Cycle Detection in Directed Graph in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the visited array for all nodes.

DSA Javascript

const visited = new Array(n).fill([1]);

Drag options to blanks, or click blank then click option'

Afalse

Btrue

Cnull

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing visited with true causes all nodes to be considered visited from the start.

Using 0 or null can cause confusion in boolean checks.

We initialize the visited array with false because no nodes have been visited at the start.

2fill in blank

medium

Complete the code to mark the current node as visited in the DFS function.

DSA Javascript

visited[node] = [1];

Drag options to blanks, or click blank then click option'

Atrue

Bfalse

Cnull

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Setting visited[node] to false again causes infinite loops.

Using null or 0 does not correctly mark the node as visited.

Marking visited[node] as true means we have now visited this node during DFS.

3fill in blank

hard

Fix the error in the condition that detects a cycle when a node is in the recursion stack.

DSA Javascript

if (recStack[neighbor] === [1]) { return true; }

Drag options to blanks, or click blank then click option'

Afalse

Bnull

Cundefined

Dtrue

Attempts:

3 left

💡 Hint

Common Mistakes

Checking for false or null will not detect cycles correctly.

Using undefined causes runtime errors.

If the neighbor is already in the recursion stack (true), it means a cycle is detected.

4fill in blank

hard

Fill both blanks to correctly remove the node from recursion stack and mark it as visited after DFS.

DSA Javascript

recStack[node] = [1];
visited[node] = [2];

Drag options to blanks, or click blank then click option'

Afalse

Btrue

Cnull

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Setting recStack[node] to true causes false cycle detection.

Marking visited[node] as false causes repeated visits.

After DFS, we remove the node from recursion stack by setting recStack[node] to false and mark it visited by setting visited[node] to true.

5fill in blank

hard

Fill all three blanks to complete the DFS function header and recursive call.

DSA Javascript

function dfs([1], graph, visited, recStack) {
  recStack[node] = true;
  for (const [2] of graph[node]) {
    if (!visited[neighbor] && dfs([3], graph, visited, recStack)) {
      return true;
    }
  }
  recStack[node] = false;
  visited[node] = true;
  return false;
}

Drag options to blanks, or click blank then click option'

Anode

Bneighbor

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'node' instead of 'neighbor' in the loop or recursive call causes errors.

Mixing variable names leads to undefined variables.

The DFS function takes 'node' as parameter, loops over neighbors using 'neighbor', and recursively calls dfs with 'neighbor'.

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