## # 1334. Find the City With the Smallest Number of Neighbors at a Threshold Distance

There are n cities numbered from 0 to n-1. Given the array edges where `edges[i] = [from_i, to_i, weight_i]` 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.
``````

### # Solution

Approach 1: Floyd-Warshall -- `O(N^3)` to find shortest distances of all pairs of nodes in a graph.

### # Code (Python)

Approach 1:

``````class Solution:
def findTheCity(self, n: int, edge_list: List[List[int]], distanceThreshold: int) -> int:
# Floyd-Warshall (N^3): input is an adj list of nodes and weights, output is the shortest distance for all pairs of nodes in the graph
# idea: for each node pair (i, j), dist[i][j] = min(dist[i][k] + dist[k][j]) for k from 0 to n-1, where k is a throughway node
# https://en.wikipedia.org/wiki/Floyd%E2%80%93Warshall_algorithm
# https://leetcode.com/problems/find-the-city-with-the-smallest-number-of-neighbors-at-a-threshold-distance/discuss/490312/Python-Floyd
dist = [[float('inf')] * n for _ in range(n)]
for i in range(n):
dist[i][i] = 0
for n1, n2, weight in edge_list:
dist[n1][n2] = weight
dist[n2][n1] = weight

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_city = -1
min_value = float('inf')
for i in range(n):
value = sum(dist[i][j] <= distanceThreshold for j in range(n))
if value <= min_value:
min_value = value
min_city = i
return min_city
``````

Approach 1:

Approach 2: