<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:
        # M(initial) = final number of nodes infected with malware after spread stops
        
        # Brute Force
        # 1. Build graph
        # 2. For each infected node
            # - Remove from initial infected nodes, count final # infected nodes
            # - Re-add to initial infected nodes
        # Minimize on # of infected nodes for each removed item
        
        
        adj = defaultdict(set)
        
        for i in range(len(graph)):
            for j in range(len(graph[i])):
                is_edge = graph[i][j] == 1
                if i == j: continue
                if is_edge:
                    adj[i].add(j)
                    adj[j].add(i)
        
        colors = {}
        
        def color_node(node, color):
            
            colors[node] = color
            
            for nxt in adj[node]:
                if nxt not in colors:
                    color_node(nxt, color)
                    
            return count
        
        
        n = len(graph)
        c = 0
        for i in range(n):
            if i not in colors:
                color_node(i, c)
                c += 1
                
        color_sizes = defaultdict(int) # number of nodes per color
        
        for node in range(n):
            color_sizes[colors[node]] += 1
            
            
        color_infected = defaultdict(int) # number of infected nodes per color
        
        for node in initial:
            color_infected[colors[node]] += 1
            
        min_i, max_num_saved = float('inf'), -float('inf')
                
        for node in initial:
            color = colors[node]
            num_infected = color_infected[color]
            num_saved = 0 if num_infected > 1 else color_sizes[color]
            if num_saved > max_num_saved or (num_saved == max_num_saved and node < min_i):
                min_i = node
                max_num_saved = num_saved
        
        return min_i
            
            
        
        
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