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

Problem

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.

Solution (Java)

class Solution {
    private List<List<Integer>> adjList;
    private List<List<Integer>> result;

    public List<List<Integer>> getAncestors(int n, int[][] edges) {
        this.adjList = new ArrayList<>();
        this.result = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            adjList.add(new ArrayList<>());
            result.add(new ArrayList<>());
        }
        for (int[] edge : edges) {
            int start = edge[0];
            int end = edge[1];
            adjList.get(start).add(end);
        }
        //  DFS for each node from 0 --> n , and add that node as root/parent into each reachable
        // node and their child
        //  Use visited[] to identify if any of the child or their childs are already visited for
        // that perticular root/parent,
        //  so will not add the root to avoid duplicacy and call reduction .
        for (int i = 0; i < n; i++) {
            boolean[] visited = new boolean[n];
            List<Integer> childList = adjList.get(i);
            for (Integer child : childList) {
                if (!visited[child]) {
                    dfs(i, child, visited);
                }
            }
        }
        return result;
    }

    private void dfs(int root, int node, boolean[] visited) {
        if (visited[node]) {
            return;
        }
        visited[node] = true;
        result.get(node).add(root);
        List<Integer> childList = adjList.get(node);
        for (Integer child : childList) {
            if (!visited[child]) {
                dfs(root, child, visited);
            }
        }
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).