LeetCode 2192: All Ancestors of a Node in a Directed Acyclic Graph Solution

Master LeetCode problem 2192 (All Ancestors of a Node in a Directed Acyclic Graph), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

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

Medium

Depth-First Search Breadth-First Search Graph Topological Sort

Problem Explanation

Explanation

To find all ancestors of a node in a Directed Acyclic Graph (DAG), we can perform a depth-first search (DFS) starting from each node and keeping track of the ancestors for each node. We can use a hashmap to store the ancestors for each node.

Create an adjacency list representation of the graph using the edges provided.
Initialize a hashmap to store the ancestors for each node.
Perform a DFS from each node to find its ancestors.
During the DFS, update the ancestors hashmap for each node visited.
Return the list of ancestors for each node in the desired format.

Time complexity: O(V + E), where V is the number of vertices and E is the number of edges. Space complexity: O(V + E) for the adjacency list and hashmap.

Solution Code

class Solution {
    public List<List<Integer>> allAncestors(int n, int[][] edges) {
        List<List<Integer>> graph = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            graph.add(new ArrayList<>());
        }
        for (int[] edge : edges) {
            graph.get(edge[0]).add(edge[1]);
        }
        
        List<List<Integer>> ancestors = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            Set<Integer> visited = new HashSet<>();
            List<Integer> currentAncestors = new ArrayList<>();
            dfs(i, graph, visited, currentAncestors);
            ancestors.add(currentAncestors);
        }
        
        return ancestors;
    }
    
    private void dfs(int node, List<List<Integer>> graph, Set<Integer> visited, List<Integer> currentAncestors) {
        if (visited.contains(node)) {
            return;
        }
        
        visited.add(node);
        for (int parent : graph.get(node)) {
            currentAncestors.add(parent);
            dfs(parent, graph, visited, currentAncestors);
        }
    }
}

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