[leetcode] 2192. All Ancestors of a Node in a Directed Acyclic Graph

문제

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.

예시

• 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]]

• 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]]

제한사항

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.

풀이1

const getAncestors = (n, edges) => {
  const inEdges = Array.from({ length: n }, () => new Set());
  const outdegrees = new Array(n);
  const finished = new Array(n);
  const dfs = (idx) => {
    if (finished[idx]) return;
    if (inEdges[idx].size === 0) return;
    for (const edge of inEdges[idx]) {
      dfs(edge);
      inEdges[idx] = new Set([...inEdges[idx], ...inEdges[edge]]);
    }
    finished[idx] = 1;
  };

  for (const [from, to] of edges) {
    inEdges[to].add(from);
    ++outdegrees[from];
  }

  for (let i = 0; i < n; ++i) {
    outdegrees[i] == 0 && dfs(i);
  }

  return inEdges.map((set) => Array.from(set).sort((a, b) => a - b));
};

하나의 node 당 몇개의 하위 원소를 갖는지를 미리 outdegrees에 체크를 해준다.
outdegree가 0인 것을 dfs를 돌리게 되면, 상위 node들은 모두 계산이 되게 되고,
[]안의 값을 오름차순으로 정렬해서 return 해 준다.