Maximum Total Importance of Roads
Medium
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Topics:
Greedy AlgorithmsArrays
You are given an integer n denoting the number of cities in a country. The cities are numbered from 0 to n - 1.
You are also given a 2D integer array roads where roads[i] = [ai, bi] denotes that there exists a bidirectional road connecting cities ai and bi.
You need to assign each city with an integer value from 1 to n, where each value can only be used once. The importance of a road is then defined as the sum of the values of the two cities it connects.
Return the maximum total importance of all roads possible after assigning the values optimally.
Example 1:
Input: n = 5, roads = [[0,1],[1,2],[2,3],[0,2],[1,3],[2,4]] Output: 43 Explanation: The figure above shows the country and the assigned values of [2,4,5,3,1]. - The road (0,1) has an importance of 2 + 4 = 6. - The road (1,2) has an importance of 4 + 5 = 9. - The road (2,3) has an importance of 5 + 3 = 8. - The road (0,2) has an importance of 2 + 5 = 7. - The road (1,3) has an importance of 4 + 3 = 7. - The road (2,4) has an importance of 5 + 1 = 6. The total importance of all roads is 6 + 9 + 8 + 7 + 7 + 6 = 43. It can be shown that we cannot obtain a greater total importance than 43.
Example 2:
Input: n = 5, roads = [[0,3],[2,4],[1,3]] Output: 20 Explanation: The figure above shows the country and the assigned values of [4,3,2,5,1]. - The road (0,3) has an importance of 4 + 5 = 9. - The road (2,4) has an importance of 2 + 1 = 3. - The road (1,3) has an importance of 3 + 5 = 8. The total importance of all roads is 9 + 3 + 8 = 20. It can be shown that we cannot obtain a greater total importance than 20.
Constraints:
2 <= n <= 5 * 1041 <= roads.length <= 5 * 104roads[i].length == 20 <= ai, bi <= n - 1ai != bi- There are no duplicate roads.
Solution
Clarifying Questions
When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:
- What are the constraints on the number of cities `n` and the number of roads (size of the `roads` array)?
- Can the road connections in the `roads` array be assumed to be bidirectional, or is a road represented by [a, b] only traversable from city a to city b?
- Are city IDs zero-based or one-based?
- If there are no roads connected to a city, what should its importance be?
- Is the input `roads` array guaranteed to be a valid graph structure (no self-loops or multi-edges), and is the graph guaranteed to be connected?
Brute Force Solution
Approach
The brute force method tries every possible assignment of importance values to the cities. It then calculates the total importance of all roads based on each assignment and chooses the assignment that gives the highest total importance. This is like trying every combination until we find the best one.
Here's how the algorithm would work step-by-step:
- Start by trying every possible way to assign a unique importance value (like 1, 2, 3, etc.) to each city.
- For each possible assignment of importance values, go through each road and calculate its importance by adding the importance values of the two cities connected by that road.
- Sum the importance of all roads for that assignment to find the total importance for that particular arrangement.
- Keep track of the highest total importance you've found so far and the specific assignment that produced it.
- Repeat steps 2 and 3 for every possible assignment of importance values to the cities.
- After you have checked every possible assignment, the assignment that gave you the highest total importance is the answer.
Code Implementation
import itertools
def maximum_total_importance_brute_force(number_of_cities, roads):
# Initialize the highest total importance found so far.
highest_total_importance = 0
best_city_importance_mapping = {}
# Generate all possible permutations of importance values.
possible_importance_values = list(range(1, number_of_cities + 1))
for city_importance_permutation in itertools.permutations(possible_importance_values):
# Create a mapping of city to importance value.
city_importance_mapping = {}
for city_index in range(number_of_cities):
city_importance_mapping[city_index] = city_importance_permutation[city_index]
# Calculate the total importance of all roads.
current_total_importance = 0
for road in roads:
city_a, city_b = road
current_total_importance += city_importance_mapping[city_a] + city_importance_mapping[city_b]
# Update the highest total importance if necessary.
if current_total_importance > highest_total_importance:
highest_total_importance = current_total_importance
return highest_total_importance
Big(O) Analysis
Time Complexity
O(n!) – The algorithm explores every possible assignment of importance values to the n cities. This means it considers all permutations of the numbers from 1 to n. Generating all permutations of n items takes O(n!) time. For each permutation, it iterates through all the roads to calculate the total importance. Although iterating through the roads might seem like another factor, the dominant factor is generating the permutations in the first place, which scales as O(n!). Therefore, the overall time complexity is O(n!).
Space Complexity
O(N!) – The brute force solution described explores every possible assignment of importance values to cities. This implies storing each permutation of importance values. Because there are N cities, the algorithm implicitly generates and potentially stores each of the N! possible mappings of cities to importance values. No other significant data structures are mentioned, so the dominant space usage comes from generating permutations, leading to O(N!).
Optimal Solution
Approach
The goal is to assign importance values to cities in a way that maximizes the total importance of roads connecting them. The clever idea is to assign higher importance to cities with more roads connected to them.
Here's how the algorithm would work step-by-step:
- First, count how many roads are connected to each city. Think of this as figuring out how 'popular' each city is.
- Next, sort the cities based on their 'popularity' from most popular to least popular.
- Then, assign the highest importance value to the most popular city, the second-highest importance value to the second most popular city, and so on, until all cities have an importance value.
- Finally, calculate the total importance of all roads by adding up the importance values of the cities at both ends of each road.
Code Implementation
def maximum_importance(number_of_cities, roads):
city_degrees = [0] * number_of_cities
for city_one, city_two in roads:
city_degrees[city_one] += 1
city_degrees[city_two] += 1
# Sort cities by their degrees to prioritize assignment.
city_degree_pairs = sorted(enumerate(city_degrees), key=lambda x: x[1], reverse=True)
city_importance = [0] * number_of_cities
# Assign importance values based on sorted order
for i, (city_index, _) in enumerate(city_degree_pairs):
city_importance[city_index] = number_of_cities - i
total_importance = 0
# Calculate the total importance of all roads.
for city_one, city_two in roads:
total_importance += city_importance[city_one] + city_importance[city_two]
return total_importanceBig(O) Analysis
Time Complexity
O(n log n) – Counting the roads connected to each city involves iterating through the 'roads' array, which takes O(R) time where R is the number of roads and is bounded by n^2 if there is a road between every city. The sorting step, based on city popularity, uses a comparison-based sorting algorithm, resulting in O(n log n) time complexity, where n is the number of cities. Assigning importance values takes O(n) time. Finally, calculating the total importance of roads requires iterating through the 'roads' array again, taking O(R) time, which can be at most O(n^2). However, assuming that the number of roads is not significantly larger than n log n or even n, the dominant term becomes O(n log n) due to the sorting step, particularly if R is sparse. Therefore, the overall time complexity is O(n log n).
Space Complexity
O(N) – The space complexity is determined by the auxiliary space used to store the degree (number of connected roads) for each city. This requires an array or hash map of size N, where N is the number of cities. Sorting the cities based on their degree might require additional space depending on the sorting algorithm implementation, but in the worst case, it can be O(N) for in-place sorting or O(N) for creating a sorted copy. Therefore, the dominant auxiliary space is O(N) to store the degrees of the cities.
Edge Cases
| Case | How to Handle |
|---|---|
| Empty roads list | If the roads list is empty, return 0 immediately since there are no roads to contribute to importance. |
| Single city (n=1) | If there is only one city, all roads are irrelevant, so return 0 immediately. |
| Maximum number of cities and roads (n and roads at their max limits) | Ensure the algorithm's time complexity is efficient enough to handle these large inputs (e.g., O(n log n) for sorting or O(n+m) for graph traversal). |
| All roads connect to the same city. | This creates a star graph; the central city will have the highest degree and thus the highest importance value. |
| Roads form a complete graph (every city connected to every other city) | The degree of each city will be n-1; assign importance values accordingly after sorting the degree counts. |
| Roads form disconnected components. | The importance within each component is independent; assign importance values within each component based on the degree of each city in that component. |
| Integer overflow when calculating total importance | Use a larger data type (e.g., long) for the total importance calculation to prevent overflow. |
| Roads list contains duplicate entries (same city pair listed multiple times) | The adjacency list or degree count should increment appropriately based on all connections, effectively giving those connections more weight |
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