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

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Problem Statement

1334. 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:

Java

Source file
class Solution {
    public int findTheCity(int n, int[][] edges, int distanceThreshold) {
        int[][] cost = new int[n][n];
        List<int[]>[] adj = new List[n];    // [] of LL of int{nbr, wt}
        for(int i=0; i<n; i++){
            Arrays.fill(cost[i], Integer.MAX_VALUE);    // INF
            cost[i][i] = 0; // self wt.
            adj[i] = new ArrayList<>(); // init LL
        }
        for(int[] e: edges){ // populate edges in adjList
            adj[e[0]].add(new int[]{e[1], e[2]});
            adj[e[1]].add(new int[]{e[0], e[2]});
        }
        for(int src=0; src<n; src++)    // djikstra on each node as source
            djikstra(n, adj, cost[src], src);
        int ans = -1, minCt = n;
        for(int i=0; i<n; i++){
            int currCt = 0;
            for(int j=0; j<n; j++)
                if(cost[i][j]<=distanceThreshold)
                    currCt++;
            if(minCt>=currCt){
                minCt = currCt;
                ans = i;
            }
        }
        return ans;
    }
    void djikstra(int n, List<int[]>[] adj, int[] cost, int src){
        Queue<int[]> pq = new PriorityQueue<>((a, b)->a[1]-b[1]);
        pq.add(new int[]{src, 0});
        Arrays.fill(cost, Integer.MAX_VALUE);
        cost[src] = 0;
        while(!pq.isEmpty()){
            int[] node = pq.poll();
            int currCity = node[0], currCost = node[1];
            if(currCost>cost[currCity]) // worse path, skip
                continue;
            for(int[] nbr: adj[currCity]){
                int nbrCity = nbr[0], nbrEdge = nbr[1];
                int newCostNbr = currCost+nbrEdge;
                if(cost[nbrCity]>newCostNbr){
                    cost[nbrCity] = newCostNbr;
                    pq.add(new int[]{nbrCity, newCostNbr});
                }
            }
        }
    }
}