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.
Solution: Brute Force
Try all possible Trios.
Time complexity: O(n^3)
Space complexity: O(n^2)
C++
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// Author: Huahua class Solution { public: int minTrioDegree(int n, vector<vector<int>>& edges) { vector<bitset<400>> g(n); for (const auto& e : edges) { g[e[0] - 1].set(e[1] - 1); g[e[1] - 1].set(e[0] - 1); } size_t 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[i][k] && g[j][k]) ans = min(ans, g[i].count() + g[j].count() + g[k].count() - 6); return ans == INT_MAX ? -1 : ans; } }; |
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