Cycle Detection in Directed Graph in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare the adjacency list for a directed graph with n vertices.

DSA C++

vector<vector<int>> graph([1]);

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

Aedges

B0

Cn

Dvisited

Attempts:

3 left

💡 Hint

Common Mistakes

Using edges instead of n for adjacency list size.

Initializing with 0 which creates an empty list with no vertices.

The adjacency list is initialized with n empty vectors, one for each vertex.

2fill in blank

medium

Complete the code to mark the current node as visited in the DFS function.

DSA C++

visited[node] = [1];

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

Atrue

B1

Cfalse

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Marking visited[node] as false which means unvisited.

Using 0 or 1 instead of true or false in boolean context.

Marking visited[node] as true indicates the node is currently being explored.

3fill in blank

hard

Fix the error in the condition to detect a cycle when a node is in the recursion stack.

DSA C++

if (recStack[[1]]) return true;

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

Aparent

Bnode

Cvisited

Dgraph

Attempts:

3 left

💡 Hint

Common Mistakes

Checking parent instead of node in recursion stack.

Using visited or graph which are not boolean arrays for recursion stack.

We check if the current node is already in the recursion stack to detect a cycle.

4fill in blank

hard

Fill both blanks to correctly remove the node from recursion stack and mark it as not visited after DFS call.

DSA C++

recStack[[1]] = false;
visited[[2]] = false;

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

Anode

Bparent

Cvisited

Dgraph

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variables for recStack and visited updates.

Marking visited[node] as false which is incorrect for cycle detection.

After exploring all neighbors, remove the current node from recursion stack and mark it as not in recursion stack.

5fill in blank

hard

Fill all three blanks to complete the DFS function header and recursive call for cycle detection.

DSA C++

bool dfs(int [1], vector<vector<int>>& graph, vector<bool>& visited, vector<bool>& [2]) {
    visited[node] = true;
    recStack[node] = true;
    for (int [3] : graph[node]) {
        if (!visited[neighbor] && dfs(neighbor, graph, visited, recStack))
            return true;
        else if (recStack[neighbor])
            return true;
    }
    recStack[node] = false;
    return false;
}

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

Anode

BrecStack

Cneighbor

Dvisited

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong variable names causing compilation errors.

Mixing up visited and recStack variable names.

The DFS function uses 'node' as current node, 'recStack' for recursion stack, and 'neighbor' for adjacent nodes.

Practice

1. What does cycle detection in a directed graph help us find?

easy

A. Whether there is a path that starts and ends at the same node following edge directions

B. The shortest path between two nodes

C. The total number of nodes in the graph

D. The maximum degree of any node

Solution

1. Step 1: Understand cycle in directed graph

   A cycle means a path where you can start from a node and follow edges in their direction to come back to the same node.

2. Step 2: Identify what cycle detection finds

   Cycle detection checks if such a path exists anywhere in the graph.

3. Final Answer:

   Whether there is a path that starts and ends at the same node following edge directions -> Option A

4. Quick Check:

   Cycle detection = find directed loops [OK]

Hint: Cycle means path back to start following edge directions [OK]

Common Mistakes:

• Confusing cycle with shortest path
• Thinking cycle ignores edge directions
• Mixing cycle with node count

2. Which of the following is the correct way to declare a graph adjacency list in C++ for cycle detection?

easy

A. vector graph(100);

B. int graph[100][100];

C. map graph;

D. vector<int> graph[100];

Solution

1. Step 1: Understand adjacency list declaration

   For cycle detection, adjacency list is often a vector of vectors to store edges from each node.

2. Step 2: Check each option

   A: vector graph(100); - standard and flexible vector of vectors for adjacency list. B: adjacency matrix, not list. C: map, valid but less common and heavier for sequential nodes. D: array of vectors, works but fixed size and less modern than vector of vectors.

3. Final Answer:

   vector graph(100); -> Option A

4. Quick Check:

   Use vector of vectors for adjacency list [OK]

Hint: Use vector for flexible adjacency list [OK]

Common Mistakes:

• Using adjacency matrix instead of list
• Confusing map with vector
• Using raw arrays limiting flexibility

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

bool dfs(int node, vector& graph, vector<int>& visited, vector<int>& recStack) {
    visited[node] = 1;
    recStack[node] = 1;
    for (int adj : graph[node]) {
        if (!visited[adj] && dfs(adj, graph, visited, recStack))
            return true;
        else if (recStack[adj])
            return true;
    }
    recStack[node] = 0;
    return false;
}

int main() {
    vector graph(3);
    graph[0].push_back(1);
    graph[1].push_back(2);
    graph[2].push_back(0);
    vector<int> visited(3, 0), recStack(3, 0);
    cout << (dfs(0, graph, visited, recStack) ? "Cycle Found" : "No Cycle") << endl;
    return 0;
}

medium

A. Compilation Error

B. Cycle Found

C. No Cycle

D. Runtime Error

Solution

1. Step 1: Analyze graph edges

   Edges: 0->1, 1->2, 2->0 form a cycle (0->1->2->0).

2. Step 2: Understand DFS cycle detection

   DFS uses recursion stack to detect back edges indicating cycle. Here, cycle exists, so dfs returns true.

3. Final Answer:

   Cycle Found -> Option B

4. Quick Check:

   Cycle in 0->1->2->0 detected [OK]

Hint: Check recursion stack for back edges to find cycle [OK]

Common Mistakes:

• Ignoring recursion stack usage
• Assuming no cycle in simple loops
• Confusing visited and recursion stack arrays

4. Find the bug in this cycle detection code snippet for a directed graph:

bool dfs(int node, vector& graph, vector<bool>& visited, vector<bool>& recStack) {
    visited[node] = true;
    for (int adj : graph[node]) {
        if (!visited[adj] && dfs(adj, graph, visited, recStack))
            return true;
        else if (recStack[adj])
            return true;
    }
    recStack[node] = false;
    return false;
}

medium

A. Graph should be vector<int> not vector

B. visited array should be int, not bool

C. Missing return statement in else block

D. recStack[node] is never set to true at start

Solution

1. Step 1: Check usage of recStack

   recStack marks nodes in current recursion path. It must be set true when node is entered.

2. Step 2: Identify missing recStack[node] = true

   Code sets recStack[node] = false at end but never sets it true at start, so cycle detection fails.

3. Final Answer:

   recStack[node] is never set to true at start -> Option D

4. Quick Check:

   Set recStack true on node entry [OK]

Hint: Always mark recStack true when entering node [OK]

Common Mistakes:

• Forgetting to set recStack true at start
• Confusing visited and recStack roles
• Wrong data types for visited array

5. Given a directed graph with nodes 0 to 4 and edges: 0->1, 1->2, 2->3, 3->1, 3->4, which nodes are part of a cycle?

hard

A. [0, 1, 2, 3]

B. [2, 3, 4]

C. [1, 2, 3]

D. [0, 4]

Solution

1. Step 1: Analyze edges for cycles

   Edges 1->2, 2->3, 3->1 form a cycle involving nodes 1, 2, 3.

2. Step 2: Identify nodes outside cycle

   Node 0 points to 1 but is not in cycle. Node 4 is only reached from 3 and has no outgoing edges back.

3. Final Answer:

   [1, 2, 3] -> Option C

4. Quick Check:

   Cycle nodes = 1, 2, 3 [OK]

Hint: Cycle nodes form closed loop edges [OK]

Common Mistakes:

• Including nodes only pointing into cycle
• Including nodes with no back edges
• Confusing cycle membership with reachability