1615. Maximal Network Rank

Problem

There is an infrastructure of n cities with some number of roads connecting these cities. Each roads[i] = [ai, bi] indicates that there is a bidirectional road between cities ai and bi.

The network rank** **of *two different cities* is defined as the total number of directly connected roads to either city. If a road is directly connected to both cities, it is only counted once.

The **maximal network rank **of the infrastructure is the maximum network rank of all pairs of different cities.

Given the integer n and the array roads, return **the *maximal network rank* of the entire infrastructure**.

  Example 1:

Input: n = 4, roads = [[0,1],[0,3],[1,2],[1,3]]
Output: 4
Explanation: The network rank of cities 0 and 1 is 4 as there are 4 roads that are connected to either 0 or 1. The road between 0 and 1 is only counted once.

Example 2:

Input: n = 5, roads = [[0,1],[0,3],[1,2],[1,3],[2,3],[2,4]]
Output: 5
Explanation: There are 5 roads that are connected to cities 1 or 2.

Example 3:

Input: n = 8, roads = [[0,1],[1,2],[2,3],[2,4],[5,6],[5,7]]
Output: 5
Explanation: The network rank of 2 and 5 is 5. Notice that all the cities do not have to be connected.

  Constraints:

• 2 <= n <= 100

• 0 <= roads.length <= n * (n - 1) / 2

• roads[i].length == 2

• 0 <= ai, bi&nbsp;<= n-1

• ai != bi

• Each pair of cities has at most one road connecting them.

Solution

/**
 * @param {number} n
 * @param {number[][]} roads
 * @return {number}
 */
var maximalNetworkRank = function(n, roads) {
    var map = {};
    for (var i = 0; i < roads.length; i++) {
        var [m, k] = roads[i];
        if (!map[m]) map[m] = {};
        if (!map[k]) map[k] = {};
        map[m][k] = true;
        map[k][m] = true;
    }
    var res = 0;
    for (var i = 0; i < n; i++) {
        for (var j = i + 1; j < n; j++) {
            res = Math.max(
                res,
                Object.keys(map[i] || {}).length
                    + Object.keys(map[j] || {}).length
                    - (map[i] && map[i][j] ? 1 : 0)
            );
        }
    }
    return res;
};

Explain:

nope.

Complexity:

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