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

Description

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.

Solutions

Solution 1: BFS

First, we construct the adjacency list $g$ based on the two-dimensional array $edges$, where $g[i]$ represents all successor nodes of node $i$.

Then, we enumerate node $i$ as the ancestor node from small to large, use BFS to search all successor nodes of node $i$, and add node $i$ to the ancestor list of these successor nodes.

The time complexity is $O(n^2)$, and the space complexity is $O(n^2)$. Where $n$ is the number of nodes.

Python3JavaC++GoTypeScriptC#

class Solution:
    def getAncestors(self, n: int, edges: List[List[int]]) -> List[List[int]]:
        def bfs(s: int):
            q = deque([s])
            vis = {s}
            while q:
                i = q.popleft()
                for j in g[i]:
                    if j not in vis:
                        vis.add(j)
                        q.append(j)
                        ans[j].append(s)

        g = defaultdict(list)
        for u, v in edges:
            g[u].append(v)
        ans = [[] for _ in range(n)]
        for i in range(n):
            bfs(i)
        return ans

class Solution {
    private int n;
    private List<Integer>[] g;
    private List<List<Integer>> ans;

    public List<List<Integer>> getAncestors(int n, int[][] edges) {
        g = new List[n];
        this.n = n;
        Arrays.setAll(g, i -> new ArrayList<>());
        for (var e : edges) {
            g[e[0]].add(e[1]);
        }
        ans = new ArrayList<>();
        for (int i = 0; i < n; ++i) {
            ans.add(new ArrayList<>());
        }
        for (int i = 0; i < n; ++i) {
            bfs(i);
        }
        return ans;
    }

    private void bfs(int s) {
        Deque<Integer> q = new ArrayDeque<>();
        q.offer(s);
        boolean[] vis = new boolean[n];
        vis[s] = true;
        while (!q.isEmpty()) {
            int i = q.poll();
            for (int j : g[i]) {
                if (!vis[j]) {
                    vis[j] = true;
                    q.offer(j);
                    ans.get(j).add(s);
                }
            }
        }
    }
}

class Solution {
public:
    vector<vector<int>> getAncestors(int n, vector<vector<int>>& edges) {
        vector<int> g[n];