Introduction

In this blog post, we will explore how to solve the LeetCode problem "1615" 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

There is an infrastructure of n cities with some number of roads connecting these cities. Each roads[i] = [ai, bi] indicates that there is a bidirectional road between cities ai and bi. The network rank of two different cities is defined as the total number of directly connected roads to either city. If a road is directly connected to both cities, it is only counted once. The maximal network rank of the infrastructure is the maximum network rank of all pairs of different cities. Given the integer n and the array roads, return the maximal network rank of the entire infrastructure. Example 1: Input: n = 4, roads = [[0,1],[0,3],[1,2],[1,3]] Output: 4 Explanation: The network rank of cities 0 and 1 is 4 as there are 4 roads that are connected to either 0 or 1. The road between 0 and 1 is only counted once. Example 2: Input: n = 5, roads = [[0,1],[0,3],[1,2],[1,3],[2,3],[2,4]] Output: 5 Explanation: There are 5 roads that are connected to cities 1 or 2. Example 3: Input: n = 8, roads = [[0,1],[1,2],[2,3],[2,4],[5,6],[5,7]] Output: 5 Explanation: The network rank of 2 and 5 is 5. Notice that all the cities do not have to be connected. Constraints: 2 <= n <= 100 0 <= roads.length <= n * (n - 1) / 2 roads[i].length == 2 0 <= ai, bi <= n-1 ai != bi Each pair of cities has at most one road connecting them.

Explanation

Here's the approach:

• Calculate Degrees: Compute the degree of each city (number of directly connected roads).
• Iterate Through Pairs: Iterate through all possible pairs of distinct cities.

• Calculate Network Rank: For each pair, calculate the network rank by summing their degrees and subtracting 1 if there's a direct road between them. Keep track of the maximum rank found.

• Runtime Complexity: O(n^2), Storage Complexity: O(n)

Code

    def maximalNetworkRank(n: int, roads: list[list[int]]) -> int:
    """
    Calculates the maximal network rank of the infrastructure.

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

    Returns:
        The maximal network rank of the infrastructure.
    """

    degrees = [0] * n
    adj_matrix = [[0] * n for _ in range(n)]

    for u, v in roads:
        degrees[u] += 1
        degrees[v] += 1
        adj_matrix[u][v] = 1
        adj_matrix[v][u] = 1

    max_rank = 0
    for i in range(n):
        for j in range(i + 1, n):
            rank = degrees[i] + degrees[j]
            if adj_matrix[i][j] == 1:
                rank -= 1
            max_rank = max(max_rank, rank)

    return max_rank