Question

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

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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

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.shift()!;
      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;
}