g1601_1700.s1615_maximal_network_rank.Solution Maven / Gradle / Ivy

Go to download

Show more of this group Show more artifacts with this name
Show all versions of leetcode-in-java21 Show documentation

Java-based LeetCode algorithm problem solutions, regularly updated

There is a newer version: 1.38

package g1601_1700.s1615_maximal_network_rank;

// #Medium #Graph #Graph_Theory_I_Day_14_Graph_Theory
// #2022_04_13_Time_3_ms_(97.34%)_Space_42.8_MB_(92.41%)

/**
 * 1615 - Maximal Network Rank\.
 *
 * Medium
 *
 * There is an infrastructure of `n` cities with some number of `roads` connecting these cities. Each roads[i] = [a_i, b_i] indicates that there is a bidirectional road between cities a_i and b_i.
 *
 * 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:**
 *
 * **![](https://assets.leetcode.com/uploads/2020/09/21/ex1.png)**
 *
 * **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:**
 *
 * **![](https://assets.leetcode.com/uploads/2020/09/21/ex2.png)**
 *
 * **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 <= a_i, b_i <= n-1
 * *   a_i != b_i
 * *   Each pair of cities has **at most one** road connecting them.
**/
public class Solution {
    public int maximalNetworkRank(int n, int[][] roads) {
        int[] degrees = new int[n];
        boolean[] connected = new boolean[40_000];
        for (int[] r : roads) {
            degrees[r[0]]++;
            degrees[r[1]]++;
            connected[(r[0] + 101) * (r[1] + 101) - 1] = true;
        }
        int max = 0;
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                if (connected[(i + 101) * (j + 101) - 1]) {
                    max = Math.max(max, degrees[i] + degrees[j] - 1);
                } else {
                    max = Math.max(max, degrees[i] + degrees[j]);
                }
            }
        }
        return max;
    }
}