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];
        for (auto& e : edges) {
            g[e[0]].push_back(e[1]);
        }
        vector<vector<int>> ans(n);
        auto bfs = [&](int s) {
            queue<int> q;
            q.push(s);
            bool vis[n];
            memset(vis, 0, sizeof(vis));
            vis[s] = true;
            while (q.size()) {
                int i = q.front();
                q.pop();
                for (int j : g[i]) {
                    if (!vis[j]) {
                        vis[j] = true;
                        ans[j].push_back(s);
                        q.push(j);
                    }
                }
            }
        };
        for (int i = 0; i < n; ++i) {
            bfs(i);
        }
        return ans;
    }
};

func getAncestors(n int, edges [][]int) [][]int {
    g := make([][]int, n)
    for _, e := range edges {
        g[e[0]] = append(g[e[0]], e[1])
    }
    ans := make([][]int, n)
    bfs := func(s int) {
        q := []int{s}
        vis := make([]bool, n)
        vis[s] = true
        for len(q) > 0 {
            i := q[0]
            q = q[1:]
            for _, j := range g[i] {
                if !vis[j] {
                    vis[j] = true
                    q = append(q, j)
                    ans[j] = append(ans[j], s)
                }
            }
        }
    }
    for i := 0; i < n; i++ {
        bfs(i)
    }
    return ans
}

function getAncestors(n: number, edges: number[][]): number[][] {
    const g: number[][] = Array.from({ length: n }, () => []);
    for (const [u, v] of edges) {
        g[u].push(v);
    }
    const ans: number[][] = Array.from({ length: n }, () => []);
    const bfs = (s: number) => {
        const q: number[] = [s];
        const vis: boolean[] = Array.from({ length: n }, () => false);
        vis[s] = true;
        while (q.length) {
            const i = q.pop()!;
            for (const j of g[i]) {
                if (!vis[j]) {
                    vis[j] = true;
                    ans[j].push(s);
                    q.push(j);
                }
            }
        }
    };
    for (let i = 0; i < n; ++i) {
        bfs(i);
    }
    return ans;
}

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

    public IList<IList<int>> GetAncestors(int n, int[][] edges) {
        g = new List<int>[n];
        this.n = n;
        for (int i = 0; i < n; i++) {
            g[i] = new List<int>();
        }
        foreach (var e in edges) {
            g[e[0]].Add(e[1]);
        }
        ans = new List<IList<int>>();
        for (int i = 0; i < n; ++i) {
            ans.Add(new List<int>());
        }
        for (int i = 0; i < n; ++i) {
            BFS(i);
        }
        return ans;
    }

    private void BFS(int s) {
        Queue<int> q = new Queue<int>();
        q.Enqueue(s);
        bool[] vis = new bool[n];
        vis[s] = true;
        while (q.Count > 0) {
            int i = q.Dequeue();
            foreach (int j in g[i]) {
                if (!vis[j]) {
                    vis[j] = true;
                    q.Enqueue(j);
                    ans[j].Add(s);
                }
            }
        }
    }
}

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