What is the time complexity of All Ancestors of a Node in a Directed Acyclic Graph?

The optimal solution has a time complexity of O(n + e).

What is the space complexity of All Ancestors of a Node in a Directed Acyclic Graph?

The space complexity is O(n^2).

What data structures are used to solve All Ancestors of a Node in a Directed Acyclic Graph?

This problem uses Depth-First Search, Breadth-First Search, Graph, Topological Sort. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving All Ancestors of a Node in a Directed Acyclic Graph?

Insight 1: The graph is a DAG, meaning there are no cycles. This allows us to use topological sorting or similar graph traversal techniques without worrying about infinite loops. Insight 2: We can efficiently find ancestors using a variation of Breadth-First Search (BFS) or Depth-First Search (DFS) combined with tracking visited nodes (ancestors) for each node.

Which companies ask LeetCode 2192. All Ancestors of a Node in a Directed Acyclic Graph in interviews?

This problem is frequently asked at Palantir Technologies.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

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

All Ancestors of a Node in a Directed Acyclic Graph - Complete Solution Guide

All Ancestors of a Node in a Directed Acyclic Graph is LeetCode problem 2192, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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 edg

Detailed Explanation

The problem asks us to find all the ancestors of each node in a directed acyclic graph (DAG). The input is the number of nodes 'n' and a list of edges 'edges', where each edge represents a directed connection from one node to another. The output should be a list of lists, where the i-th list contains all the ancestors of node i, sorted in ascending order. An ancestor of a node 'v' is any node 'u' that can reach 'v' through a path of edges.

Solution Approach

The provided code implements a Breadth-First Search (BFS) approach with topological sorting. It first builds an adjacency list representation of the graph and calculates the in-degree of each node (number of incoming edges). Then, it uses a queue to process nodes with an in-degree of 0 (source nodes). As it processes each node, it updates the ancestor sets of its neighbors, propagating the ancestors of the current node to its neighbors. Finally, it converts the ancestor sets to sorted lists and returns the result.

Step-by-Step Algorithm

Step 1: Build the Adjacency List: Create an adjacency list `adj` to represent the graph, where `adj[u]` stores the list of nodes that `u` points to.
Step 2: Calculate In-Degrees: Calculate the in-degree of each node and store it in the `in_degree` array.
Step 3: Initialize Ancestor Sets: Create a list of sets `ancestors_sets`, where `ancestors_sets[i]` will store the set of ancestors of node `i`.
Step 4: Initialize Queue: Add all nodes with an in-degree of 0 to a queue `queue`. These are the source nodes.
Step 5: BFS Traversal: While the queue is not empty, dequeue a node `u`. For each neighbor `v` of `u`, add `u` to the ancestor set of `v`, and also add all the ancestors of `u` to the ancestor set of `v`. Decrement the in-degree of `v`. If the in-degree of `v` becomes 0, enqueue `v`.
Step 6: Convert Sets to Sorted Lists: Convert each set of ancestors to a sorted list and store it in the `result` list.
Step 7: Return Result: Return the `result` list.

Key Insights

Insight 1: The graph is a DAG, meaning there are no cycles. This allows us to use topological sorting or similar graph traversal techniques without worrying about infinite loops.
Insight 2: We can efficiently find ancestors using a variation of Breadth-First Search (BFS) or Depth-First Search (DFS) combined with tracking visited nodes (ancestors) for each node.
Insight 3: Using sets to store ancestors for each node helps avoid duplicate entries and automatically facilitates the 'sorted in ascending order' requirement when converting to a list.

Complexity Analysis

Time Complexity: O(n + e)

Space Complexity: O(n^2)

Topics

This problem involves: Depth-First Search, Breadth-First Search, Graph, Topological Sort.

Companies

Asked at: Palantir Technologies.

All Ancestors of a Node in a Directed Acyclic Graph | LeetCode 2192

Medium

Depth-First Search Breadth-First Search Graph Topological Sort

Time: O(n + e)Space: O(n^2)

Solution Code

Available in:

import java.util.*;

class Solution {
    public List<List<Integer>> getAncestors(int n, int[][] edges) {
        List<Integer>[] adj = new List[n];
        for (int i = 0; i < n; i++) {
            adj[i] = new ArrayList<>();
        }
        int[] in_degree = new int[n];
        for (int[] edge : edges) {
            int u = edge[0];
            int v = edge[1];
            adj[u].add(v);
            in_degree[v]++;
        }

        List<Set<Integer>> ancestors_sets = new List[n];
        for (int i = 0; i < n; i++) {
            ancestors_sets[i] = new HashSet<>();
        }

        Queue<Integer> queue = new LinkedList<>();
        for (int i = 0; i < n; i++) {
            if (in_degree[i] == 0) {
                queue.offer(i);
            }
        }

        while (!queue.isEmpty()) {
            int u = queue.poll();

            for (int v : adj[u]) {
                ancestors_sets[v].add(u);
                ancestors_sets[v].addAll(ancestors_sets[u]);

                in_degree[v]--;
                if (in_degree[v] == 0) {
                    queue.offer(v);
                }
            }
        }

        List<List<Integer>> result = new ArrayList<>();
        for (Set<Integer> s : ancestors_sets) {
            List<Integer> sortedList = new ArrayList<>(s);
            Collections.sort(sortedList);
            result.add(sortedList);
        }

        return result;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 2192: All Ancestors of a Node in a Directed Acyclic Graph - Detailed explanation and solution approaches.

Topics

Depth-First Search Breadth-First Search Graph Topological Sort

Companies

Palantir Technologies

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2192

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

7 steps

1
Step 1: Build the Adjacency List: Create an adjacency list `adj` to represent the graph, where `adj[u]` stores the list of nodes that `u` points to.
2
Step 2: Calculate In-Degrees: Calculate the in-degree of each node and store it in the `in_degree` array.
3
Step 3: Initialize Ancestor Sets: Create a list of sets `ancestors_sets`, where `ancestors_sets[i]` will store the set of ancestors of node `i`.
4
Step 4: Initialize Queue: Add all nodes with an in-degree of 0 to a queue `queue`. These are the source nodes.
5
Step 5: BFS Traversal: While the queue is not empty, dequeue a node `u`. For each neighbor `v` of `u`, add `u` to the ancestor set of `v`, and also add all the ancestors of `u` to the ancestor set of `v`. Decrement the in-degree of `v`. If the in-degree of `v` becomes 0, enqueue `v`.
6
Step 6: Convert Sets to Sorted Lists: Convert each set of ancestors to a sorted list and store it in the `result` list.
7
Step 7: Return Result: Return the `result` list.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n + e)

💾

Space Complexity

O(n^2)

📚 Quadratic space - Uses lots of memory!

📝

Why?

The time complexity is dominated by the BFS traversal and building of the adjacency list. The space complexity is determined by the `ancestors_sets`, where each node can potentially store n-1 ancestors.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Depth-First Search (DFS)

A DFS approach can also be used to solve this problem. For each node, perform a DFS to find all reachable nodes, which are its descendants. Then, reverse the perspective: for each node, find all nodes that can reach it via DFS, these are its ancestors.

Time: O(n * (n + e)) in the worst case. For each node, we are running a DFS with O(n+e).Space: O(n^2). Similar to the BFS approach, each node can potentially have n-1 ancestors, thus requiring O(n^2) space to store the ancestor sets.

Trade-offs: DFS might be more intuitive for some, but it can be less efficient in dense graphs due to potentially revisiting nodes multiple times. BFS is generally preferred for finding shortest paths, and here provides a good balance of performance.

Transitive Closure

Calculate the transitive closure of the graph. The transitive closure matrix indicates which nodes are reachable from other nodes. The ancestors of a node can then be determined by looking at the column of the transitive closure matrix corresponding to that node.

Time: O(n^3) using Floyd-Warshall Algorithm, or O(n*e) using repeated DFS/BFS.Space: O(n^2) to store the transitive closure matrix.

Trade-offs: Transitive closure provides a direct lookup, but building it requires significant time and space, making it less efficient than the BFS approach for this specific problem, unless you need to perform many ancestor queries on the same graph.

✨

Key Insights

1Insight 1: The graph is a DAG, meaning there are no cycles. This allows us to use topological sorting or similar graph traversal techniques without worrying about infinite loops.
2Insight 2: We can efficiently find ancestors using a variation of Breadth-First Search (BFS) or Depth-First Search (DFS) combined with tracking visited nodes (ancestors) for each node.
3Insight 3: Using sets to store ancestors for each node helps avoid duplicate entries and automatically facilitates the 'sorted in ascending order' requirement when converting to a list.

📚

Key Terms Explained

For Students

Graph

Nodes connected by edges. Like a social network or city map!

Edge Cases

Edge case 1: Graph with no edges. The algorithm correctly handles this case because the in-degree of all nodes will be 0, and they will all be added to the queue initially. The ancestor lists will remain empty, which is the correct output.
Edge case 2: Node with no ancestors. The algorithm correctly handles this case because such nodes will either have an in-degree of 0 and be processed first, or their in-degree will become 0 after processing their predecessors. Their ancestor sets will remain empty.
Edge case 3: A long chain of nodes where one node is an ancestor of many other nodes. The BFS correctly propagates the ancestors along the chain.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling duplicate ancestors. Using a list instead of a set to store ancestors can lead to duplicate entries in the output. Using sets is crucial for efficiently removing duplicates.

✗Common mistake 2: Incorrectly updating in-degrees during BFS. If in-degrees are not updated correctly, the BFS traversal might not process all nodes and lead to incomplete ancestor lists.

✗Common mistake 3: Forgetting to sort the ancestor list before returning it. The problem specifically requires the ancestors to be sorted in ascending order.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Dependency resolution in software projects. Identifying the dependencies (ancestors) of a particular module or component is crucial for building and deploying software systems. This algorithm can determine the complete dependency chain.

🏦Real-world scenario 2: Genealogy and ancestry tracking. Representing family relationships as a directed graph can allow this algorithm to identify all ancestors of a particular individual in the family tree.

🎮Real-world scenario 3: Task scheduling and workflow management. Determining the prerequisites (ancestors) of a task in a workflow is essential for proper scheduling and execution. This algorithm can identify all tasks that must be completed before a given task can start.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: July 3, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

❓

Frequently Asked Questions

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "All Ancestors of a Node in a Directed Acyclic Graph"?

For this medium-level Depth-First Search problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether breadth-first search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?