2192. All Ancestors of a Node in a Directed Acyclic Graph
Tags: php, leetcode, algorithms, programming
Published: 2024-06-29 18:40:12

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 <code>edges[i] = [from<sub>i</sub>, to<sub>i</sub>]</code> denotes that there is a **unidirectional** edge from <code>from<sub>i</sub></code> to <code>to<sub>i</sub></code> in the graph.

Return a list `answer`, where `answer[i]` is the **list of ancestors** of the <code>i<sup>th</sup></code> 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](https://assets.leetcode.com/uploads/2019/12/12/e1.png)

- **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](https://assets.leetcode.com/uploads/2019/12/12/e2.png)

- **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:**


- <code>1 <= n <= 1000</code>
- <code>0 <= edges.length <= min(2000, n * (n - 1) / 2)</code>
- <code>edges[i].length == 2</code>
- <code>0 <= from<sub>i</sub>, to<sub>i</sub> <= n - 1</code>
- <code>from<sub>i</sub> != to<sub>i</sub></code>
- 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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