Maximum Total Importance of Roads

class Solution {
public:
    long long maxImportance(int N, vector<vector<int>>& Routes) {
        vector<long long> degrees(N, 0);

        for (auto& route : Routes) {
            degrees[route[0]]++;
            degrees[route[1]]++;
        }

        sort(degrees.begin(), degrees.end());

        long long importance = 1;
        long long finalImportance = 0;
        for (auto& degree : degrees) {
            finalImportance += degree * importance;
            importance++;
        }

        return finalImportance;
    }
};

To solve the problem of maximizing the total importance of roads, the given C++ solution employs a straightforward method involving sorting and calculating weighted sums based on node degrees. Here's how the code works:

1. Initialize a vector degrees of size N to keep a count of connections (roads) each city (node) has. Initialize all values to zero, ensuring each city starts with no connections.

2. Iterate over the Routes, which contains pairs of cities connected by roads. For each pair, increment the degree of both cities involved in the route. This loop effectively counts how many roads each city is connected to.

3. Sort the degrees vector in ascending order. Sorting allows for assigning the highest importance to cities with the most roads, maximizing the total importance.

4. Initialize two variables, importance starting at 1 and finalImportance starting at 0. importance represents the importance value that will be assigned to each city, incrementing by 1 after each assignment.

5. Loop through the sorted degrees array. For each degree (i.e., each city's number of connections), calculate the contribution to the total importance by multiplying the degree by its corresponding importance. Accumulate these values in finalImportance.

6. The loop's end result in finalImportance is the maximum total importance based on the given connectivity of the cities.

By sorting the cities by their connections and weighting them accordingly, this approach ensures that the assignment of importance values is optimal for achieving the maximum total importance. Return finalImportance as the solution.

class Solution {

    public long maxImportance(int cityCount, int[][] connections) {
        long[] connectionsCount = new long[cityCount];

        for (int[] road : connections) {
            connectionsCount[road[0]]++;
            connectionsCount[road[1]]++;
        }

        Arrays.sort(connectionsCount);

        long importanceValue = 1;
        long importanceSum = 0;
        for (long count : connectionsCount) {
            importanceSum += (importanceValue * count);
            importanceValue++;
        }

        return importanceSum;
    }
}

This Java solution addresses the problem of calculating the maximum total importance of roads based on a given number of cities and their connections. Here's how the solution is structured:

• Initialize a connectionsCount array to store the number of roads connected to each city.
• Increment the road count for both cities involved in each connection within the connections array.
• Sort the connectionsCount array to facilitate the calculation of maximum importance.
• Initialize variables importanceValue starting from 1, and importanceSum to accumulate the total importance.
• Iterate through each count in the sorted connectionsCount:
  • Multiply the current importanceValue by the count of connections for the city.
  • Add the result to importanceSum.
  • Increment importanceValue for the next city.
• Return the accumulated importanceSum as the result.

This method ensures each city's importance is weighed according to the number of connections, prioritizing cities with fewer roads by multiplying them with lower importance values. This is why sorting is crucial as it aligns lesser connected cities with lower importance values to maximize the overall importance sum.

class Solution:
    def computeMaxImportance(self, number_nodes: int, connect_list: List[List[int]]) -> int:
        node_degrees = [0] * number_nodes

        for connection in connect_list:
            node_degrees[connection[0]] += 1
            node_degrees[connection[1]] += 1

        node_degrees.sort()

        importance_value = 1
        final_importance = 0
        for degree in node_degrees:
            final_importance += importance_value * degree
            importance_value += 1

        return final_importance

The Python program provided is designed to compute the maximum total importance of roads in a graph, where the graph is represented by nodes connected by edges. The function computeMaxImportance accepts two parameters: number_nodes, which is the number of nodes in the graph, and connect_list, a list of lists representing pairs of connected nodes.

Here's a summary of how the code accomplishes this task:

• Initialize Node Degrees: A list called node_degrees is initialized to zero for each node. This list will keep track of how many connections (or edges) each node has.

• Tabulating Each Connection: The program iterates through each connection in connect_list, incrementing the degree in node_degrees for the nodes involved in each connection.

• Sorting Node Degrees: The node_degrees list is sorted in ascending order to prepare for importance calculation.

• Calculating Importance: The program initializes variables importance_value and final_importance. It then iterates through the sorted node_degrees, calculating the final importance by multiplying each degree by an increasing importance_value and summing up these products to final_importance.

• Return Result: Finally, the function returns final_importance, which represents the maximum total importance of the roads based on the node connection degrees.

The result is a straightforward yet efficient way of determining the total importance based on the degree of connections each node has, exploiting the principle that more connected nodes contribute more significantly to the network's overall importance.