<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
        
        # [[1,1,0],
        #  [1,1,0],
        #  [0,0,1]]
        
        #  [0,1,2]
        #   * 
        # seen = { 1 }
        
        # adj
        # 0: {1}
        # 1: {0}
        # 2: {}
        
        
        
        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)
        
        seen = set()
        
        def count_infected(node):
            if node in seen: return 0
            
            seen.add(node)
            
            count = 1
            
            for nxt in adj[node]:
                if nxt not in seen:
                    count += count_infected(nxt)
            print(node, count)
            return count
        
        min_i, min_infected = float('inf'), float('inf')
        initial.sort()
        
        for i in range(len(initial)):
            cur_infected = 0
            for j in range(len(initial)):
                if i == j: continue
                cur_infected += count_infected(initial[j])
                
            if cur_infected < min_infected or (cur_infected == min_infected and initial[i] < min_i):
                min_infected = cur_infected
                min_i = initial[i]
                
            seen.clear()
                
        return min_i
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