How do you solve Maximum Total Importance of Roads on LeetCode?

Maximum Total Importance of Roads (LeetCode #2285) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n + m log m) time complexity and O(n) space complexity. Key insight: Cities with higher degrees should receive higher values to maximize road importance.

What is the brute force solution for Maximum Total Importance of Roads?

The brute force approach for Maximum Total Importance of Roads is Brute Force, with O(n!) time complexity and O(n) space complexity. In the brute force approach, we try all possible assignments of values to cities and calculate the total importance for each assignment. This method is straightforward but inefficient due to the large number of permutations. The optimal Optimal Solution approach improves this to O(n + m log m).

What algorithm or data structure is used to solve Maximum Total Importance of Roads?

Maximum Total Importance of Roads uses the following concepts: Greedy, Graph Theory, Sorting, Heap (Priority Queue). The recommended patterns to study are: Greedy Algorithms, Graph Theory, Sorting.

What are the interview tips for Maximum Total Importance of Roads?

Always consider the properties of the problem, such as degrees in graph problems. Discuss your thought process out loud; it helps the interviewer understand your reasoning. If you get stuck, think about simpler cases or related problems to gain insights.

What are common mistakes when solving Maximum Total Importance of Roads?

Failing to account for the degree of cities when assigning values. Overcomplicating the problem by trying to brute-force permutations instead of using a greedy approach.

#2285

Maximum Total Importance of Roads

Medium

GreedyGraph TheorySortingHeap (Priority Queue)Greedy AlgorithmsGraph TheorySorting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n!)	O(n + m log m)
Space	O(n)	O(n)

💡

Intuition

Time O(n + m log m)Space O(n)

Instead of checking all permutations, we can focus on the degree of each city. Cities with more connections (higher degrees) should get higher values to maximize the importance of the roads they connect.

⚙️

Algorithm

4 steps

1Step 1: Count the degree of each city based on the roads provided.
2Step 2: Sort the cities by their degree in descending order.
3Step 3: Assign values from n down to 1 based on the sorted order of cities.
4Step 4: Calculate the total importance using the assigned values.

solution.py10 lines

1def maxTotalImportance(n, roads):
2    degree = [0] * n
3    for a, b in roads:
4        degree[a] += 1
5        degree[b] += 1
6    sorted_cities = sorted(range(n), key=lambda x: degree[x], reverse=True)
7    total_importance = 0
8    for i, city in enumerate(sorted_cities):
9        total_importance += (i + 1) * degree[city]
10    return total_importance

ℹ

Complexity note: The time complexity is O(n + m log m), where n is the number of cities and m is the number of roads. The sorting step dominates the complexity. The space complexity is O(n) for storing the degree of each city.

1Cities with higher degrees should receive higher values to maximize road importance.
2Sorting cities by their degree allows for optimal value assignment.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.