components = []
        
        # Step 1: Find all connected components
        for i in range(n):
            if i not in visited:
                components.append(dfs(i, visited, graph))
        
        # Step 2: Map which component each node belongs to
        node_to_component = {}
        for idx, component in enumerate(components):
            for node in component:
                node_to_component[node] = idx
        
        # Step 3: Count the number of initially infected nodes in each component
        component_infected_count = [0] * len(components)
        for node in initial:
            component_infected_count[node_to_component[node]] += 1
        
        # Step 4: Evaluate the impact of removing each node in initial
        result = (-1, min(initial))  # Stores (reduction in infections, node index)
        for node in initial:
            component_idx = node_to_component[node]
            if component_infected_count[component_idx] == 1:  # Only this node is infected in its component
                reduction = len(components[component_idx])  # Removing this node saves the entire component
            else:
                reduction = 0  # Removing this node won't save the component since others are infected
            
            # Choose the node that gives the maximum reduction in infections
            if reduction > result[0] or (reduction == result[0] and node < result[1]):
                result = (reduction, node)
        
        return result[1]
19400