2192. All Ancestors of a Node in a Directed Acyclic Graph

Attempted

Medium

Topics

Companies

Hint

You are given a positive integer n representing the number of nodes of a Directed Acyclic Graph (DAG). The nodes are numbered from 0 to n - 1 (inclusive).

You are also given a 2D integer array edges, where edges[i] = [fromi, toi] denotes that there is a unidirectional edge from fromi to toi in the graph.

Return a list answer, where answer[i] is the list of ancestors of the ith node, sorted in ascending order.

A node u is an ancestor of another node v if u can reach v via a set of edges.

Example 1:

Input: n = 8, edgeList = [[0,3],[0,4],[1,3],[2,4],[2,7],[3,5],[3,6],[3,7],[4,6]]
Output: [[],[],[],[0,1],[0,2],[0,1,3],[0,1,2,3,4],[0,1,2,3]]
Explanation:
The above diagram represents the input graph.
- Nodes 0, 1, and 2 do not have any ancestors.
- Node 3 has two ancestors 0 and 1.
- Node 4 has two ancestors 0 and 2.
- Node 5 has three ancestors 0, 1, and 3.
- Node 6 has five ancestors 0, 1, 2, 3, and 4.
- Node 7 has four ancestors 0, 1, 2, and 3.

Example 2:

Input: n = 5, edgeList = [[0,1],[0,2],[0,3],[0,4],[1,2],[1,3],[1,4],[2,3],[2,4],[3,4]]
Output: [[],[0],[0,1],[0,1,2],[0,1,2,3]]
Explanation:
The above diagram represents the input graph.
- Node 0 does not have any ancestor.
- Node 1 has one ancestor 0.
- Node 2 has two ancestors 0 and 1.
- Node 3 has three ancestors 0, 1, and 2.
- Node 4 has four ancestors 0, 1, 2, and 3.

Constraints:

• 1 <= n <= 1000
• 0 <= edges.length <= min(2000, n * (n - 1) / 2)
• edges[i].length == 2
• 0 <= fromi, toi <= n - 1
• fromi != toi
• There are no duplicate edges.
• The graph is directed and acyclic.

class Solution {

    /** Naive Thought & will not work */
    public List<List<Integer>> _getAncestors(int n, int[][] edges) {
        List<Set<Integer>> ans= new ArrayList<>();

        for( int i=0; i<n ;i++){
            ans.add(new TreeSet<>() );
        }

        for( int[] edge : edges){
            ans.get(edge[1]).add( edge[0]);
            for( int i : ans.get(edge[0])){
                ans.get(edge[1]).add(i);
            }
        }

        List<List<Integer>> finalAns= new ArrayList<>();

         for( Set<Integer> set : ans){
            List<Integer> member= new ArrayList<>();
            for( int j: set) member.add(j);
            finalAns.add(member);
         }

        return finalAns;
    }



    void dfs(  List<List<Integer>> adjList, int currNode,  Set<Integer> visited){
                for( int nei : adjList.get(currNode) ){
                        dfs(adjList, nei, visited);
                }
                if(!visited.contains(currNode))  visited.add(currNode); 
    }

 /**  T : O(n×(n+e))+O(n^2l)  S: O(n)*/
    public List<List<Integer>> getAncestors( int n, int[][] edges){

        /** construct the adjacency list by reversing the edges*/
        List<List<Integer>> adjList= new ArrayList<>();
        for( int i=0;i<n;i++) adjList.add( new ArrayList<>());

        for( int[] edge : edges) adjList.get(edge[1]).add(edge[0]);
        System.out.println(adjList);


        List<List<Integer>> ans = new ArrayList<>();

        /**perfrom dfs on the each edge & make list of all nodes in path */
        for( int i=0; i< n; i++){
            Set<Integer> visited= new HashSet<>();
             dfs( adjList, i, visited);
            
                List<Integer> ansistors=new ArrayList<>();
                for( int j=0; j< n; j++){
                    if( visited.contains(j) && i!=j ) ansistors.add(j);
                }
                 ans.add(ansistors);
   
        }

        return ans;
    }


   
}