1761. Minimum Degree of a Connected Trio in a Graph

Problem

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

/**
 * @param {number} n
 * @param {number[][]} edges
 * @return {number}
 */
var minTrioDegree = function(n, edges) {
    var { map, lenMap } = buildMap(n, edges);
    var res = Number.MAX_SAFE_INTEGER;
    for (var i = 0; i < map.length; i++) {
        var arr = Object.keys(map[i]);
        for (var j = 0; j < arr.length; j++) {
            var m = arr[j];
            if (m < i) continue;
            for (var k = j + 1; k < arr.length; k++) {
                var n = arr[k];
                if (n < i) continue;
                if (map[m][n]) {
                    res = Math.min(
                        res,
                        arr.length - 2 + lenMap[m] - 2 + lenMap[n] - 2,
                    );
                }
            }
        }
    }
    return res === Number.MAX_SAFE_INTEGER ? -1 : res;
};

var buildMap = function(n, edges) {
    var map = Array(n + 1).fill(0).map(() => ({}));
    var lenMap = Array(n + 1).fill(0);
    for (var i = 0; i < edges.length; i++) {
        var [m, n] = edges[i];
        map[m][n] = 1;
        map[n][m] = 1;
        lenMap[n]++;
        lenMap[m]++;
    }
    return { map, lenMap };
};

Explain:

Brute-force with hashmap.

Complexity:

• Time complexity : O(n ^ 3).
• Space complexity : O(n ^ 2).