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 long it takes to find a cycle in a directed graph.

The question is: how does the time needed grow as the graph gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

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

This code checks if a directed graph has a cycle using depth-first search and recursion.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Visiting each node and exploring its neighbors recursively.
• How many times: Each node is visited once, and each edge is checked once during DFS.

How Execution Grows With Input

As the graph grows, the time to check all nodes and edges grows roughly in proportion to their count.

Input Size (n nodes)	Approx. Operations (nodes + edges)
10	About 10 nodes + edges checked once each
100	About 100 nodes + edges checked once each
1000	About 1000 nodes + edges checked once each

Pattern observation: The work grows linearly with the number of nodes and edges combined.

Final Time Complexity

Time Complexity: O(n + e)

This means the time grows linearly with the number of nodes (n) plus the number of edges (e) in the graph.

Common Mistake

[X] Wrong: "The algorithm checks every possible path, so it takes exponential time."

[OK] Correct: Each node and edge is visited only once due to marking visited nodes, so the time is linear, not exponential.

Interview Connect

Understanding this time complexity helps you explain how graph traversal algorithms scale, a key skill in many coding challenges.

Self-Check

"What if we changed the graph representation from adjacency list to adjacency matrix? 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 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