Maximal Network Rank

🔗 Leetcode Link: https://leetcode.com/problems/maximal-network-rank

⏰ Time to complete: 6 mins

1. U-nderstand

   • How can you efficiently check if there is a road connecting two different cities?
   • Would it be efficient to use a matrix to save all the edges between vertex?
   • If the edge is close to city_1 and city_2, then during counting, are they counted twice?

   HAPPY CASE
   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. 
   
   EDGE CASE 
   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.

2. M-atch

   • Use a hashtable of size n which stores sets for each city. Members of the sets are cities which are directly connected to the city the set corresponds to. Then, examine all unique pairs (city_1, city_2) and sum up the edges of both cities. If there is an edge between both cities, the sum needs to be reduced by one since the edge is counted twice. We store the highest sum of edges and return it.

3. P-lan

   General Description of plan (1-2 sentences)

   1) Use a hashtable of size `n` which stores sets for each city. Members of the sets are cities which are directly connected to the city the set corresponds to.
   2) Then, examine all unique pairs `(city_1, city_2)`
   3) Sum up the edges of both cities.
   4) If there is an edge between both cities, the sum needs to be reduces by 1 since the edge is counted 2x.
   4) Then, store the highest sum of edges and return it.
   
   Time Complexity: O(M + N)
   Space Complexity: O(N)
   

   Common Mistakes

   • A common mistake would be iterating 2 times, in other words, 2 for loops, to find the 1st and 2nd maximum values while traversing the loop.
   • The 2 cities with most connections may not be necessarily connected with each other, and if they are connected, the common connection is counted only once.

4. I-mplement

   class Solution {
       public int maximalNetworkRank(int n, int[][] roads) {
           Map<Integer, Set<Integer>> map = buildMap(n, roads);
           int ans = 0;
           for (int i = 0; i < n; i++) {
               for (int j = 0; j < n; j++) {
                   if (i != j) {
                       int sum = map.get(i).size() + map.get(j).size();
                       if (map.get(i).contains(j)) {
                           sum--;
                       }
                       ans = Math.max(ans, sum);
                   }
               }
           }
           return ans;
       }
       
       private Map<Integer, Set<Integer>> buildMap(int n, int[][] roads) {
           Map<Integer, Set<Integer>> map = new HashMap<>();
           for (int i = 0; i < n; i++) {
               map.put(i, new HashSet<>());
           }
           for (int[] road : roads) {
               map.get(road[0]).add(road[1]);
               map.get(road[1]).add(road[0]);
           }
           return map;
       }
   }

   class Solution:
       def maximalNetworkRank(self, n: int, roads: List[List[int]]) -> int:
           city_to_cities = [ set() for i in range( n ) ]
           max_network_rank = 0
           for road in roads:
               city_to_cities[ road[0] ].add( road[1] )
               city_to_cities[ road[1] ].add( road[0] )
           for city_1 in range( n ):
               for city_2 in range( city_1 + 1, n ):
                   network_rank = len( city_to_cities[city_1] ) + len( city_to_cities[city_2] )
                   if ( city_1 in city_to_cities[city_2] ):
                       network_rank -= 1
                   max_network_rank = max(max_network_rank, network_rank)
           return max_network_rank

5. R-eview

   Verify the code works for the happy and edge cases you created in the “Understand” section

6. E-valuate

   • Time Complexity: O(M+N)
   • Space Complexity: O(N)