Find the City With the Smallest Number of

Tested Python solution for LeetCode 1334 with 16 pytest cases. Generate a practice environment with lcpy.

LeetCode 1334, Medium. Topics: Dynamic Programming, Graph Theory, Shortest Path, Dijkstra’s Algorithm, Bellman-Ford Algorithm, Floyd-Warshall Algorithm. View on LeetCode.

Generate this problem as a practice environment: tested reference solution, 16 parametrized pytest cases, and a playground notebook:

lcpy gen -n 1334   # by problem number
lcpy gen -s find_the_city_with_the_smallest_number_of_neighbors_at_a_threshold_distance   # by problem name

​

Problem

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.

​

Examples

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.

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.

​

Solution

Reference implementation from solution.py on GitHub, full suite in test_solution.py:

class Solution:
    def find_the_city(self, n: int, edges: list[list[int]], distance_threshold: int) -> int:
        inf = 10**9
        dist = [[inf] * n for _ in range(n)]
        for i in range(n):
            dist[i][i] = 0
        for u, v, w in edges:
            dist[u][v] = w
            dist[v][u] = w
        for k in range(n):
            for i in range(n):
                for j in range(n):
                    if dist[i][k] + dist[k][j] < dist[i][j]:
                        dist[i][j] = dist[i][k] + dist[k][j]
        best_city, best_count = -1, n + 1
        for i in range(n):
            count = sum(1 for j in range(n) if j != i and dist[i][j] <= distance_threshold)
            if count <= best_count:
                best_count = count
                best_city = i
        return best_city

​

Complexity

Time	Space
-	-

​

Tags

NeetCode All.