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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

• Java
• C++
• Python
• Go

• class Solution {
      private List<Integer>[] g;
      private boolean[] vis;
  
      public int countCompleteComponents(int n, int[][] edges) {
          g = new List[n];
          vis = new boolean[n];
          Arrays.setAll(g, k -> new ArrayList<>());
          for (int[] e : edges) {
              int a = e[0], b = e[1];
              g[a].add(b);
              g[b].add(a);
          }
          int ans = 0;
          for (int i = 0; i < n; ++i) {
              if (!vis[i]) {
                  int[] t = dfs(i);
                  if (t[0] * (t[0] - 1) == t[1]) {
                      ++ans;
                  }
              }
          }
          return ans;
      }
  
      private int[] dfs(int i) {
          vis[i] = true;
          int x = 1, y = g[i].size();
          for (int j : g[i]) {
              if (!vis[j]) {
                  int[] t = dfs(j);
                  x += t[0];
                  y += t[1];
              }
          }
          return new int[] {x, y};
      }
  }
  

• class Solution {
  public:
      int countCompleteComponents(int n, vector<vector<int>>& edges) {
          vector<vector<int>> g(n);
          bool vis[n];
          memset(vis, false, sizeof(vis));
          for (auto& e : edges) {
              int a = e[0], b = e[1];
              g[a].push_back(b);
              g[b].push_back(a);
          }
          function<pair<int, int>(int)> dfs = [&](int i) -> pair<int, int> {
              vis[i] = true;
              int x = 1, y = g[i].size();
              for (int j : g[i]) {
                  if (!vis[j]) {
                      auto [a, b] = dfs(j);
                      x += a;
                      y += b;
                  }
              }
              return make_pair(x, y);
          };
          int ans = 0;
          for (int i = 0; i < n; ++i) {
              if (!vis[i]) {
                  auto [a, b] = dfs(i);
                  if (a * (a - 1) == b) {
                      ++ans;
                  }
              }
          }
          return ans;
      }
  };
  

• class Solution:
      def countCompleteComponents(self, n: int, edges: List[List[int]]) -> int:
          def dfs(i: int) -> (int, int):
              vis[i] = True
              x, y = 1, len(g[i])
              for j in g[i]:
                  if not vis[j]:
                      a, b = dfs(j)
                      x += a
                      y += b
              return x, y
  
          g = defaultdict(list)
          for a, b in edges:
              g[a].append(b)
              g[b].append(a)
          vis = [False] * n
          ans = 0
          for i in range(n):
              if not vis[i]:
                  a, b = dfs(i)
                  ans += a * (a - 1) == b
          return ans
  
  

• func countCompleteComponents(n int, edges [][]int) (ans int) {
  	g := make([][]int, n)
  	vis := make([]bool, n)
  	for _, e := range edges {
  		a, b := e[0], e[1]
  		g[a] = append(g[a], b)
  		g[b] = append(g[b], a)
  	}
  	var dfs func(int) (int, int)
  	dfs = func(i int) (int, int) {
  		vis[i] = true
  		x, y := 1, len(g[i])
  		for _, j := range g[i] {
  			if !vis[j] {
  				a, b := dfs(j)
  				x += a
  				y += b
  			}
  		}
  		return x, y
  	}
  	for i := range vis {
  		if !vis[i] {
  			a, b := dfs(i)
  			if a*(a-1) == b {
  				ans++
  			}
  		}
  	}
  	return
  }
  

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