Question

1761. Minimum Degree of a Connected Trio in a Graph

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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

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

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;
  }
}

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;
  }
};

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
}

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;
}