Introduction

In this blog post, we will explore how to solve the LeetCode problem "2285" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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 <= 5 104 1 <= roads.length <= 5 104 roads[i].length == 2 0 <= ai, bi <= n - 1 ai != bi There are no duplicate roads.

Explanation

Here's the solution:

• Calculate Degrees: Compute the degree of each city, which represents the number of roads connected to that city.
• Sort by Degree: Sort cities based on their degree in descending order.

• Assign Values: Assign the largest value (n) to the city with the highest degree, the second largest value (n-1) to the city with the second highest degree, and so on. This maximizes the contribution of the most connected cities to the total importance.

• Runtime Complexity: O(n + m log m), where n is the number of cities and m is the number of roads. Storage Complexity: O(n)

Code

    def maximum_importance(n: int, roads: list[list[int]]) -> int:
    """
    Calculates the maximum total importance of all roads after assigning values to cities optimally.

    Args:
        n: The number of cities.
        roads: A list of roads, where each road is a list of two city indices.

    Returns:
        The maximum total importance of all roads.
    """

    degrees = [0] * n
    for road in roads:
        degrees[road[0]] += 1
        degrees[road[1]] += 1

    cities = list(range(n))
    cities.sort(key=lambda city: degrees[city], reverse=True)

    city_values = {}
    for i in range(n):
        city_values[cities[i]] = n - i

    total_importance = 0
    for road in roads:
        total_importance += city_values[road[0]] + city_values[road[1]]

    return total_importance