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 mostdistanceThreshold, 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.
ac1: Floyd-Warshall, O(V^3)
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.
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:
void spfa(int n, List<int[]>[] adj, int[] dist, int src) {
Deque<Integer> 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)
class Solution {
public int findTheCity(int n, int[][] edges, int distanceThreshold) {
Map<Integer, Map<Integer, Integer>> 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<Integer, Map<Integer, Integer>> graph) {
PriorityQueue<int[]> 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;
