Cycle Detection in Undirected Graph in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the adjacency list for the graph.

DSA Javascript

const graph = Array.from({ length: n }, () => [1]);

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

Anull

B{}

C[]

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using an object {} instead of an array for adjacency list.

Initializing with null or 0 which cannot hold neighbors.

We use an array of empty arrays to represent adjacency lists for each node.

2fill in blank

medium

Complete the code to add an edge between two nodes u and v in the undirected graph.

DSA Javascript

graph[u].push([1]);
graph[v].push(u);

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Av

Bu

Cgraph

Dnull

Attempts:

3 left

💡 Hint

Common Mistakes

Adding the same node to its own adjacency list.

Adding u to u's adjacency list instead of v.

We add v to u's adjacency list and u to v's adjacency list to represent an undirected edge.

3fill in blank

hard

Fix the error in the DFS function to detect a cycle by correctly checking the parent node.

DSA Javascript

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

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

Aneighbor

Bnode

Cparent

Dvisited

Attempts:

3 left

💡 Hint

Common Mistakes

Passing parent instead of node causes wrong cycle detection.

Passing neighbor causes infinite recursion.

When calling dfs recursively, the current node becomes the parent for the neighbor.

4fill in blank

hard

Fill both blanks to complete the cycle detection function that iterates over all nodes.

DSA Javascript

for (let i = 0; i < n; i++) {
  if (!visited[i] && dfs([1], [2])) {
    return true;
  }
}
return false;

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

Ai

B0

C-1

Dnull

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 as parent causes false cycle detection.

Passing null instead of -1 breaks parent check.

Start DFS from node i with parent -1 (no parent) to detect cycles in all components.

5fill in blank

hard

Fill all three blanks to complete the full cycle detection code snippet.

DSA Javascript

const visited = new Array(n).fill(false);

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

for (let i = 0; i < n; i++) {
  if (!visited[i] && dfs([3], -1)) {
    console.log('Cycle detected');
    break;
  }
}

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

Anode

Bparent

Ci

Dvisited

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing up node and parent in DFS calls.

Starting DFS from wrong index.

The DFS call passes current node as parent, checks neighbor not equal to parent, and starts DFS from i.

Practice

1. What is the main idea behind detecting a cycle in an undirected graph using DFS?

easy

A. If we find a node with no neighbors, a cycle exists.

B. If we visit a node that is already visited and is not the parent, a cycle exists.

C. If the graph has more edges than nodes, it always has a cycle.

D. If all nodes are visited exactly once, a cycle exists.

Solution

1. Step 1: Understand DFS traversal in undirected graph

   DFS visits nodes and tracks the parent node to avoid false cycle detection.

2. Step 2: Detect cycle condition

   If during DFS, a visited node is found that is not the parent, it means a cycle exists.

3. Final Answer:

   If we visit a node that is already visited and is not the parent, a cycle exists. -> Option B

4. Quick Check:

   Cycle detection in undirected graph = check visited node not parent [OK]

Hint: Check visited neighbors excluding parent to find cycles [OK]

Common Mistakes:

• Ignoring the parent node in cycle check
• Assuming more edges than nodes always means cycle
• Confusing cycle detection logic with directed graphs

2. Which of the following is the correct way to represent an undirected graph adjacency list in JavaScript?

easy

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

B. const graph = [[0,1], [1,2], [2,3]];

C. const graph = {0: 1, 1: 2, 2: 3};

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

Solution

1. Step 1: Understand adjacency list format

   An adjacency list maps each node to an array of its neighbors.

2. Step 2: Check each option

   const graph = {0: [1, 2], 1: [0, 3], 2: [0], 3: [1]}; correctly maps nodes to arrays of neighbors. const graph = [[0,1], [1,2], [2,3]]; is edge list, not adjacency list. const graph = {0: 1, 1: 2, 2: 3}; maps nodes to single values, not arrays. const graph = [[1, 2], [0], [0], [1]]; is an array of arrays but connections are not symmetric (node 1 missing neighbor 3).

3. Final Answer:

   const graph = {0: [1, 2], 1: [0, 3], 2: [0], 3: [1]}; -> Option A

4. Quick Check:

   Adjacency list = node keys with neighbor arrays [OK]

Hint: Adjacency list uses node keys with arrays of neighbors [OK]

Common Mistakes:

• Using edge list instead of adjacency list
• Mapping nodes to single values, not arrays
• Using arrays without keys for adjacency list

3. What will be the output of the following code snippet detecting cycle in an undirected graph?

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

function dfs(node, parent, visited) {
  visited[node] = true;
  for (const neighbor of graph[node]) {
    if (!visited[neighbor]) {
      if (dfs(neighbor, node, visited)) return true;
    } else if (neighbor !== parent) {
      return true;
    }
  }
  return false;
}

const visited = {};
const hasCycle = dfs(0, -1, visited);
console.log(hasCycle);

medium

A. true

B. undefined

C. false

D. TypeError

Solution

1. Step 1: Trace DFS calls and visited nodes

   Start at node 0, mark visited. Visit neighbors 1 and 2. Both lead to nodes without cycles.

2. Step 2: Check for cycle condition

   No neighbor is visited again except parent, so no cycle detected. DFS returns false.

3. Final Answer:

   false -> Option C

4. Quick Check:

   No back edge found = false [OK]

Hint: Cycle if visited neighbor not parent found during DFS [OK]

Common Mistakes:

• Confusing parent node with cycle neighbor
• Assuming any visited node means cycle
• Not initializing visited properly

4. Find the bug in the following cycle detection code for an undirected graph:

function dfs(node, parent, visited) {
  visited[node] = true;
  for (const neighbor of graph[node]) {
    if (!visited[neighbor]) {
      if (dfs(neighbor, node, visited)) return true;
    } else if (neighbor === parent) {
      return true;
    }
  }
  return false;
}

medium

A. The function should return false inside the for loop.

B. The visited array is not initialized before DFS call.

C. The graph variable is not defined inside the function.

D. The condition 'neighbor === parent' should be 'neighbor !== parent' to detect cycles.

Solution

1. Step 1: Analyze cycle detection condition

   The code returns true if neighbor equals parent, which is incorrect because parent is allowed to be visited.

2. Step 2: Correct condition for cycle detection

   It should return true if neighbor is visited and neighbor is NOT the parent, indicating a cycle.

3. Final Answer:

   The condition 'neighbor === parent' should be 'neighbor !== parent' to detect cycles. -> Option D

4. Quick Check:

   Cycle if visited neighbor not parent [OK]

Hint: Cycle check: visited neighbor must not be parent [OK]

Common Mistakes:

• Returning true when neighbor is parent
• Not initializing visited before DFS
• Returning false too early inside loop

5. Given the undirected graph below, which edges form a cycle?

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

Select the correct set of edges that create a cycle.

hard

A. Edges (0-1), (1-2), (2-0)

B. Edges (1-3), (3-4), (4-2)

C. Edges (0-2), (2-4), (4-3)

D. Edges (0-1), (1-3), (3-0)

Solution

1. Step 1: Identify cycles by checking connected edges

   Edges (0-1), (1-2), and (2-0) connect nodes in a triangle, forming a cycle.

2. Step 2: Verify other edges

   Edges involving nodes 3 and 4 do not form cycles as they are connected linearly.

3. Final Answer:

   Edges (0-1), (1-2), (2-0) -> Option A

4. Quick Check:

   Triangle edges form cycle [OK]

Hint: Look for triangle or loop edges to find cycles [OK]

Common Mistakes:

• Assuming linear chains form cycles
• Ignoring the triangle formed by nodes 0,1,2
• Confusing edge directions in undirected graph