Skip to content

1761. Minimum Degree of a Connected Trio in a Graph

Description

You are given an undirected graph. You are given an integer n which is the number of nodes in the graph and an array edges, where each edges[i] = [ui, vi] indicates that there is an undirected edge between ui and vi.

A connected trio is a set of three nodes where there is an edge between every pair of them.

The degree of a connected trio is the number of edges where one endpoint is in the trio, and the other is not.

Return the minimum degree of a connected trio in the graph, or -1 if the graph has no connected trios.

 

Example 1:

Input: n = 6, edges = [[1,2],[1,3],[3,2],[4,1],[5,2],[3,6]]
Output: 3
Explanation: There is exactly one trio, which is [1,2,3]. The edges that form its degree are bolded in the figure above.

Example 2:

Input: n = 7, edges = [[1,3],[4,1],[4,3],[2,5],[5,6],[6,7],[7,5],[2,6]]
Output: 0
Explanation: There are exactly three trios:
1) [1,4,3] with degree 0.
2) [2,5,6] with degree 2.
3) [5,6,7] with degree 2.

 

Constraints:

  • 2 <= n <= 400
  • edges[i].length == 2
  • 1 <= edges.length <= n * (n-1) / 2
  • 1 <= ui, vi <= n
  • ui != vi
  • There are no repeated edges.

Solutions

Solution 1: Brute Force Enumeration

Thinking

The degree of a trio is the sum of the three vertex degrees minus \(6\). The graph is small enough to enumerate triangles in \(O(n^3)\).

An adjacency matrix tests edges and \(\textit{deg}\) stores degrees. For \(i<j<k\) with all three edges, update \(\textit{deg}[i]+\textit{deg}[j]+\textit{deg}[k]-6\). Return \(-1\) if none exist.

We first store all edges in the adjacency matrix \(\textit{g}\), and then store the degree of each node in the array \(\textit{deg}\). Initialize the answer \(\textit{ans} = +\infty\).

Then enumerate all triplets \((i, j, k)\), where \(i \lt j \lt k\). If \(\textit{g}[i][j] = \textit{g}[j][k] = \textit{g}[i][k] = 1\), it means these three nodes form a connected trio. In this case, update the answer to \(\textit{ans} = \min(\textit{ans}, \textit{deg}[i] + \textit{deg}[j] + \textit{deg}[k] - 6)\).

After enumerating all triplets, if the answer is still \(+\infty\), it means there is no connected trio in the graph, return \(-1\). Otherwise, return the answer.

The time complexity is \(O(n^3)\), and the space complexity is \(O(n^2)\). Here, \(n\) is the number of nodes.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
def min(a: int, b: int) -> int:
    return a if a < b else b


class Solution:
    def minTrioDegree(self, n: int, edges: List[List[int]]) -> int:
        g = [[False] * n for _ in range(n)]
        deg = [0] * n
        for u, v in edges:
            u, v = u - 1, v - 1
            g[u][v] = g[v][u] = True
            deg[u] += 1
            deg[v] += 1
        ans = inf
        for i in range(n):
            for j in range(i + 1, n):
                if g[i][j]:
                    for k in range(j + 1, n):
                        if g[i][k] and g[j][k]:
                            ans = min(ans, deg[i] + deg[j] + deg[k] - 6)
        return -1 if ans == inf else ans

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
class Solution {
    public int minTrioDegree(int n, int[][] edges) {
        boolean[][] g = new boolean[n][n];
        int[] deg = new int[n];
        for (var e : edges) {
            int u = e[0] - 1, v = e[1] - 1;
            g[u][v] = true;
            g[v][u] = true;
            ++deg[u];
            ++deg[v];
        }
        int ans = 1 << 30;
        for (int i = 0; i < n; ++i) {
            for (int j = i + 1; j < n; ++j) {
                if (g[i][j]) {
                    for (int k = j + 1; k < n; ++k) {
                        if (g[i][k] && g[j][k]) {
                            ans = Math.min(ans, deg[i] + deg[j] + deg[k] - 6);
                        }
                    }
                }
            }
        }
        return ans == 1 << 30 ? -1 : ans;
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
class Solution {
public:
    int minTrioDegree(int n, vector<vector<int>>& edges) {
        bool g[n][n];
        memset(g, 0, sizeof g);
        int deg[n];
        memset(deg, 0, sizeof deg);
        for (auto& e : edges) {
            int u = e[0] - 1, v = e[1] - 1;
            g[u][v] = g[v][u] = true;
            deg[u]++, deg[v]++;
        }
        int ans = INT_MAX;
        for (int i = 0; i < n; ++i) {
            for (int j = i + 1; j < n; ++j) {
                if (g[i][j]) {
                    for (int k = j + 1; k < n; ++k) {
                        if (g[j][k] && g[i][k]) {
                            ans = min(ans, deg[i] + deg[j] + deg[k] - 6);
                        }
                    }
                }
            }
        }
        return ans == INT_MAX ? -1 : ans;
    }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
func minTrioDegree(n int, edges [][]int) int {
    g := make([][]bool, n)
    deg := make([]int, n)
    for i := range g {
        g[i] = make([]bool, n)
    }
    for _, e := range edges {
        u, v := e[0]-1, e[1]-1
        g[u][v], g[v][u] = true, true
        deg[u]++
        deg[v]++
    }
    ans := 1 << 30
    for i := 0; i < n; i++ {
        for j := i + 1; j < n; j++ {
            if g[i][j] {
                for k := j + 1; k < n; k++ {
                    if g[i][k] && g[j][k] {
                        ans = min(ans, deg[i]+deg[j]+deg[k]-6)
                    }
                }
            }
        }
    }
    if ans == 1<<30 {
        return -1
    }
    return ans
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
function minTrioDegree(n: number, edges: number[][]): number {
    const g = Array.from({ length: n }, () => Array(n).fill(false));
    const deg: number[] = Array(n).fill(0);
    for (let [u, v] of edges) {
        u--;
        v--;
        g[u][v] = g[v][u] = true;
        ++deg[u];
        ++deg[v];
    }
    let ans = Infinity;
    for (let i = 0; i < n; ++i) {
        for (let j = i + 1; j < n; ++j) {
            if (g[i][j]) {
                for (let k = j + 1; k < n; ++k) {
                    if (g[i][k] && g[j][k]) {
                        ans = Math.min(ans, deg[i] + deg[j] + deg[k] - 6);
                    }
                }
            }
        }
    }
    return ans === Infinity ? -1 : ans;
}

Comments