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] = [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.
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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