Given a directed acyclic graph (DAG) of n nodes labeled from 0 to n - 1, find all possible paths from node 0 to node n - 1 and return them in any order.
The graph is given as follows: graph[i] is a list of all nodes you can visit from node i (i.e., there is a directed edge from node i to node graph[i][j]).
Example:
Input: graph = [[1,2],[3],[3],[]]
Output: [[0,1,3],[0,2,3]]
Explanation: There are two paths: 0 -> 1 -> 3 and 0 -> 2 -> 3.
class Solution {
public List<List<Integer>> allPathsSourceTarget(int[][] graph) {
// we have to find all the possible path from source to destination.
// we can do that with depth first search.
//we can keep track of visited nodes in a list in a depth first search manner.
//if we reach the target node we will store list 'l', and we will backtrack to adjacent nodes of the previous node to check all the possible paths
List<List<Integer>> list = new ArrayList<>();
dfs(0,graph,new ArrayList<>(Arrays.asList(0)),list); //initially we are also adding 0 in the list l, as the starting node
return list;
}
public void dfs(int node, int[][] graph,List<Integer> l, List<List<Integer>> list){
if(node ==graph.length-1){
list.add(new ArrayList<>(l));
return;
}
// below loop will give array of all adjacent nodes of current node i.e 'node'
for(int i : graph[node]){
//take
l.add(i);
dfs(i,graph,l,list);
//don't take
l.remove(l.size()-1);
}
}
}
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