Cycle Detection in Undirected Graph in DSA Typescript - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of DFS-based cycle detection in undirected graph

What is the output of the following TypeScript code that detects a cycle in an undirected graph using DFS?

DSA Typescript

class Graph {
  adjacencyList: Map<number, number[]> = new Map();

  addEdge(u: number, v: number) {
    if (!this.adjacencyList.has(u)) this.adjacencyList.set(u, []);
    if (!this.adjacencyList.has(v)) this.adjacencyList.set(v, []);
    this.adjacencyList.get(u)!.push(v);
    this.adjacencyList.get(v)!.push(u);
  }

  hasCycle(): boolean {
    const visited = new Set<number>();

    const dfs = (node: number, parent: number): boolean => {
      visited.add(node);
      for (const neighbor of this.adjacencyList.get(node) || []) {
        if (!visited.has(neighbor)) {
          if (dfs(neighbor, node)) return true;
        } else if (neighbor !== parent) {
          return true;
        }
      }
      return false;
    };

    for (const node of this.adjacencyList.keys()) {
      if (!visited.has(node)) {
        if (dfs(node, -1)) return true;
      }
    }
    return false;
  }
}

const g = new Graph();
g.addEdge(1, 2);
g.addEdge(2, 3);
g.addEdge(3, 4);
g.addEdge(4, 2);
console.log(g.hasCycle());

A true

B false

C undefined

D SyntaxError

💡 Hint

Think about whether the graph has a loop formed by edges.

The graph edges form a cycle: 2 → 3 → 4 → 2. The DFS detects this cycle and returns true.

🧠 Conceptual

intermediate

Cycle detection condition in undirected graph DFS

In DFS cycle detection for an undirected graph, which condition correctly identifies a cycle?

A If a neighbor is visited and is the parent of the current node, a cycle exists.

B If a visited neighbor is not the parent of the current node, a cycle exists.

C If a neighbor is unvisited, it means a cycle exists.

D If the current node has no neighbors, a cycle exists.

💡 Hint

Think about how back edges indicate cycles in undirected graphs.

In undirected graphs, a cycle is detected if during DFS we find a neighbor that is already visited and is not the parent node. This means we found a back edge.

🔧 Debug

advanced

Identify the error in cycle detection code

What error will this TypeScript code produce when detecting cycles in an undirected graph?

DSA Typescript

class Graph {
  adjacencyList: Map<number, number[]> = new Map();

  addEdge(u: number, v: number) {
    if (!this.adjacencyList.has(u)) this.adjacencyList.set(u, []);
    if (!this.adjacencyList.has(v)) this.adjacencyList.set(v, []);
    this.adjacencyList.get(u)!.push(v);
    this.adjacencyList.get(v)!.push(u);
  }

  hasCycle(): boolean {
    const visited = new Set<number>();

    const dfs = (node: number): boolean => {
      visited.add(node);
      for (const neighbor of this.adjacencyList.get(node) || []) {
        if (!visited.has(neighbor)) {
          if (dfs(neighbor)) return true;
        } else {
          return true;
        }
      }
      return false;
    };

    for (const node of this.adjacencyList.keys()) {
      if (!visited.has(node)) {
        if (dfs(node)) return true;
      }
    }
    return false;
  }
}

A The code will incorrectly detect cycles due to missing parent tracking.

B The code will cause a syntax error due to missing return statements.

C The code will never detect any cycle because it returns false always.

D The code will cause a runtime error due to undefined adjacency list.

💡 Hint

Check how the parent node is handled in DFS.

The code does not track the parent node, so it treats the immediate parent as a cycle, causing false positives.

❓ Predict Output

advanced

Output of BFS-based cycle detection in undirected graph

What is the output of this TypeScript code that detects a cycle in an undirected graph using BFS?

DSA Typescript

class Graph {
  adjacencyList: Map<number, number[]> = new Map();

  addEdge(u: number, v: number) {
    if (!this.adjacencyList.has(u)) this.adjacencyList.set(u, []);
    if (!this.adjacencyList.has(v)) this.adjacencyList.set(v, []);
    this.adjacencyList.get(u)!.push(v);
    this.adjacencyList.get(v)!.push(u);
  }

  hasCycle(): boolean {
    const visited = new Set<number>();

    for (const start of this.adjacencyList.keys()) {
      if (!visited.has(start)) {
        const queue: Array<[number, number]> = [[start, -1]];
        visited.add(start);

        while (queue.length > 0) {
          const [node, parent] = queue.shift()!;
          for (const neighbor of this.adjacencyList.get(node) || []) {
            if (!visited.has(neighbor)) {
              visited.add(neighbor);
              queue.push([neighbor, node]);
            } else if (neighbor !== parent) {
              return true;
            }
          }
        }
      }
    }
    return false;
  }
}

const g = new Graph();
g.addEdge(1, 2);
g.addEdge(2, 3);
g.addEdge(3, 4);
g.addEdge(4, 5);
console.log(g.hasCycle());

A true

B TypeError

C false

D undefined

💡 Hint

Check if the graph edges form any cycle.

The graph is a simple chain 1-2-3-4-5 with no cycles, so BFS cycle detection returns false.

🧠 Conceptual

expert

Why is parent tracking necessary in undirected graph cycle detection?

Why must we track the parent node when detecting cycles in an undirected graph using DFS or BFS?

A To detect cycles only in directed graphs.

B To improve the runtime complexity from exponential to linear.

C To ensure all nodes are visited exactly once.

D To avoid falsely detecting the immediate parent as a cycle neighbor.

💡 Hint

Think about the bidirectional edges in undirected graphs.

In undirected graphs, edges go both ways. Without parent tracking, the algorithm treats the edge back to the parent as a cycle, causing false positives.

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.