<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>

from collections import defaultdict

class DSU:
    def __init__(self, n):
        self.rootOf = [i for i in range(n)]
        self.rankOf = [0 for i in range(n)]

    def find(self, x):
        while self.rootOf[x] != x:
            x = self.rootOf[x]
        return self.rootOf[x]

    def union(self, x, y):
        rX, rY = self.find(x), self.find(y)
        if rX != rY:
            if self.rankOf[rX] > self.rankOf[rY]:
                self.rootOf[rY] = rX
            elif self.rankOf[rX] < self.rankOf[rY]:
                self.rootOf[rX] = rY
            else:
                self.rootOf[rX] = rY
                self.rankOf[rY] += 1


class Solution:
    def findConnectedComponents(self, n, edges):
        dsu = DSU(n)
        graph = defaultdict(list)

        for u, v in edges:
            if dsu.find(u) != dsu.find(v):
                dsu.union(u, v)
            graph[u].append(v)
            graph[v].append(u)

        print(dsu.rootOf)
        print(dsu.rankOf)
        for x in range(n):
            dsu.rootOf[x] = dsu.find(x)

        components = defaultdict(list)
        for i, root in enumerate(dsu.rootOf):
            components[root].append(i)
        return components, graph

    def countCompleteComponents(self, n: int, edges: List[List[int]]) -> int:
        components, graph = self.findConnectedComponents(n, edges)
        print(components)
        res = 0
        for root, nodes in components.items():
            should_break = False
            for u in nodes:
                for v in nodes:
                    if v != u and v not in graph[u]:
                        should_break = True
                        break
                if should_break:
                    break
            if not should_break:
                res += 1
        return res

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