What is the time complexity of Count Visited Nodes in a Directed Graph?

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

What data structures are used to solve Count Visited Nodes in a Directed Graph?

This problem uses Dynamic Programming, Graph, Memoization. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Count Visited Nodes in a Directed Graph?

Insight 1: The graph structure guarantees that starting from any node, you will eventually enter a cycle. Insight 2: The problem can be solved by detecting cycles and tail lengths associated with each starting node.

Which companies ask LeetCode 2876. Count Visited Nodes in a Directed Graph in interviews?

This problem is frequently asked at BNY Mellon.

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Count Visited Nodes in a Directed Graph - LeetCode 2876 Solution

Count Visited Nodes in a Directed Graph - Complete Solution Guide

Count Visited Nodes in a Directed Graph is LeetCode problem 2876, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

There is a directed graph consisting of n nodes numbered from 0 to n - 1 and n directed edges. You are given a 0-indexed array edges where edges[i] indicates that there is an edge from node i to node edges[i] . Consider the following process on the graph: You start from a node x and keep visiting other nodes through edges until you reach a node that you have already visited before on this same process. Return an array answer where answer[i] is the number of different nodes that you will visit if

Detailed Explanation

The problem describes a directed graph where each node has exactly one outgoing edge. Given an array `edges` representing these edges, we need to find, for each starting node `i`, the number of distinct nodes visited before encountering a previously visited node *in the same traversal*. The result is an array `answer` where `answer[i]` is the number of visited nodes starting from node `i`.

Solution Approach

The solution iterates through each node of the graph. For each node, if the answer is already calculated, the solution continues to the next node. Otherwise, it starts traversing the graph from the current node, keeping track of the visited nodes in the current path and their distances from the starting node. If a cycle is detected or a previously computed node is encountered, the solution uses this information to compute the result for all nodes on the current path, and stores the result in the `answer` array.

Step-by-Step Algorithm

Step 1: Initialize an `answer` array of size `n` with all values set to 0. This array will store the number of visited nodes for each starting node.
Step 2: Iterate through each node `i` from 0 to `n-1`. If `answer[i]` is not 0, it means the result for node `i` is already computed, so continue to the next node.
Step 3: For each unvisited node `i`, start traversing the graph. Use a `path` list to store the nodes visited in the current traversal and a `dist` dictionary/map to store the distance of each node from the starting node `i`.
Step 4: Continue traversing until a cycle is detected (a node is revisited in the current path) or a node with a non-zero value in the `answer` array is encountered.
Step 5: If a node `curr` with a non-zero value in the `answer` array is encountered, it means a previously computed path has been reached. The number of visited nodes for each node in the current path is then calculated by adding the remaining path length from that node to the value of `answer[curr]` (the already computed length).
Step 6: If a cycle is detected, identify the starting node of the cycle `cycle_start_index`. The length of the cycle `cycle_len` is the difference between the current path length and `cycle_start_index`.
Step 7: Set `answer[node]` to `cycle_len` for each node in the cycle.
Step 8: For nodes in the tail (nodes before the start of the cycle), set `answer[node]` to `(cycle_start_index - j) + cycle_len`, where `j` is the index of the node in the `path` list.

Key Insights

Insight 1: The graph structure guarantees that starting from any node, you will eventually enter a cycle.
Insight 2: The problem can be solved by detecting cycles and tail lengths associated with each starting node.
Insight 3: Memoization (dynamic programming) is crucial to avoid redundant calculations when multiple starting nodes lead to the same cycle.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Dynamic Programming, Graph, Memoization.

Companies

Asked at: BNY Mellon.

Count Visited Nodes in a Directed Graph | LeetCode 2876

Hard

Dynamic Programming Graph Memoization

Time: O(n)Space: O(n)

Solution Code

Available in:

import java.util.List;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.Map;

class Solution {
    public int[] countVisitedNodes(List<Integer> edges) {
        int n = edges.size();
        int[] answer = new int[n];

        for (int i = 0; i < n; i++) {
            if (answer[i] != 0) {
                continue;
            }

            List<Integer> path = new ArrayList<>();
            Map<Integer, Integer> dist = new HashMap<>();
            int curr = i;

            while (!dist.containsKey(curr) && answer[curr] == 0) {
                dist.put(curr, path.size());
                path.add(curr);
                curr = edges.get(curr);
            }

            if (answer[curr] != 0) {
                int pathLen = path.size();
                for (int j = 0; j < pathLen; j++) {
                    int node = path.get(j);
                    answer[node] = (pathLen - j) + answer[curr];
                }
            } else { // curr is in dist
                int cycleStartIndex = dist.get(curr);
                int cycleLen = path.size() - cycleStartIndex;

                // Fill answers for nodes in the cycle
                for (int j = cycleStartIndex; j < path.size(); j++) {
                    int node = path.get(j);
                    answer[node] = cycleLen;
                }

                // Fill answers for nodes in the tail
                for (int j = 0; j < cycleStartIndex; j++) {
                    int node = path.get(j);
                    answer[node] = (cycleStartIndex - j) + cycleLen;
                }
            }
        }

        return answer;
    }
}

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Problem Info

Description

Problem 2876: Count Visited Nodes in a Directed Graph - Detailed explanation and solution approaches.

Topics

Dynamic Programming Graph Memoization

Companies

BNY Mellon

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:2876

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

8 steps

1
Step 1: Initialize an `answer` array of size `n` with all values set to 0. This array will store the number of visited nodes for each starting node.
2
Step 2: Iterate through each node `i` from 0 to `n-1`. If `answer[i]` is not 0, it means the result for node `i` is already computed, so continue to the next node.
3
Step 3: For each unvisited node `i`, start traversing the graph. Use a `path` list to store the nodes visited in the current traversal and a `dist` dictionary/map to store the distance of each node from the starting node `i`.
4
Step 4: Continue traversing until a cycle is detected (a node is revisited in the current path) or a node with a non-zero value in the `answer` array is encountered.
5
Step 5: If a node `curr` with a non-zero value in the `answer` array is encountered, it means a previously computed path has been reached. The number of visited nodes for each node in the current path is then calculated by adding the remaining path length from that node to the value of `answer[curr]` (the already computed length).
6
Step 6: If a cycle is detected, identify the starting node of the cycle `cycle_start_index`. The length of the cycle `cycle_len` is the difference between the current path length and `cycle_start_index`.
7
Step 7: Set `answer[node]` to `cycle_len` for each node in the cycle.
8
Step 8: For nodes in the tail (nodes before the start of the cycle), set `answer[node]` to `(cycle_start_index - j) + cycle_len`, where `j` is the index of the node in the `path` 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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The algorithm avoids redundant calculations by storing the results for each node in the `answer` array. The use of `path` and `dist` allow for efficient cycle detection and length determination within each traversal. Because each node's answer is only computed once directly through traversal or derivation from known neighbors, the algorithm has a time complexity of O(n). Similarly, the space complexity derives from the potential of path, dist and answer all using space proportional to n.

📖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) with cycle detection using three states

Another approach could use DFS with color coding (white, gray, black) to detect cycles. White means unvisited, gray means currently in the recursion stack (i.e., part of the current path), and black means completely visited. On finding a gray node, a cycle is detected.

Time: O(n)Space: O(n)

Trade-offs: This approach is conceptually similar but might be slightly more complex to implement than the iterative approach. The iterative approach explicitly maintains the path and distance, making it easier to calculate the tail lengths.

✨

Key Insights

1Insight 1: The graph structure guarantees that starting from any node, you will eventually enter a cycle.
2Insight 2: The problem can be solved by detecting cycles and tail lengths associated with each starting node.
3Insight 3: Memoization (dynamic programming) is crucial to avoid redundant calculations when multiple starting nodes lead to the same cycle.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Graph

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

Edge Cases

Edge case 1: Acyclic graph (though not possible given problem constraints). The given code correctly handles the creation of a 'cycle' that is the length of entire path when there is no real cycle and another already processed node is found. This is implicitly handled by the initial `if answer[curr]:` check.
Edge case 2: A single cycle including all nodes. The algorithm correctly identifies the cycle and assigns the cycle length to each node.
Edge case 3: Very long tail with a short cycle. The algorithm efficiently calculates the path from the tail and cycle and applies proper calculations based on node locations.
Edge case 4: Self loops are disallowed, so we don't have to handle cases where edges[i] == i.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the cycle detection properly, leading to infinite loops.

✗Common mistake 2: Not memoizing the results, leading to redundant calculations and exceeding the time limit.

✗Common mistake 3: Incorrect calculation of the number of visited nodes in the tail of the path.

✅

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: Detecting loops in data processing pipelines. For example, identifying infinite loops in a system where data flows through different processing stages.

🏦Real-world scenario 2: Analyzing network traffic patterns. This algorithm can be used to find cycles in network traffic flow to detect potential network congestion or malicious activities.

💡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

Start with recursion, then memoize. The hardest part is identifying the state. Ask yourself: "What information do I need to make the current decision?"

🎯 Interview Tip

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ Watch Out

Forgetting base cases or initializing the DP array incorrectly. Start by solving the smallest subproblems.

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: November 24, 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

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 13

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

@InterviewProJul 6

❓

Frequently Asked Questions

How do I identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

How do I represent a graph in code?

Use adjacency list (HashMap or array of lists) for sparse graphs - it's space-efficient O(V+E). Use adjacency matrix for dense graphs or when you need O(1) edge lookup. For weighted graphs, store (neighbor, weight) pairs in adjacency lists.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

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 "Count Visited Nodes in a Directed Graph"?

For this hard-level Dynamic Programming problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether graph techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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