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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

Solution 1

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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

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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};
    }
}

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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;
    }
};

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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
}

Solution 2: Simple Method

Problems needed to solve：

  1. How do we maintain the link state between each node and the others? 如
  2. 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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class Solution {
public:
    int countCompleteComponents(int n, vector<vector<int>>& edges) {
        int ans = 0;
        vector<set<int>> m(n + 1, set<int>());
        for (int i = 0; i < n; i++) {
            m[i].insert(i);
        }
        for (auto x : edges) {
            m[x[0]].insert(x[1]);
            m[x[1]].insert(x[0]);
        }
        map<set<int>, int> s;
        for (int i = 0; i < n; i++) {
            s[m[i]]++;
        }
        for (auto& [x, y] : s) {
            if (y == x.size()) {
                ans++;
            }
        }
        return ans;
    }
};

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