Introduction

In this blog post, we will explore how to solve the LeetCode problem "1334" 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 are n cities numbered from 0 to n-1. Given the array edges where edges[i] = [fromi, toi, weighti] represents a bidirectional and weighted edge between cities fromi and toi, and given the integer distanceThreshold. Return the city with the smallest number of cities that are reachable through some path and whose distance is at most distanceThreshold, If there are multiple such cities, return the city with the greatest number. Notice that the distance of a path connecting cities i and j is equal to the sum of the edges' weights along that path. Example 1: Input: n = 4, edges = [[0,1,3],[1,2,1],[1,3,4],[2,3,1]], distanceThreshold = 4 Output: 3 Explanation: The figure above describes the graph. The neighboring cities at a distanceThreshold = 4 for each city are: City 0 -> [City 1, City 2] City 1 -> [City 0, City 2, City 3] City 2 -> [City 0, City 1, City 3] City 3 -> [City 1, City 2] Cities 0 and 3 have 2 neighboring cities at a distanceThreshold = 4, but we have to return city 3 since it has the greatest number. Example 2: Input: n = 5, edges = [[0,1,2],[0,4,8],[1,2,3],[1,4,2],[2,3,1],[3,4,1]], distanceThreshold = 2 Output: 0 Explanation: The figure above describes the graph. The neighboring cities at a distanceThreshold = 2 for each city are: City 0 -> [City 1] City 1 -> [City 0, City 4] City 2 -> [City 3, City 4] City 3 -> [City 2, City 4] City 4 -> [City 1, City 2, City 3] The city 0 has 1 neighboring city at a distanceThreshold = 2. Constraints: 2 <= n <= 100 1 <= edges.length <= n * (n - 1) / 2 edges[i].length == 3 0 <= fromi < toi < n 1 <= weighti, distanceThreshold <= 10^4 All pairs (fromi, toi) are distinct.

Explanation

• Floyd-Warshall Algorithm: Use the Floyd-Warshall algorithm to compute the shortest distances between all pairs of cities. This algorithm is well-suited for finding shortest paths in dense graphs.
  • Counting Reachable Cities: After computing all-pairs shortest paths, iterate through each city and count the number of cities reachable from it within the given distanceThreshold.
  • Finding the Optimal City: Maintain track of the city with the smallest number of reachable cities. If multiple cities have the same smallest number, choose the city with the largest index.

• Runtime Complexity: O(n^3), Storage Complexity: O(n^2)

Code

    def findTheCity(n: int, edges: list[list[int]], distanceThreshold: int) -> int:
    """
    Finds the city with the smallest number of reachable cities within a distance threshold.

    Args:
        n: The number of cities.
        edges: A list of edges, where each edge is a list of [from_city, to_city, weight].
        distanceThreshold: The distance threshold.

    Returns:
        The city with the smallest number of reachable cities.
    """

    dist = [[float('inf')] * n for _ in range(n)]
    for i in range(n):
        dist[i][i] = 0

    for u, v, w in edges:
        dist[u][v] = w
        dist[v][u] = w

    for k in range(n):
        for i in range(n):
            for j in range(n):
                dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j])

    min_reachable = float('inf')
    result_city = -1

    for city in range(n):
        reachable_count = 0
        for neighbor in range(n):
            if dist[city][neighbor] <= distanceThreshold:
                reachable_count += 1

        if reachable_count <= min_reachable:
            min_reachable = reachable_count
            result_city = city

    return result_city