Cycle Detection in Undirected Graph in DSA Typescript - Build from Scratch

Cycle Detection in Undirected Graph

📖 Scenario: Imagine you are working on a network of friends where each friend is connected to others. You want to check if there is any circle of friends where you can start from one friend and come back to the same friend by following connections without repeating edges.

🎯 Goal: You will build a program to detect if there is a cycle in an undirected graph using adjacency lists and depth-first search.

📋 What You'll Learn

Create an adjacency list to represent the graph

Use a helper array to track visited nodes

Implement a depth-first search function to detect cycles

Print true if a cycle exists, otherwise false

💡 Why This Matters

🌍 Real World

Cycle detection helps find loops in networks like social connections, computer networks, or road maps to avoid repeated paths or errors.

💼 Career

Understanding cycle detection is important for software engineers working on graph algorithms, network analysis, and debugging complex data structures.

1

Create the graph adjacency list

Create a variable called graph as an object representing the adjacency list with these exact entries: 0: [1, 2], 1: [0, 2], 2: [0, 1, 3], 3: [2]

DSA Typescript

# Your code here

Hint

Use an object where keys are node numbers and values are arrays of connected nodes.

2

Create a visited array

Create a variable called visited as an array of booleans with length 4, initialized with false values

DSA Typescript

const graph: { [key: number]: number[] } = {
  0: [1, 2],
  1: [0, 2],
  2: [0, 1, 3],
  3: [2]
};
# Your code here

Hint

Use an array with four false values to mark nodes as not visited yet.

3

Implement DFS function to detect cycle

Write a function called dfs that takes node and parent as parameters. Inside, mark visited[node] as true. Then loop over neighbors in graph[node]. If a neighbor is not visited, recursively call dfs with neighbor and current node. If a neighbor is visited and not equal to parent, return true. Return false if no cycle found.

DSA Typescript

const graph: { [key: number]: number[] } = {
  0: [1, 2],
  1: [0, 2],
  2: [0, 1, 3],
  3: [2]
};
const visited: boolean[] = [false, false, false, false];
// Write dfs function here
# Your code here

Hint

Use recursion to visit neighbors and check if a visited neighbor is not the parent to detect a cycle.

4

Check for cycle and print result

Write code to loop over nodes 0 to 3. If visited[node] is false, call dfs(node, -1). If dfs returns true, print true and stop. If no cycle found after loop, print false.

DSA Typescript

const graph: { [key: number]: number[] } = {
  0: [1, 2],
  1: [0, 2],
  2: [0, 1, 3],
  3: [2]
};
const visited: boolean[] = [false, false, false, 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;
}

// Check for cycle and print result
# Your code here

Hint

Loop through all nodes and call dfs if not visited. Print true if cycle found, else false.

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.