Introduction

In this blog post, we will explore how to solve the LeetCode problem "2492" 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 a positive integer n representing n cities numbered from 1 to n. You are also given a 2D array roads where roads[i] = [ai, bi, distancei] indicates that there is a bidirectional road between cities ai and bi with a distance equal to distancei. The cities graph is not necessarily connected. The score of a path between two cities is defined as the minimum distance of a road in this path. Return the minimum possible score of a path between cities 1 and n. Note: A path is a sequence of roads between two cities. It is allowed for a path to contain the same road multiple times, and you can visit cities 1 and n multiple times along the path. The test cases are generated such that there is at least one path between 1 and n. Example 1: Input: n = 4, roads = [[1,2,9],[2,3,6],[2,4,5],[1,4,7]] Output: 5 Explanation: The path from city 1 to 4 with the minimum score is: 1 -> 2 -> 4. The score of this path is min(9,5) = 5. It can be shown that no other path has less score. Example 2: Input: n = 4, roads = [[1,2,2],[1,3,4],[3,4,7]] Output: 2 Explanation: The path from city 1 to 4 with the minimum score is: 1 -> 2 -> 1 -> 3 -> 4. The score of this path is min(2,2,4,7) = 2. Constraints: 2 <= n <= 105 1 <= roads.length <= 105 roads[i].length == 3 1 <= ai, bi <= n ai != bi 1 <= distancei <= 104 There are no repeated edges. There is at least one path between 1 and n.

Explanation

Here's the breakdown of the solution:

• Find Connected Component: Use Depth-First Search (DFS) or Breadth-First Search (BFS) to identify all cities connected to city 1. The problem allows revisiting cities, so we only need to find the connected component containing city 1.

• Find Minimum Edge: Iterate through the roads and find the minimum distance among the roads where both endpoints belong to the connected component. This minimum distance represents the minimum possible score for any path between cities 1 and n, since any path must consist of edges between connected cities.

• Time and Space Complexity: The time complexity is O(n + m), where n is the number of cities and m is the number of roads. This is because we traverse the roads to build the graph and use DFS to find the connected component. The space complexity is O(n + m) because we store the graph in an adjacency list and the visited set for DFS.

Code

    def minScore(n: int, roads: list[list[int]]) -> int:
    """
    Finds the minimum possible score of a path between cities 1 and n.

    Args:
        n: The number of cities.
        roads: A list of roads where each road is represented as [city1, city2, distance].

    Returns:
        The minimum possible score of a path between cities 1 and n.
    """

    graph = [[] for _ in range(n + 1)]
    for u, v, distance in roads:
        graph[u].append((v, distance))
        graph[v].append((u, distance))

    visited = set()
    def dfs(city):
        visited.add(city)
        for neighbor, _ in graph[city]:
            if neighbor not in visited:
                dfs(neighbor)

    dfs(1)

    min_score = float('inf')
    for u, v, distance in roads:
        if u in visited and v in visited:
            min_score = min(min_score, distance)

    return min_score