Leetcode_Solutions/2685. Count_the_Number_of_Complete_Components.py at main · Aniketsy/Leetcode_Solutions · GitHub

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You are given an integer n. There is an undirected graph with n vertices, numbered from 0 to n - 1. You are given a 2D integer array edges where edges[i] = [ai, bi] denotes that there exists an undirected edge connecting vertices ai and bi.

Return the number of complete connected components of the graph.

A connected component 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.

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


Example 1:


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


Input: n = 6, edges = [[0,1],[0,2],[1,2],[3,4],[3,5]]
Output: 1
Explanation: 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.


Constraints:

1 <= n <= 50
0 <= edges.length <= n * (n - 1) / 2
edges[i].length == 2
0 <= ai, bi <= n - 1
ai != bi
There are no repeated edges.


#############################   Solution  ###################################

class Solution:
    def countCompleteComponents(self, n: int, edges: List[List[int]]) -> int:
        def dfs(v, res):
            if v in visit:
                return
            visit.add(v)
            res.append(v)
            for nei in adj[v]:
                dfs(nei,res)
            return res

        res =0
        adj = defaultdict(list)
        for v1,v2 in edges:
            adj[v1].append(v2)
            adj[v2].append(v1)

        res =0
        visit = set()
        for v in range(n):
            if v in visit:
                continue
            Component = dfs(v,[])
            if all ([len(Component) -1 == len(adj[v2]) for v2 in Component]):
                res +=1
        return res