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Diffstat (limited to 'find_city_smallest_neighbour.py')
-rw-r--r-- | find_city_smallest_neighbour.py | 50 |
1 files changed, 50 insertions, 0 deletions
diff --git a/find_city_smallest_neighbour.py b/find_city_smallest_neighbour.py new file mode 100644 index 0000000..9fae45a --- /dev/null +++ b/find_city_smallest_neighbour.py @@ -0,0 +1,50 @@ +# 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 'from_i' and 'to_i' 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 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 of the edges' weights along that path +""" + + +class Solution(object): + def findTheCity(self, n, edges, distanceThreshold): + """ + :type n: int + :type edges: List[List[int]] + :type distanceThreshold: int + :rtype: int + """ + dist = [[float("inf")] * n for _ in range(n)] + for i, j, k in edges: + dist[i][j] = dist[j][i] = k + for i in range(n): + dist[i][i] = 0 + 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]) + ans = {sum(d <= distanceThreshold for d in dist[i]): i for i in range(n)} + return ans[min(ans)] + + +if __name__ == "__main__": + obj = Solution() + print( + obj.findTheCity( + n=4, edges=[[0, 1, 3], [1, 2, 1], [1, 3, 4], [2, 3, 1]], distanceThreshold=4 + ) + ) + print( + obj.findTheCity( + n=5, + edges=[[0, 1, 2], [0, 4, 8], [1, 2, 3], [1, 4, 2], [2, 3, 1], [3, 4, 1]], + distanceThreshold=2, + ) + ) |