Solution Explanation

This problem asks to find the city with the smallest number of neighbors within a given distance threshold. The solution involves finding the shortest paths between all pairs of cities and then counting the number of cities reachable within the threshold from each city.

The most efficient approach uses Floyd-Warshall algorithm or Dijkstra's algorithm. Both algorithms are used to compute the shortest paths between all pairs of nodes in a graph.

Floyd-Warshall Algorithm: This algorithm is simpler to implement but has a higher time complexity of O(N³), where N is the number of cities. It works by iteratively considering all possible intermediate nodes in paths between pairs of cities.

Dijkstra's Algorithm: This algorithm is more efficient for sparse graphs (graphs with relatively few edges) with a time complexity of O(E log V), where E is the number of edges and V is the number of vertices (cities). It explores the graph by iteratively selecting the closest unvisited city and updating distances to its neighbors. However, in this problem, since we need the shortest path between all pairs, the solution will involve repeatedly running Dijkstra's Algorithm.

Both approaches will be explained below.

Approach 1: Floyd-Warshall Algorithm

1. Initialization: Create an adjacency matrix g representing the graph where g[i][j] stores the shortest distance between city i and city j. Initialize g[i][j] to infinity if there's no direct edge between i and j, and to the weight of the edge if there is one. Set the diagonal elements g[i][i] to 0.

2. Floyd-Warshall: Iterate through all possible intermediate nodes k and update the shortest distances: g[i][j] = min(g[i][j], g[i][k] + g[k][j]). This step ensures that we consider all possible paths between cities i and j, including those that pass through intermediate nodes k.

3. Counting Reachable Cities: For each city i, count the number of cities j such that g[i][j] is less than or equal to the distanceThreshold.

4. Finding the City: Iterate through the counts, keeping track of the city with the minimum number of reachable cities and choosing the one with the largest index if there's a tie.

Approach 2: Repeated Dijkstra's Algorithm

1. Initialization: Create an adjacency list or matrix representing the graph.

2. Repeated Dijkstra's: Run Dijkstra's algorithm for each city i to find the shortest distances from city i to all other cities.

3. Counting Reachable Cities: For each city i, count the number of cities j such that the shortest distance from i to j is less than or equal to the distanceThreshold.

4. Finding the City: Iterate through the counts, finding the city with the minimum number of reachable cities, and choosing the one with the largest index in case of a tie.

Time Complexity Analysis

• Floyd-Warshall: O(N³). The dominant operation is the triple nested loop in the Floyd-Warshall algorithm.
• Repeated Dijkstra's: O(N * (E log V)). Dijkstra's algorithm is run N times, once for each starting city.

Space Complexity Analysis

Both algorithms have a space complexity of O(N²), primarily due to the adjacency matrix used to store shortest paths. In the Dijkstra method, the adjacency list might use slightly less space in case of sparse graphs, however, the dominant factor is the distance matrix needed to store the results.

Code Examples (Python - Approach 1: Floyd-Warshall)

import sys
 
class Solution:
    def findTheCity(self, n: int, edges: List[List[int]], distanceThreshold: int) -> int:
        g = [[sys.maxsize] * n for _ in range(n)]  # Initialize with infinity
        for u, v, w in edges:
            g[u][v] = g[v][u] = w  # Bi-directional edges
        for i in range(n):
            g[i][i] = 0  # Distance to self is 0
 
        # Floyd-Warshall algorithm
        for k in range(n):
            for i in range(n):
                for j in range(n):
                    g[i][j] = min(g[i][j], g[i][k] + g[k][j])
 
        min_neighbors = float('inf')
        result_city = -1
        for i in range(n):
            reachable_neighbors = sum(1 for dist in g[i] if dist <= distanceThreshold)
            if reachable_neighbors <= min_neighbors:
                min_neighbors = reachable_neighbors
                result_city = i
        return result_city
 

The other languages (Java, C++, Go, TypeScript, JavaScript) would follow similar logic, but with their respective syntax and data structures. You would replace sys.maxsize with the appropriate representation of infinity for each language. The core Floyd-Warshall algorithm remains the same. The code provided in the original response shows both Floyd-Warshall and repeated Dijkstra solutions, choose the one that better suits your needs based on the expected size of the graph. For smaller graphs, Floyd-Warshall might be simpler to implement; for larger and sparser graphs, repeated Dijkstra is likely to be more efficient.