<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:
        # connection = [[] for _ in len(graph)]
        # for i in range(graph):
        #     for j in range(graph[0]):
        #         connection[i].append(j)
        
        islands = []
        visited = [False] * len(graph)
        def dfs(index, connect):
            if visited[index]:
                return
            connect.append(index)
            visited[index] = True
            for i, con in enumerate(graph[index]):
                if con == 1 and i != index:
                    dfs(i, connect)
        
        for i in range(len(graph)):
            if not visited[i]:
                island = []
                dfs(i, island)
                islands.append(island)

        islands.sort(key=lambda x: -len(x))
        
        # print(islands)
        new_islands = []
        for island in islands:
            infect = False
            for i in island:
                if i in set(initial):
                    infect = True
                    break
            if infect:
                new_islands.append(island)
        islands = new_islands
        # print(islands)

        result = 0
        idx = min(initial)
        for remove in range(len(graph)):
            island = []
            for isld in islands:
                if remove in set(isld):
                    island = isld
                    break
            find = False
            for el in initial:
                if el != remove and el in island:
                    find = True
            if not find:
                if len(island) > result:
                    result = len(island)
                    idx = remove
            


        # initial = set(initial)
        # for island in islands:
        #     for i in island:
        #         if i in initial:
        #             return i
        
        return idx
25000