<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 {
public:
    map<int, int> par; // Parent map for union-find
    map<int, int> sz;  // Size map to store the size of each component
    map<int, int> cnt; // Count map to store the number of malware nodes in each component

    // Find with path compression
    int find(int a) {
        if (par[a] == a) {
            return a;
        }
        return par[a] = find(par[a]);
    }

    // Union by size
    void unite(int a, int b) {
        int par_a = find(a);
        int par_b = find(b);
        if (par_a != par_b) {
            par[par_a] = par_b;
            sz[par_b] += sz[par_a];
        }
    }

    int minMalwareSpread(vector<vector<int>>& graph, vector<int>& initial) {
        int n = graph.size();

        // Initialize union-find structures
        for (int i = 0; i < n; i++) {
            par[i] = i;
            sz[i] = 1;
        }

        // Build the union-find structure based on the graph
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (graph[i][j] && i != j) {
                    unite(i, j);
                }
            }
        }

        // Count the number of malware nodes in each connected component
        for (auto it : initial) {
            int root = find(it);
            cnt[root]++;
        }

        int mx_size = -1;
        int id = -1;

        // Find the best node to remove
        for (auto it : initial) {
            int root = find(it);
            if (cnt[root] == 1) { // Only consider components with exactly one malware
                if (sz[root] > mx_size || (sz[root] == mx_size && it < id)) {
                    mx_size = sz[root];
                    id = it;
                }
            }
        }
   if (id == -1) {
            id = *min_element(initial.begin(), initial.end());
        }

        return id;
    }
};

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