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LeetCode 1334 Find The City With The Smallest Number Of Neighbors At A Threshold Distance - Medium

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

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1334. Find The City With The Smallest Number Of Neighbors At A Threshold Distance — Medium

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Problem

  1. Find the City With the Smallest Number of Neighbors at a Threshold Distance -- Medium

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.

Solution

# Floyd Alg DP:
class Solution:
    def findTheCity(self, n: int, edges: List[List[int]], distanceThreshold: int) -> int:
        # Floyd Alg to find minimum distance between two nodes in directed weighted graph
        dp = [[float('inf')] * n for _ in range(n)]
        for edge in edges:
            dp[edge[0]][edge[1]] = edge[2]
            dp[edge[1]][edge[0]] = edge[2]
        for i in range(n):
            dp[i][i] = 0
        for k in range(n):
            for i in range(n):
                for j in range(n):
                    dp[i][j] = min(dp[i][j], dp[i][k] + dp[k][j])
 
        city = -1
        minCount = float('inf')
        for i in range(n):
            count = 0
            for j in range(n):
                if dp[i][j] <= distanceThreshold:
                    count += 1
            if count <= minCount:
                city = i
                minCount = count
        return city
        
# BFS + Heap:
class Solution:
    def findTheCity(self, n: int, edges: List[List[int]], distanceThreshold: int) -> int:
        graph = collections.defaultdict(list)
        for u, v, w in edges:
            graph[u].append((v, w))
            graph[v].append((u, w))
        def getNumberOfNeighbors(city):
            heap = [(0, city)]
            dist = {}
            while heap:
                currW, u = heapq.heappop(heap)
                if u in dist:
                    continue
                if u != city:    
                    dist[u] = currW
                for v, w in graph[u]:
                    if v in dist:
                        continue
                    if currW + w <= distanceThreshold:
                        heapq.heappush(heap, (currW + w, v))
            return len(dist)
        return max([(getNumberOfNeighbors(city), city) for city in range(n)], key=lambda x: (-x[0], x[1]))[-1]

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