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MD ARIFUL HAQUE
MD ARIFUL HAQUE

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

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

Medium

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:

e1

  • 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:

e2

  • 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.

Solution:

class Solution { /** * @param Integer $n * @param Integer[][] $edges * @return Integer[][] */ function getAncestors($n, $edges) { $adjacencyList = array_fill(0, $n, []); foreach ($edges as $edge) { $from = $edge[0]; $to = $edge[1]; $adjacencyList[$to][] = $from; } $ancestorsList = []; for ($i = 0; $i < $n; $i++) { $ancestors = []; $visited = []; $this->findChildren($i, $adjacencyList, $visited); for ($node = 0; $node < $n; $node++) { if ($node == $i) continue; if (in_array($node, $visited)) $ancestors[] = $node; } $ancestorsList[] = $ancestors; } return $ancestorsList; } private function findChildren($currentNode, &$adjacencyList, &$visitedNodes) { $visitedNodes[] = $currentNode; foreach ($adjacencyList[$currentNode] as $neighbour) { if (!in_array($neighbour, $visitedNodes)) { $this->findChildren($neighbour, $adjacencyList, $visitedNodes); } } } }
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