<p>You are given an integer <code>n</code>. There is an <strong>undirected</strong> graph with <code>n</code> vertices, numbered from <code>0</code> to <code>n - 1</code>. You are given a 2D integer array <code>edges</code> where <code>edges[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> denotes that there exists an <strong>undirected</strong> edge connecting vertices <code>a<sub>i</sub></code> and <code>b<sub>i</sub></code>.</p>

<p>Return <em>the number of <strong>complete connected components</strong> of the graph</em>.</p>

<p>A <strong>connected component</strong> is a subgraph of a graph in which there exists a path between any two vertices, and no vertex of the subgraph shares an edge with a vertex outside of the subgraph.</p>

<p>A connected component is said to be <b>complete</b> if there exists an edge between every pair of its vertices.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<p><strong class="example"><img alt="" src="https://assets.leetcode.com/uploads/2023/04/11/screenshot-from-2023-04-11-23-31-23.png" style="width: 671px; height: 270px;" /></strong></p>

<pre>
<strong>Input:</strong> n = 6, edges = [[0,1],[0,2],[1,2],[3,4]]
<strong>Output:</strong> 3
<strong>Explanation:</strong> From the picture above, one can see that all of the components of this graph are complete.
</pre>

<p><strong class="example">Example 2:</strong></p>

<p><strong class="example"><img alt="" src="https://assets.leetcode.com/uploads/2023/04/11/screenshot-from-2023-04-11-23-32-00.png" style="width: 671px; height: 270px;" /></strong></p>

<pre>
<strong>Input:</strong> n = 6, edges = [[0,1],[0,2],[1,2],[3,4],[3,5]]
<strong>Output:</strong> 1
<strong>Explanation:</strong> The component containing vertices 0, 1, and 2 is complete since there is an edge between every pair of two vertices. On the other hand, the component containing vertices 3, 4, and 5 is not complete since there is no edge between vertices 4 and 5. Thus, the number of complete components in this graph is 1.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 50</code></li>
	<li><code>0 &lt;= edges.length &lt;= n * (n - 1) / 2</code></li>
	<li><code>edges[i].length == 2</code></li>
	<li><code>0 &lt;= a<sub>i</sub>, b<sub>i</sub> &lt;= n - 1</code></li>
	<li><code>a<sub>i</sub> != b<sub>i</sub></code></li>
	<li>There are no repeated edges.</li>
</ul>

class UnionFind:
    def __init__(self, n):
        self.parent = list(range(n))
        self.rank = [1] * n
        self.size = [1] * n  # Track the size of each connected component
        self.edge_count = [0] * n  # Track the number of edges in each component

    def find(self, x):
        if self.parent[x] != x:
            self.parent[x] = self.find(self.parent[x])  # Path compression
        return self.parent[x]

    def union(self, x, y):
        rootX = self.find(x)
        rootY = self.find(y)

        if rootX != rootY:
            # Union by rank
            if self.rank[rootX] > self.rank[rootY]:
                self.parent[rootY] = rootX
                self.size[rootX] += self.size[rootY]
                self.edge_count[rootX] += self.edge_count[rootY] + 1
            elif self.rank[rootX] < self.rank[rootY]:
                self.parent[rootX] = rootY
                self.size[rootY] += self.size[rootX]
                self.edge_count[rootY] += self.edge_count[rootX] + 1
            else:
                self.parent[rootY] = rootX
                self.rank[rootX] += 1
                self.size[rootX] += self.size[rootY]
                self.edge_count[rootX] += self.edge_count[rootY] + 1
        else:
            # If already in the same component, just increment the edge count
            self.edge_count[rootX] += 1

class Solution:
    def countCompleteComponents(self, n: int, edges: List[List[int]]) -> int:
        uf = UnionFind(n)
        
        # Step 1: Union the nodes based on edges
        for u, v in edges:
            uf.union(u, v)

        # Step 2: Count the components that form complete graphs
        component_roots = set(uf.find(i) for i in range(n))
        complete_components = 0

        for root in component_roots:
            size = uf.size[root]
            # A complete graph has exactly (v * (v - 1)) / 2 edges
            if uf.edge_count[root] == size * (size - 1) // 2:
                complete_components += 1

        return complete_components

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