# GeetCode Hub

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.

class Solution { public int findTheCity(int n, int[][] edges, int distanceThreshold) { } }