2685. Count the Number of Complete Components
Description
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.
Solutions
Solution 1
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Solution 2: Simple Method
Problems needed to solve:
- How do we maintain the link state between each node and the others? 如
- How can one determine whether multiple points form a connected graph?
For the first one: we can maintain each node's connection set(including itself).
For the second one: After solving the first one, we can see:
- the node itself includes every node in the connected graph(including itself).
- and only connected to the nodes in the connected graph.
Take example 1 to explain:
- Node 5's connected node is itself, so it is a connected graph.
- Node 0's connected 0, 1, 2. Same as nodes 1, 2.
- Nodes 3 and 4 also include themselves and each other.
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