2285. Maximum Total Importance of Roads
Explanation:
To solve this problem, we can use a greedy algorithm approach. We want to assign values to cities in such a way that the total importance of all roads is maximized. We can achieve this by assigning the highest values to cities that are most frequently connected by roads.
- Create a map to store the frequency of each city appearing in the roads array.
- Sort the cities based on their frequency in descending order.
- Assign values starting from the highest value to the most frequent city, then decrement the value for subsequent cities.
- Calculate the total importance of roads by summing the importance of each road.
Time Complexity:
The time complexity of this approach is O(n log n) where n is the number of cities.
Space Complexity:
The space complexity is O(n) to store the frequency map.
:
class Solution {
public int maxTotalImportance(int n, int[][] roads) {
Map<Integer, Integer> freqMap = new HashMap<>();
for (int[] road : roads) {
freqMap.put(road[0], freqMap.getOrDefault(road[0], 0) + 1);
freqMap.put(road[1], freqMap.getOrDefault(road[1], 0) + 1);
}
List<Integer> cities = new ArrayList<>(freqMap.keySet());
cities.sort((a, b) -> freqMap.get(b) - freqMap.get(a));
int[] values = new int[n];
int value = n;
for (int city : cities) {
values[city] = value;
value--;
}
int totalImportance = 0;
for (int[] road : roads) {
totalImportance += values[road[0]] + values[road[1]];
}
return totalImportance;
}
}Code Editor (Testing phase)
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