Cycle Detection in Directed Graph in DSA Javascript

Mental Model

We want to find if there is a path that starts and ends at the same node following the direction of edges.

Analogy: Imagine a one-way street map where you want to know if you can start driving from a place and come back to it by following the one-way roads.

0 -> 1 -> 2
↑         ↓
└─────────

Dry Run Walkthrough

Input: Graph with edges: 0->1, 1->2, 2->0 (cycle exists)

Goal: Detect if the graph contains a cycle following edge directions

Step 1: Start DFS from node 0, mark 0 as visiting

visiting: {0}, visited: {}

Why: We begin exploring from node 0 to find cycles

Step 2: From 0, go to neighbor 1, mark 1 as visiting

visiting: {0,1}, visited: {}

Why: We follow edges to explore deeper nodes

Step 3: From 1, go to neighbor 2, mark 2 as visiting

visiting: {0,1,2}, visited: {}

Why: Continue exploring neighbors to detect cycle

Step 4: From 2, neighbor is 0 which is already visiting -> cycle found

visiting: {0,1,2}, visited: {}

Why: Finding a node already in visiting means a cycle

Result:

Cycle detected because node 0 is revisited while still visiting

Annotated Code

DSA Javascript

class Graph {
  constructor(vertices) {
    this.V = vertices;
    this.adj = new Array(vertices).fill(0).map(() => []);
  }

  addEdge(u, v) {
    this.adj[u].push(v);
  }

  isCyclicUtil(v, visited, recStack) {
    if (!visited[v]) {
      visited[v] = true;
      recStack[v] = true;

      for (const neighbor of this.adj[v]) {
        if (!visited[neighbor] && this.isCyclicUtil(neighbor, visited, recStack)) {
          return true;
        } else if (recStack[neighbor]) {
          return true;
        }
      }
    }
    recStack[v] = false;
    return false;
  }

  isCyclic() {
    const visited = new Array(this.V).fill(false);
    const recStack = new Array(this.V).fill(false);

    for (let i = 0; i < this.V; i++) {
      if (this.isCyclicUtil(i, visited, recStack)) {
        return true;
      }
    }
    return false;
  }
}

// Driver code
const g = new Graph(3);
g.addEdge(0, 1);
g.addEdge(1, 2);
g.addEdge(2, 0);

console.log(g.isCyclic() ? "Cycle detected" : "No cycle detected");

visited[v] = true; recStack[v] = true;

Mark current node as visited and add to recursion stack to track path

for (const neighbor of this.adj[v]) {

Explore all neighbors of current node

if (!visited[neighbor] && this.isCyclicUtil(neighbor, visited, recStack)) {

If neighbor not visited, recursively check for cycle

else if (recStack[neighbor]) {

If neighbor is in recursion stack, cycle found

recStack[v] = false;

Remove current node from recursion stack after exploring

Output

Cycle detected

Complexity Analysis

Time: O(V + E) because we visit each vertex and edge once during DFS

Space: O(V) for recursion stack and visited arrays

vs Alternative: Compared to checking all paths naively, DFS with recursion stack is efficient and avoids repeated work

Edge Cases

Empty graph with no vertices

No cycle detected as there are no nodes

DSA Javascript

for (let i = 0; i < this.V; i++) {

Graph with single node and no edges

No cycle detected because no edges to form cycle

DSA Javascript

for (const neighbor of this.adj[v]) {

Graph with multiple disconnected components

Cycle detection runs on each component independently

DSA Javascript

for (let i = 0; i < this.V; i++) {

When to Use This Pattern

When you need to find if a directed graph has a cycle, use DFS with a recursion stack to track nodes in the current path.

Common Mistakes

Mistake: Not using a recursion stack to track nodes in the current path, causing false negatives.
Fix: Use a separate recursion stack array to mark nodes currently being explored.

Summary

Detects if a directed graph contains a cycle by tracking nodes in the current DFS path.

Use when you need to know if following directed edges can lead back to the start node.

The key insight is that revisiting a node in the current path means a cycle exists.

Practice

1. What is the main purpose of cycle detection in a directed graph?

easy

A. To find if there is a path that starts and ends at the same node following the direction of edges

B. To count the total number of edges in the graph

C. To find the shortest path between two nodes

D. To check if the graph is connected

Solution

1. Step 1: Understand cycle in directed graph

   A cycle means a path that starts and ends at the same node following edge directions.

2. Step 2: Purpose of cycle detection

   Cycle detection checks if such a path exists in the graph.

3. Final Answer:

   To find if there is a path that starts and ends at the same node following the direction of edges -> Option A

4. Quick Check:

   Cycle detection = find directed loops [OK]

Hint: Cycle means a directed loop returning to start [OK]

Common Mistakes:

• Confusing cycle detection with counting edges
• Thinking cycle detection finds shortest paths
• Assuming cycle detection checks connectivity

2. Which of the following JavaScript code snippets correctly initializes a graph as an adjacency list for cycle detection?

easy

A. const graph = {0: [1, 2], 1: [2], 2: [0]};

B. const graph = [[0,1], [1,2], [2,0]];

C. const graph = new Map([0, [1, 2], 1, [2], 2, [0]]);

D. const graph = {0 => [1, 2], 1 => [2], 2 => [0]};

Solution

1. Step 1: Check adjacency list syntax in JavaScript

   An adjacency list is an object with keys as nodes and values as arrays of neighbors.

2. Step 2: Validate each option

   const graph = {0: [1, 2], 1: [2], 2: [0]}; uses correct object syntax with arrays as values. const graph = [[0,1], [1,2], [2,0]]; is an edge list, not adjacency list. const graph = new Map([0, [1, 2], 1, [2], 2, [0]]); uses Map incorrectly (missing array of pairs). const graph = {0 => [1, 2], 1 => [2], 2 => [0]}; uses invalid arrow syntax in object keys.

3. Final Answer:

   const graph = {0: [1, 2], 1: [2], 2: [0]}; -> Option A

4. Quick Check:

   Object with arrays = adjacency list [OK]

Hint: Adjacency list uses object keys with array values [OK]

Common Mistakes:

• Using edge list instead of adjacency list
• Incorrect Map initialization syntax
• Using invalid object key syntax

3. Consider the following JavaScript code for cycle detection in a directed graph using DFS:

const graph = {0: [1], 1: [2], 2: [0]};
const visited = new Set();
const recStack = new Set();

function dfs(node) {
  if (recStack.has(node)) return true;
  if (visited.has(node)) return false;
  visited.add(node);
  recStack.add(node);
  for (const neighbor of graph[node] || []) {
    if (dfs(neighbor)) return true;
  }
  recStack.delete(node);
  return false;
}

console.log(dfs(0));

What will be printed when this code runs?

medium

A. false

B. true

C. undefined

D. TypeError

Solution

1. Step 1: Trace DFS calls starting from node 0

   Start at 0: visited={0}, recStack={0}. Visit neighbor 1: visited={0,1}, recStack={0,1}. Visit neighbor 2: visited={0,1,2}, recStack={0,1,2}. Visit neighbor 0 again: recStack has 0, so return true (cycle found).

2. Step 2: Return true up the call stack

   Since cycle detected, dfs returns true and console.log prints true.

3. Final Answer:

   true -> Option B

4. Quick Check:

   Cycle detected = true [OK]

Hint: Cycle if node revisited in recursion stack [OK]

Common Mistakes:

• Ignoring recursion stack and returning false
• Confusing visited set with recursion stack
• Expecting false because all nodes visited

4. The following code attempts to detect a cycle in a directed graph but does not work correctly. What is the main error?

const graph = {0: [1], 1: [2], 2: [0]};
const visited = new Set();

function dfs(node) {
  if (visited.has(node)) return false;
  visited.add(node);
  for (const neighbor of graph[node] || []) {
    if (dfs(neighbor)) return true;
  }
  return false;
}

console.log(dfs(0));

medium

A. Visited set is not cleared after recursion

B. Function does not return any value

C. Graph adjacency list is incorrectly defined

D. Missing recursion stack to detect back edges

Solution

1. Step 1: Analyze cycle detection logic

   The code uses only a visited set but no recursion stack to track nodes in current path.

2. Step 2: Understand why recursion stack is needed

   Without recursion stack, the code cannot detect back edges that form cycles, so it misses cycles.

3. Final Answer:

   Missing recursion stack to detect back edges -> Option D

4. Quick Check:

   Recursion stack needed for cycle detection [OK]

Hint: Use recursion stack to detect cycles in DFS [OK]

Common Mistakes:

• Assuming visited set alone detects cycles
• Not tracking nodes in current recursion path
• Confusing visited set clearing with cycle detection

5. You have a directed graph represented as an adjacency list:

const graph = {
  0: [1],
  1: [2, 3],
  2: [4],
  3: [4],
  4: [1]
};

Which of the following modifications will break the cycle detection algorithm that uses DFS with a recursion stack?

hard

A. Starting DFS from node 0 instead of node 4

B. Adding a visited set to track visited nodes

C. Removing the recursion stack and using only a visited set

D. Using an iterative BFS instead of DFS

Solution

1. Step 1: Understand the role of recursion stack in cycle detection

   The recursion stack tracks nodes in the current path to detect cycles (back edges).

2. Step 2: Effect of removing recursion stack

   Without recursion stack, the algorithm cannot detect cycles correctly, even if visited set is used.

3. Step 3: Other options explained

   Adding visited set helps. Starting DFS from different nodes does not break detection if all nodes are visited. Using BFS does not detect cycles in directed graphs reliably.

4. Final Answer:

   Removing the recursion stack and using only a visited set -> Option C

5. Quick Check:

   Recursion stack needed for cycle detection [OK]

Hint: Recursion stack is essential for detecting cycles in DFS [OK]

Common Mistakes:

• Thinking visited set alone detects cycles
• Believing BFS can replace DFS for cycle detection
• Assuming start node affects cycle detection correctness