Formatted question description: https://leetcode.ca/all/1334.html

# 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:

• 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.

Related Topics:
Graph

## Solution 1.

// OJ: https://leetcode.com/problems/find-the-city-with-the-smallest-number-of-neighbors-at-a-threshold-distance/
// Time: O(V * (E + VlogV))
// Space: O(E)
class Solution {
typedef pair<int, int> iPair;
unordered_map<int, vector<iPair>> graph;
int th, N, minNum = INT_MAX;
int dijkstra(int i) {
priority_queue<iPair, vector<iPair>, greater<iPair>> q;
vector<int> dist(N, INT_MAX);
q.push(make_pair(0, i));
dist[i] = 0;
while (q.size()) {
int u = q.top().second;
q.pop();
for (auto ne : graph[u]) {
int v = ne.first;
int w = ne.second;
if (dist[v] > dist[u] + w) {
dist[v] = dist[u] + w;
q.push(make_pair(dist[v], v));
}
}
}
int cnt = 0;
for (int j = 0; j < N; ++j) {
if (j != i && dist[j] <= th) ++cnt;
}
return cnt;
}
public:
int findTheCity(int n, vector<vector<int>>& edges, int distanceThreshold) {
th = distanceThreshold;
N = n;
for (auto e : edges) {
m[e].push_back(make_pair(e, e));
m[e].push_back(make_pair(e, e));
}
int ans = 0;
for (int i = 0; i < n; ++i) {
int cnt = dijkstra(i);
if (cnt <= minNum) {
minNum = cnt;
ans = i;
}
}
return ans;
}
};


## Solution 2. Floyd-Warshall

// OJ: https://leetcode.com/problems/find-the-city-with-the-smallest-number-of-neighbors-at-a-threshold-distance/
// Time: O(N^3)
// Space: O(N^2)
class Solution {
public:
int findTheCity(int n, vector<vector<int>>& E, int k) {
vector<vector<int>> d(n, vector<int>(n, (int) 1e6));
for (int i = 0; i < n; ++i) d[i][i] = 0;
for (auto &e : E) {
d[e][e] = d[e][e] = e;
}
for (int i = 0; i < n; ++i) {
for (int u = 0; u < n; ++u) {
for (int v = 0; v < n; ++v) {
d[u][v] = min(d[u][v], d[u][i] + d[i][v]);
}
}
}
int ans = 0, minCnt = n;
for (int i = 0; i < n; ++i) {
int cnt = 0;
for (int j = 0; j < n; ++j) {
if (d[i][j] <= k) ++cnt;
}
if (cnt <= minCnt) {
minCnt = cnt;
ans = i;
}
}
return ans;
}
};


Java

• class Solution {
public int findTheCity(int n, int[][] edges, int distanceThreshold) {
Map<Integer, List<int[]>> distancesMap = new HashMap<Integer, List<int[]>>();
for (int[] edge : edges) {
int city1 = edge, city2 = edge, distance = edge;
List<int[]> list1 = distancesMap.getOrDefault(city1, new ArrayList<int[]>());
List<int[]> list2 = distancesMap.getOrDefault(city2, new ArrayList<int[]>());
distancesMap.put(city1, list1);
distancesMap.put(city2, list2);
}
int[] reachableCounts = new int[n];
for (int i = 0; i < n; i++) {
int[] distances = getDistance(n, i, distancesMap);
for (int distance : distances) {
if (distance > 0 && distance <= distanceThreshold)
reachableCounts[i]++;
}
}
int minReachable = n - 1;
for (int i = 0; i < n; i++)
minReachable = Math.min(minReachable, reachableCounts[i]);
int cityIndex = 0;
for (int i = n - 1; i >= 0; i--) {
if (reachableCounts[i] == minReachable) {
cityIndex = i;
break;
}
}
return cityIndex;
}

public int[] getDistance(int n, int source, Map<Integer, List<int[]>> distancesMap) {
int[] distances = new int[n];
for (int i = 0; i < n; i++)
distances[i] = Integer.MAX_VALUE;
distances[source] = 0;
PriorityQueue<CityDistance> priorityQueue = new PriorityQueue<CityDistance>();
priorityQueue.offer(new CityDistance(source, 0));
while (!priorityQueue.isEmpty()) {
CityDistance cityDistance = priorityQueue.poll();
int city = cityDistance.city, distance = cityDistance.distance;
if (distance > distances[city])
continue;
List<int[]> adjacentCities = distancesMap.getOrDefault(city, new ArrayList<int[]>());
int newDistance = distance + nextDistance;
if (newDistance < distances[nextCity]) {
distances[nextCity] = newDistance;
priorityQueue.offer(new CityDistance(nextCity, newDistance));
}
}
}
return distances;
}
}

class CityDistance implements Comparable<CityDistance> {
public int city;
public int distance;

public CityDistance(int city, int distance) {
this.city = city;
this.distance = distance;
}

public int compareTo(CityDistance cityDistance2) {
return this.distance - cityDistance2.distance;
}
}

• // OJ: https://leetcode.com/problems/find-the-city-with-the-smallest-number-of-neighbors-at-a-threshold-distance/
// Time: O(V * (E + VlogV))
// Space: O(E)
class Solution {
typedef pair<int, int> iPair;
unordered_map<int, vector<iPair>> graph;
int th, N, minNum = INT_MAX;
int dijkstra(int i) {
priority_queue<iPair, vector<iPair>, greater<iPair>> q;
vector<int> dist(N, INT_MAX);
q.push(make_pair(0, i));
dist[i] = 0;
while (q.size()) {
int u = q.top().second;
q.pop();
for (auto ne : graph[u]) {
int v = ne.first;
int w = ne.second;
if (dist[v] > dist[u] + w) {
dist[v] = dist[u] + w;
q.push(make_pair(dist[v], v));
}
}
}
int cnt = 0;
for (int j = 0; j < N; ++j) {
if (j != i && dist[j] <= th) ++cnt;
}
return cnt;
}
public:
int findTheCity(int n, vector<vector<int>>& edges, int distanceThreshold) {
th = distanceThreshold;
N = n;
for (auto e : edges) {
m[e].push_back(make_pair(e, e));
m[e].push_back(make_pair(e, e));
}
int ans = 0;
for (int i = 0; i < n; ++i) {
int cnt = dijkstra(i);
if (cnt <= minNum) {
minNum = cnt;
ans = i;
}
}
return ans;
}
};

• print("Todo!")