<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:
        
        initial = set(initial)

        # 1. traverse each connected component and track # initial inf in component and total size

        # 2. map each node in initial to initial inf and total connected component size

        inf_node_map = defaultdict(list)  # key = inf node, value = [# initial inf, # of connected component size]

        def helper(node):
            
            inf_size = 1
            seen = set([node])
            queue = deque([node])

            while len(queue) > 0:
                
                curr = queue.popleft()

                for nei in range(len(graph)):
                    if nei == curr:
                        continue
                    elif graph[curr][nei] == 0:
                        continue
                    elif nei in seen:
                        continue
                    
                    seen.add(nei)
                    queue.append(nei)

                    if nei in initial:
                        inf_size += 1

            return inf_size, len(seen)

        for node in initial:
            inf_node_map[node] = helper(node)

        # 3. If each connected component has >= 2 initial inf, return smallest index
        if all([inf_size >= 2 for inf_size, _ in inf_node_map.values()]) is True:
            return min(initial)

        # 4. If there's a component with exactly 1 initial inf, iterate through node in initial and find largest 
        # number of component size (take smallest index)
        ansIdx = len(graph) 
        largest_size = -1

        for inf_node, (inf_size, connected_size) in inf_node_map.items():

            if inf_size >= 2:
                continue
            elif largest_size < connected_size:
                largest_size = connected_size
                ansIdx = inf_node
            elif largest_size == connected_size:
                ansIdx = min(ansIdx, inf_node)

        return ansIdx

19000