Cycle Detection in Undirected Graph in DSA Javascript - Time & Space Complexity

Time Complexity: Cycle Detection in Undirected Graph

O(n + m)

Understanding Time Complexity

We want to know how the time needed to find a cycle in an undirected graph changes as the graph grows.

How does the number of nodes and edges affect the work done?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function hasCycle(graph) {
  const visited = new Set();

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

  for (const node in graph) {
    if (!visited.has(node)) {
      if (dfs(node, null)) return true;
    }
  }
  return false;
}
    

This code checks if an undirected graph has a cycle by visiting nodes using depth-first search.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The depth-first search (DFS) visits each node and its neighbors.
• How many times: Each node and edge is visited once during the DFS traversal.

How Execution Grows With Input

As the graph grows with more nodes and edges, the DFS explores more connections.

Input Size (n nodes, m edges)	Approx. Operations
10 nodes, 15 edges	About 40 visits (nodes + edges)
100 nodes, 200 edges	About 500 visits
1000 nodes, 3000 edges	About 4000 visits

Pattern observation: The work grows roughly with the sum of nodes and edges, increasing linearly.

Final Time Complexity

Time Complexity: O(n + m)

This means the time grows linearly with the number of nodes plus edges in the graph.

Common Mistake

[X] Wrong: "The DFS visits every edge multiple times, so time is much higher than linear."

[OK] Correct: Each edge is checked only twice (once from each node), so total work stays proportional to edges plus nodes.

Interview Connect

Understanding this time complexity helps you explain how graph traversal scales, a key skill for many coding challenges.

Self-Check

"What if the graph was directed instead of undirected? How would the time complexity change?"

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