m1334.py

"""Find the City With the Smallest Number of Neighbors at a Threshold Distance
TBD: graph

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.
"""