Cycle Detection in Directed Graph in DSA C - Time & Space Complexity

Time Complexity: Cycle Detection in Directed Graph

O(n + e)

Understanding Time Complexity

We want to understand how the time needed to find a cycle in a directed graph changes as the graph grows.

Specifically, how does the process of checking each node and its connections scale?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

// Detect cycle using DFS in directed graph
bool dfs(int node, int visited[], int recStack[], List* graph) {
    visited[node] = 1;
    recStack[node] = 1;
    for (int i = 0; i < graph[node].size; i++) {
        int neighbor = graph[node].edges[i];
        if (!visited[neighbor] && dfs(neighbor, visited, recStack, graph))
            return true;
        else if (recStack[neighbor])
            return true;
    }
    recStack[node] = 0;
    return false;
}

bool isCyclic(int n, List* graph) {
    int visited[n];
    int recStack[n];
    for (int i = 0; i < n; i++) {
        visited[i] = 0;
        recStack[i] = 0;
    }
    for (int i = 0; i < n; i++) {
        if (!visited[i] && dfs(i, visited, recStack, graph))
            return true;
    }
    return false;
}
    

This code checks each node using depth-first search to find if a cycle exists in a directed graph.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: DFS visits each node and explores all its edges recursively.
• How many times: Each node is visited once, and each edge is checked once during the DFS calls.

How Execution Grows With Input

As the number of nodes and edges increase, the DFS explores more connections.

Input Size (n nodes, e edges)	Approx. Operations
10 nodes, 15 edges	About 25 operations (visiting nodes + edges)
100 nodes, 200 edges	About 300 operations
1000 nodes, 5000 edges	About 6000 operations

Pattern observation: The operations grow roughly proportional to the sum of nodes and edges.

Final Time Complexity

Time Complexity: O(n + e)

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

Common Mistake

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

[OK] Correct: Each node is marked visited and explored only once, so repeated visits do not happen, keeping the time linear.

Interview Connect

Understanding this time complexity helps you explain how graph traversal scales and why DFS is efficient for cycle detection.

Self-Check

"What if the graph was represented as an adjacency matrix instead of adjacency lists? How would the time complexity change?"

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 that starts and ends at the same node following the direction of edges.

2. Step 2: Identify what cycle detection finds

   Cycle detection checks if such a path exists 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 = path starting and ending at same node [OK]

Hint: Cycle means path loops back to start following edges [OK]

Common Mistakes:

• Confusing cycle with shortest path
• Thinking cycle ignores edge directions
• Mixing cycle detection with counting nodes

2. Which of the following is the correct way to declare a visited array for cycle detection in C for a graph with n nodes?

easy

A. int visited = new int[n];

B. int visited[n] = {0};

C. int visited[] = {0};

D. int visited[n];

Solution

1. Step 1: Understand array declaration in C

   In C, to declare an array of size n, use int visited[n]; but it is uninitialized.

2. Step 2: Initialize array to zero

   To initialize all elements to zero, use int visited[n] = {0}; which sets all to zero.

3. Final Answer:

   int visited[n] = {0}; -> Option B

4. Quick Check:

   Array declaration with zero init = int visited[n] = {0}; [OK]

Hint: Use int arr[n] = {0} to initialize all zeros in C [OK]

Common Mistakes:

• Using 'new' keyword which is C++ not C
• Declaring array without size
• Incorrect initialization syntax

3. Consider the following C code snippet for cycle detection in a directed graph using DFS. What will be the output?

void dfs(int node, int visited[], int recStack[], int graph[][3]) {
    visited[node] = 1;
    recStack[node] = 1;
    for (int i = 0; i < 3; i++) {
        int adj = graph[node][i];
        if (adj == -1) continue;
        if (!visited[adj]) dfs(adj, visited, recStack, graph);
        else if (recStack[adj]) printf("Cycle detected\n");
    }
    recStack[node] = 0;
}

int main() {
    int graph[3][3] = {{1, -1, -1}, {2, -1, -1}, {0, -1, -1}};
    int visited[3] = {0}, recStack[3] = {0};
    dfs(0, visited, recStack, graph);
    return 0;
}

medium

A. No output

B. Infinite loop

C. Segmentation fault

D. Cycle detected

Solution

1. Step 1: Analyze graph edges

   Graph edges: 0->1, 1->2, 2->0 form a cycle.

2. Step 2: Trace DFS calls and cycle detection

   DFS visits 0, then 1, then 2. When 2 tries to visit 0 again which is in recursion stack, it prints "Cycle detected".

3. Final Answer:

   Cycle detected -> Option D

4. Quick Check:

   Cycle in graph triggers print "Cycle detected" [OK]

Hint: Cycle prints when node revisited in recursion stack [OK]

Common Mistakes:

• Ignoring recursion stack for cycle detection
• Assuming no output means no cycle
• Confusing visited and recursion stack arrays

4. The following code attempts to detect a cycle in a directed graph but does not work correctly. What is the main error?

void dfs(int node, int visited[], int graph[][3]) {
    visited[node] = 1;
    for (int i = 0; i < 3; i++) {
        int adj = graph[node][i];
        if (adj == -1) continue;
        if (!visited[adj]) dfs(adj, visited, graph);
        else printf("Cycle detected\n");
    }
}

int main() {
    int graph[3][3] = {{1, -1, -1}, {2, -1, -1}, {0, -1, -1}};
    int visited[3] = {0};
    dfs(0, visited, graph);
    return 0;
}

medium

A. Graph adjacency matrix is incorrect

B. Visited array is not initialized

C. Missing recursion stack to track nodes in current path

D. DFS function missing return type

Solution

1. Step 1: Understand cycle detection logic

   Cycle detection in directed graph needs to track nodes in current recursion path using a recursion stack.

2. Step 2: Identify missing recursion stack

   This code only uses visited array, so it cannot distinguish back edges causing cycles from cross edges.

3. Final Answer:

   Missing recursion stack to track nodes in current path -> Option C

4. Quick Check:

   Recursion stack needed for correct cycle detection [OK]

Hint: Use recursion stack to detect back edges in DFS [OK]

Common Mistakes:

• Assuming visited array alone detects cycles
• Ignoring need to track current DFS path
• Confusing undirected and directed cycle detection

5. You have a directed graph represented as adjacency lists:

0: [1]
1: [2]
2: [3, 4]
3: [1]
4: []

Which nodes are part of a cycle detected by DFS cycle detection algorithm?

hard

A. Nodes 1, 2, 3

B. Nodes 0, 1, 4

C. Nodes 0, 2, 4

D. Nodes 3, 4

Solution

1. Step 1: Trace the graph edges

   Edges: 0->1, 1->2, 2->3 and 2->4, 3->1.

2. Step 2: Identify cycle nodes

   Cycle is 1->2->3->1. Nodes 1, 2, 3 form the cycle.

3. Final Answer:

   Nodes 1, 2, 3 -> Option A

4. Quick Check:

   Cycle nodes = 1, 2, 3 [OK]

Hint: Follow edges to find loops including nodes [OK]

Common Mistakes:

• Including nodes not in cycle like 0 or 4
• Missing node 3 in cycle
• Confusing cycle with path