1059. All Paths from Source Lead to Destination
1059. All Paths from Source Lead to Destination
Given the edges of a directed graph, and two nodes source and destination of this graph, determine whether
or not all paths starting from source eventually end at destination, that is:
At least one path exists from the source node to the destination node
If a path exists from the source node to a node with no outgoing edges, then that node is equal to
destination.
The number of possible paths from source to destination is a finite number.
Return true if and only if all roads from source lead to destination.
Example 1:
Input: n = 3, edges = [[0,1],[0,2]], source = 0, destination = 2
Output: false
Explanation: It is possible to reach and get stuck on both node 1 and node 2.
Example 2:
Input: n = 4, edges = [[0,1],[0,3],[1,2],[2,1]], source = 0, destination = 3
Output: false
Explanation: We have two possibilities: to end at node 3, or to loop over node 1 and node 2 indefinitely.
Example 3:
Input: n = 4, edges = [[0,1],[0,2],[1,3],[2,3]], source = 0, destination = 3
Output: true
Example 4:
Input: n = 3, edges = [[0,1],[1,1],[1,2]], source = 0, destination = 2
Output: false
Explanation: All paths from the source node end at the destination node, but there are an infinite number
of paths, such as 0-1-2, 0-1-1-2, 0-1-1-1-2, 0-1-1-1-1-2, and so on.
Example 5:
Input: n = 2, edges = [[0,1],[1,1]], source = 0, destination = 1
Output: false
Explanation: There is infinite self-loop at destination node.
Note:
The given graph may have self loops and parallel edges.
The number of nodes n in the graph is between 1 and 10000
The number of edges in the graph is between 0 and 10000
0 <= edges.length <= 10000
edges[i].length == 2
0 <= source <= n - 1
0 <= destination <= n - 1
- DFS
class Solution {
int src, dst;
boolean flag;
public boolean leadsToDestination(int n, int[][] edges, int source, int destination) {
this.src = source;
this.dst = destination;
this.flag = true;
ArrayList<Integer>[] graph = new ArrayList[n];
HashSet<Integer>[] used = new HashSet[n];
boolean[] visited = new boolean[n];
for(int i = 0; i < n; i++) {
graph[i] = new ArrayList<>();
used[i] = new HashSet<>();
}
for(int[] e : edges) {
int i = e[0], j = e[1];
graph[i].add(j);
}
if(graph[this.dst].size() > 0)
return false;
boolean result = dfs(graph, used, visited, this.src);
return result;
}
public boolean dfs(ArrayList<Integer>[] graph, HashSet<Integer>[] used, boolean[] visited, int curr) {
// System.out.println(curr);
if(visited[curr] && curr != this.dst) { // circle
return false;
}
visited[curr] = true;
if(graph[curr].size() == used[curr].size()) { // check destination
return curr == this.dst;
}
for(int next : graph[curr]) {
if(used[curr].contains(next)) continue;
used[curr].add(next);
if(!dfs(graph, used, visited, next)) {
return false;
}
used[curr].remove(next);
}
visited[curr] = false;
return true;
}
}