LeetCode #2192 — MEDIUM

All Ancestors of a Node in a Directed Acyclic Graph

Move from brute-force thinking to an efficient approach using topological sort strategy.

The Problem

Problem Statement

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] = [from_i, to_i] denotes that there is a unidirectional edge from from_i to to_i in the graph.

Return a list answer, where answer[i] is the list of ancestors of the i^th 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 <= from_i, to_i <= n - 1
from_i != to_i
There are no duplicate edges.
The graph is directed and acyclic.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Topological Sort

Example 1

8
[[0,3],[0,4],[1,3],[2,4],[2,7],[3,5],[3,6],[3,7],[4,6]]

Example 2

5
[[0,1],[0,2],[0,3],[0,4],[1,2],[1,3],[1,4],[2,3],[2,4],[3,4]]

Core Insight

What unlocks the optimal approach

Consider how reversing each edge of the graph can help us.
How can performing BFS/DFS on the reversed graph help us find the ancestors of every node?

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2192: All Ancestors of a Node in a Directed Acyclic Graph
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);
                }
            }
        }
    }
}

// Accepted solution for LeetCode #2192: All Ancestors of a Node in a Directed Acyclic Graph
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
}

# Accepted solution for LeetCode #2192: All Ancestors of a Node in a Directed Acyclic Graph
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

// Accepted solution for LeetCode #2192: All Ancestors of a Node in a Directed Acyclic Graph
/**
 * [2192] All Ancestors of a Node in a Directed Acyclic Graph
 *
 * 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 i^th 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:
 * <img alt="" src="https://assets.leetcode.com/uploads/2019/12/12/e1.png" style="width: 322px; height: 265px;" />
 * 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:
 * <img alt="" src="https://assets.leetcode.com/uploads/2019/12/12/e2.png" style="width: 343px; height: 299px;" />
 * 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.
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/all-ancestors-of-a-node-in-a-directed-acyclic-graph/
// discuss: https://leetcode.com/problems/all-ancestors-of-a-node-in-a-directed-acyclic-graph/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    pub fn get_ancestors(n: i32, edges: Vec<Vec<i32>>) -> Vec<Vec<i32>> {
        vec![]
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    #[ignore]
    fn test_2192_example_1() {
        let n = 8;
        let edges = vec![
            vec![0, 3],
            vec![0, 4],
            vec![1, 3],
            vec![2, 4],
            vec![2, 7],
            vec![3, 5],
            vec![3, 6],
            vec![3, 7],
            vec![4, 6],
        ];

        let result = vec![
            vec![],
            vec![],
            vec![],
            vec![0, 1],
            vec![0, 2],
            vec![0, 1, 3],
            vec![0, 1, 2, 3, 4],
            vec![0, 1, 2, 3],
        ];

        assert_eq!(Solution::get_ancestors(n, edges), result);
    }

    #[test]
    #[ignore]
    fn test_2192_example_2() {
        let n = 5;
        let edges = vec![
            vec![0, 1],
            vec![0, 2],
            vec![0, 3],
            vec![0, 4],
            vec![1, 2],
            vec![1, 3],
            vec![1, 4],
            vec![2, 3],
            vec![2, 4],
            vec![3, 4],
        ];

        let result = vec![vec![], vec![0], vec![0, 1], vec![0, 1, 2], vec![0, 1, 2, 3]];

        assert_eq!(Solution::get_ancestors(n, edges), result);
    }
}

// Accepted solution for LeetCode #2192: All Ancestors of a Node in a Directed Acyclic Graph
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;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n^2)

Space

O(n^2)

Approach Breakdown

REPEATED DFS

O(V × E) time

O(V) space

Repeatedly find a vertex with no incoming edges, remove it and its outgoing edges, and repeat. Finding the zero-in-degree vertex scans all V vertices, and we do this V times. Removing edges touches E edges total. Without an in-degree array, this gives O(V × E).

TOPOLOGICAL SORT

O(V + E) time

O(V + E) space

Build an adjacency list (O(V + E)), then either do Kahn's BFS (process each vertex once + each edge once) or DFS (visit each vertex once + each edge once). Both are O(V + E). Space includes the adjacency list (O(V + E)) plus the in-degree array or visited set (O(V)).

Shortcut: Process each vertex once + each edge once → O(V + E). Same as BFS/DFS on a graph.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.