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MD ARIFUL HAQUE

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

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:

• 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.

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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