# 1334. Find the City With the Smallest Number of Neighbors at a Threshold Distance ## Description 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:** ![](https://assets.leetcode.com/uploads/2020/01/16/find_the_city_01.png) ``` **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:** ![](https://assets.leetcode.com/uploads/2020/01/16/find_the_city_02.png) ``` **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. ## ac1: Floyd-Warshall, O(V^3) ```java class Solution { public int findTheCity(int n, int[][] edges, int distanceThreshold) { int[][] dist = new int[n][n]; for (int i = 0; i < n; i++) { Arrays.fill(dist[i], Integer.MAX_VALUE / 2); // MAX_VALUE will cause overflow downstream dist[i][i] = 0; } for (int[] edge : edges) { int u = edge[0]; int v = edge[1]; int weight = edge[2]; dist[u][v] = weight; dist[v][u] = weight; } // Standard Floyd-Warshall step: 1) iterate each V; 2) assume k is the mid point from i to j, relax dist[i][j]; for (int k = 0; k < n; k++) { for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { if (dist[i][k] + dist[k][j] < dist[i][j]) { dist[i][j] = dist[i][k] + dist[k][j]; } } } } int res = -1; int minReach = Integer.MAX_VALUE; for (int i = 0; i < n; i++) { int reach = 0; for (int j = 0; j < n; j++) { if (i == j) continue; // Same city if (dist[i][j] <= distanceThreshold) { reach++; } } if (reach <= minReach) { res = i; minReach = reach; } } return res; } } // Floyd-Warshall, O(V^3) time, O(V^2) space. Tricky: use "Integer.MAX_VALUE / 2" to avoid overflow. ``` ## ac2: Bellman-Ford, O(VEV) ```java class Solution { public int findTheCity(int n, int[][] edges, int distanceThreshold) { int[][] dist = new int[n][n]; for (int i = 0; i < n; i++) { Arrays.fill(dist[i], Integer.MAX_VALUE / 2); // MAX_VALUE will cause overflow downstream dist[i][i] = 0; } for (int i = 0; i < n; i++) { // Do the following for each V to get a matrix of distance. // Bellman-Ford: 1) iterate each V; 2) iterate each edge try to relax u-v, tricky: it's undirected graph, so do u->v and v->u. for (int j = 0; j < n; j++) { for (int[] edge : edges) { int u = edge[0]; int v = edge[1]; int weight = edge[2]; if (dist[i][u] + weight < dist[i][v]) { dist[i][v] = dist[i][u] + weight; } if (dist[i][v] + weight < dist[i][u]) { dist[i][u] = dist[i][v] + weight; } } } } int res = -1; int minReach = Integer.MAX_VALUE; for (int i = 0; i < n; i++) { int reach = 0; for (int j = 0; j < n; j++) { if (i == j) continue; // Same city if (dist[i][j] <= distanceThreshold) { reach++; } } if (reach <= minReach) { res = i; minReach = reach; } } return res; } } // Bellman-Ford, O(V*EV) time, O(V^2) space. E = V^2, so it's O(V^4), which is worse than Floyd-Warshall. ``` Shortest Path Faster Algorithm (SPFA) is an improvement of Bellman-Ford: ```java void spfa(int n, List[] adj, int[] dist, int src) { Deque q = new ArrayDeque<>(); int[] updateTimes = new int[n]; q.add(src); while (!q.isEmpty()) { int u = q.removeFirst(); for (int[] next : adj[u]) { int v = next[0]; int duv = next[1]; if (dist[v] > dist[u] + duv) { dist[v] = dist[u] + duv; updateTimes[v]++; q.add(v); if (updateTimes[v] > n) {System.out.println("wrong");} } } } } ``` ## ac3: Dijkstra, O(VElogV) ```java class Solution { public int findTheCity(int n, int[][] edges, int distanceThreshold) { Map> graph = new HashMap<>(); int[][] dist = new int[n][n]; for (int i = 0; i < n; i++) { Arrays.fill(dist[i], Integer.MAX_VALUE / 2); // MAX_VALUE will cause overflow downstream dist[i][i] = 0; graph.put(i, new HashMap<>()); } for (int[] edge : edges) { int u = edge[0]; int v = edge[1]; int weight = edge[2]; graph.get(u).put(v, weight); graph.get(v).put(u, weight); } for (int i = 0; i < n; i++) { // Do the following for each V to get a matrix of distance. dijkstra(i, dist[i], graph); } int res = -1; int minReach = Integer.MAX_VALUE; for (int i = 0; i < n; i++) { int reach = 0; for (int j = 0; j < n; j++) { if (i == j) continue; // Same city if (dist[i][j] <= distanceThreshold) { reach++; } } if (reach <= minReach) { res = i; minReach = reach; } } return res; } private void dijkstra(int src, int[] dist, Map> graph) { PriorityQueue q = new PriorityQueue<>((a, b) -> a[1] - b[1]); q.offer(new int[]{src, 0}); while (!q.isEmpty()) { int[] curr = q.poll(); int u = curr[0]; int srcToU = curr[1]; if (dist[u] < srcToU) continue; // src to u(dist[u]) is already smaller, no point to keep going. for (int v : graph.get(u).keySet()) { if (srcToU + graph.get(u).get(v) < dist[v]) { dist[v] = srcToU + graph.get(u).get(v); q.offer(new int[]{v, dist[v]}); } } } } } // Dijkstra, O(V*ElogV) time, O(V^2) space. E = V^2, so it's O(V^3logV), which is worse than Floyd-Warshall. // Tricky: you can use boolean[] visited to avoid loop, but it's better to rely on their weight, i.e. if (dist[u] < srcToU) continue; ``` --- # Agent Instructions: Querying This Documentation If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question. Perform an HTTP GET request on the current page URL with the `ask` query parameter: ``` GET https://jaywin.gitbook.io/leetcode/solutions/1334-find-the-city-with-the-smallest-number-of-neighbors-at-a-threshold-distance.md?ask= ``` The question should be specific, self-contained, and written in natural language. The response will contain a direct answer to the question and relevant excerpts and sources from the documentation. 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