Cycle Detection in Undirected Graph in DSA Typescript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

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

DSA Typescript

const graph: Map<number, number[]> = new Map();
for (let i = 0; i < n; i++) {
  graph.set(i, [1]);
}

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

A[]

B0

Cnull

D{}

Attempts:

3 left

💡 Hint

Common Mistakes

Using an object {} instead of an array for neighbors.

Initializing with null or 0 which are not iterable.

We use an empty array [] to store neighbors for each node.

2fill in blank

medium

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

DSA Typescript

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

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

Agraph

Bv

Cu

Dnull

Attempts:

3 left

💡 Hint

Common Mistakes

Adding u to u's own list.

Adding graph or null instead of the node number.

We add v to u's neighbors to represent the edge.

3fill in blank

hard

Fix the error in the DFS function to avoid revisiting the parent node.

DSA Typescript

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

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

Anode

Bvisited

Cneighbor

Dparent

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing neighbor with node instead of parent.

Not checking for parent at all.

We check if the neighbor is not the parent to detect a cycle properly.

4fill in blank

hard

Fill both blanks to complete the cycle detection loop over all nodes.

DSA Typescript

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

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 which is a valid node and causes wrong cycle detection.

Using null which is not a number.

Start DFS from node i with parent set to -1 (no parent).

5fill in blank

hard

Fill all three blanks to create a function that detects cycles in an undirected graph.

DSA Typescript

function hasCycle(n: number, edges: number[][]): boolean {
  const graph: Map<number, number[]> = new Map();
  const visited: boolean[] = new Array(n).fill(false);

  for (let i = 0; i < n; i++) {
    graph.set(i, [1]);
  }

  for (const [u, v] of edges) {
    graph.get(u)!.push([2]);
    graph.get(v)!.push(u);
  }

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

  for (let i = 0; i < n; i++) {
    if (!visited[i] && dfs(i, [3])) {
      return true;
    }
  }
  return false;
}

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

A[]

Bv

C-1

Du

Attempts:

3 left

💡 Hint

Common Mistakes

Using objects instead of arrays for adjacency lists.

Adding wrong nodes to adjacency lists.

Starting DFS with parent 0 or null instead of -1.

Initialize adjacency lists with empty arrays, add v to u's neighbors, and start DFS with parent -1.

Practice

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

easy

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

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

C. If all nodes are visited exactly once, there is a cycle.

D. If the graph is disconnected, it contains a cycle.

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 back edge exists, indicating a cycle.

3. Final Answer:

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

4. Quick Check:

   Cycle detection = visit visited node not parent [OK]

Hint: Check visited nodes excluding parent during DFS for cycle [OK]

Common Mistakes:

• Ignoring the parent node check causes false positives.
• Assuming more edges than nodes always means cycle.
• Confusing disconnected graph with cycle presence.

2. Which of the following TypeScript function signatures correctly represents a DFS function for cycle detection in an undirected graph?

easy

A. function dfs(node: number, parent: number, visited: boolean[], graph: number[][]): boolean

B. function dfs(node: string, graph: number[][]): void

C. function dfs(visited: boolean[], graph: number[][]): number

D. function dfs(node: number, graph: number[][]): boolean

Solution

1. Step 1: Identify required parameters for DFS cycle detection

   We need current node, parent node, visited array, and graph adjacency list.

2. Step 2: Match function signature with parameters and return type

   function dfs(node: number, parent: number, visited: boolean[], graph: number[][]): boolean has all parameters and returns boolean indicating cycle presence.

3. Final Answer:

   function dfs(node: number, parent: number, visited: boolean[], graph: number[][]): boolean -> Option A

4. Quick Check:

   DFS needs node, parent, visited, graph, returns boolean [OK]

Hint: DFS cycle detection needs parent node parameter [OK]

Common Mistakes:

• Missing parent parameter causes incorrect cycle detection.
• Wrong return type (void instead of boolean).
• Using wrong node type (string instead of number).

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

const graph = [[1,2],[0,2],[0,1,3],[2]];
const visited = new Array(4).fill(false);
function dfs(node: number, parent: number): boolean {
  visited[node] = true;
  for (const neighbor of graph[node]) {
    if (!visited[neighbor]) {
      if (dfs(neighbor, node)) return true;
    } else if (neighbor !== parent) {
      return true;
    }
  }
  return false;
}
console.log(dfs(0, -1));

medium

A. undefined

B. true

C. false

D. Runtime Error

Solution

1. Step 1: Analyze graph structure and DFS traversal

   Graph edges: 0-1, 0-2, 1-2, 2-3. Nodes 0,1,2 form a triangle (cycle).

2. Step 2: Trace DFS calls from node 0

   DFS visits 0, then 1, then 2. At 2, neighbor 0 is visited and not parent, cycle detected, returns true.

3. Final Answer:

   true -> Option B

4. Quick Check:

   Triangle cycle detected = true [OK]

Hint: Triangle in graph means cycle detected true [OK]

Common Mistakes:

• Ignoring parent check causes false cycle detection.
• Assuming no cycle in triangle graph.
• Confusing return values of DFS function.

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

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

medium

A. Visited array is not updated correctly.

B. Parent parameter is not used in the function.

C. Missing return statement for recursive dfs call result.

D. Graph adjacency list is not passed correctly.

Solution

1. Step 1: Check recursive DFS call handling

   The recursive call dfs(neighbor, node, visited, graph) is made but its boolean result is ignored.

2. Step 2: Identify missing return for cycle detection

   If recursive call returns true (cycle found), it should be returned immediately to propagate detection.

3. Final Answer:

   Missing return statement for recursive dfs call result. -> Option C

4. Quick Check:

   Return recursive DFS result to detect cycle [OK]

Hint: Always return recursive DFS result for cycle detection [OK]

Common Mistakes:

• Ignoring return value of recursive DFS calls.
• Not updating visited array before recursion.
• Misusing parent parameter in cycle check.

5. Given an undirected graph represented as an adjacency list, how can you modify the cycle detection algorithm to also return the nodes forming the cycle?

hard

A. Mark nodes visited twice to identify cycle nodes.

B. Use BFS instead of DFS to find cycles and record nodes.

C. Count edges and nodes; if edges > nodes, return all nodes as cycle.

D. Track the path stack during DFS and return the cycle path when detected.

Solution

1. Step 1: Understand cycle detection with path tracking

   Maintain a stack or list of nodes visited in current DFS path to track traversal.

2. Step 2: Return cycle nodes when back edge found

   When a visited node (not parent) is found, extract nodes from stack forming the cycle path.

3. Final Answer:

   Track the path stack during DFS and return the cycle path when detected. -> Option D

4. Quick Check:

   Use DFS path stack to get cycle nodes [OK]

Hint: Use DFS path stack to capture cycle nodes [OK]

Common Mistakes:

• Using BFS which is less natural for cycle path retrieval.
• Assuming edge count alone identifies cycle nodes.
• Marking nodes visited twice without path context.