<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:
        # 1. build the adj_list 
        adj_list = collections.defaultdict(list)
        n = len(graph)
        for i in range(n):
            for j in range(n):
                if graph[i][j] == 1: 
                    adj_list[i].append(j)
                    adj_list[j].append(i)

        # 3. fixe length arr 
        # -1: unexplored yet, never visit 
        # -2: this node shouldn't be considered, it has more than 
            # 1 sources 
        # any other value: valid color 
        colors = [-1] * n 


        def dfs(i, visited, source):
            # dfs to collect how many nodes it goes to 
            # we need to calulate how many nodes it can reach 
            # from the index i 

            """
            2 
            |  \
            0 - 1 
            (1,2)
            colors = [1,-1,]
            """
            # -1 means never visited 
            if colors[i] == -1: 
                colors[i] = source 
            # -2 means this node is never going to be 
            # uninfected by just dropping 1 index node 
            elif colors[i] != -1 and colors[i] != -2: 
                colors[i] = -2 
            for nei in adj_list[i]:
                if nei not in visited: 
                    visited.add(nei)
                    dfs(nei, visited, source)
            


        # 3. iterate through each index in initial 
        for start in initial: 
            # 4. dfs starting from the index 
            dfs(start, {start}, start)
        
        print(colors)
        # iterate throught the color 
        hashmap = collections.defaultdict(int)
        max_count = 0 
        max_count_index = float("inf")
        for val in colors: 
            if val != -1 and val != -2: 
                hashmap[val] += 1 
                if hashmap[val] > max_count or (hashmap[val] == max_count and val < max_count_index): 
                    max_count = hashmap[val]
                    max_count_index = val 
        if max_count_index != float("inf"):
            return max_count_index 
        return min(initial)
20000