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

Time Complexity: Cycle Detection in Directed Graph

O(n + e)

Understanding Time Complexity

We want to know how long it takes to find a cycle in a directed graph.

Specifically, how the time grows as the graph gets bigger.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

boolean dfs(int node, boolean[] visited, boolean[] recStack, List<List<Integer>> graph) {
    visited[node] = true;
    recStack[node] = true;
    for (int neighbor : graph.get(node)) {
        if (!visited[neighbor] && dfs(neighbor, visited, recStack, graph))
            return true;
        else if (recStack[neighbor])
            return true;
    }
    recStack[node] = false;
    return false;
}

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

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

Identify Repeating Operations

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

How Execution Grows With Input

As the number of nodes and edges grows, the time to check all connections grows roughly in proportion.

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

Pattern observation: The work grows roughly linearly with the number of nodes plus edges.

Final Time Complexity

Time Complexity: O(n + e)

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

Common Mistake

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

[OK] Correct: The algorithm marks nodes visited and does not revisit them, so each edge and node is processed only once, keeping time linear.

Interview Connect

Understanding this time complexity helps you explain how your cycle detection scales with graph size, showing you know efficient graph traversal.

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. The maximum degree of any node

B. The shortest path between two nodes

C. The total number of nodes in the graph

D. Whether there is a path that starts and ends at the same node following the direction of edges

Solution

1. Step 1: Understand cycle in directed graph

   A cycle means a path that starts and ends at the same node following edge directions.

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 the direction of edges -> Option D

4. Quick Check:

   Cycle detection = finding directed cycles [OK]

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

Common Mistakes:

• Confusing cycle with shortest path
• Ignoring edge directions
• Thinking cycle means any repeated node without direction

2. Which Java data structure is commonly used to keep track of nodes currently in the recursion stack during DFS for cycle detection?

easy

A. Stack

B. HashSet

C. Queue

D. ArrayList

Solution

1. Step 1: Identify recursion stack tracking

   During DFS, nodes in the current path are tracked to detect back edges indicating cycles.

2. Step 2: Choose suitable data structure

   A HashSet is used to quickly check if a node is in the current recursion stack.

3. Final Answer:

   HashSet -> Option B

4. Quick Check:

   Use HashSet to track recursion stack nodes [OK]

Hint: Use HashSet for O(1) check if node is in recursion stack [OK]

Common Mistakes:

• Using Queue which is FIFO, not suitable here
• Using Stack but not checking membership efficiently
• Using ArrayList which is slower for membership check

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

import java.util.*;

public class GraphCycle {
    private final int V = 3;
    private final List adj = new ArrayList<>();
    private final boolean[] visited = new boolean[V];
    private final boolean[] recStack = new boolean[V];

    public GraphCycle() {
        for (int i = 0; i < V; i++) adj.add(new ArrayList<>());
        adj.get(0).add(1);
        adj.get(1).add(2);
        adj.get(2).add(0);
    }

    private boolean dfs(int node) {
        if (recStack[node]) return true;
        if (visited[node]) return false;
        visited[node] = true;
        recStack[node] = true;
        for (int neighbor : adj.get(node)) {
            if (dfs(neighbor)) return true;
        }
        recStack[node] = false;
        return false;
    }

    public boolean hasCycle() {
        for (int i = 0; i < V; i++) {
            if (dfs(i)) return true;
        }
        return false;
    }

    public static void main(String[] args) {
        GraphCycle graph = new GraphCycle();
        System.out.println(graph.hasCycle());
    }
}

medium

A. true

B. false

C. Compilation error

D. Runtime exception

Solution

1. Step 1: Analyze graph edges

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

2. Step 2: Trace DFS cycle detection

   DFS from node 0 visits 1, then 2, then back to 0 which is in recursion stack, so returns true.

3. Final Answer:

   true -> Option A

4. Quick Check:

   Cycle exists in graph = true [OK]

Hint: Cycle if DFS revisits node in recursion stack [OK]

Common Mistakes:

• Ignoring recursion stack check
• Assuming no cycle because visited[] is true
• Confusing visited[] and recStack[] roles

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

boolean dfs(int node, boolean[] visited, boolean[] recStack, List adj) {
    visited[node] = true;
    recStack[node] = true;
    for (int neighbor : adj.get(node)) {
        if (!visited[neighbor]) {
            dfs(neighbor, visited, recStack, adj);
        } else if (recStack[neighbor]) {
            return true;
        }
    }
    recStack[node] = false;
    return false;
}

medium

A. Loop iterates over wrong adjacency list

B. recStack is not reset to false

C. Missing return statement in recursive call

D. visited array is not updated

Solution

1. Step 1: Check recursive call handling

   Recursive call dfs(neighbor, ...) is made but its return value is ignored.

2. Step 2: Understand impact on cycle detection

   If dfs(neighbor) returns true (cycle found), this should be returned immediately to propagate detection.

3. Final Answer:

   Missing return statement in recursive call -> Option C

4. Quick Check:

   Return recursive call result to detect cycles [OK]

Hint: Return recursive dfs result to catch cycles early [OK]

Common Mistakes:

• Ignoring return value of recursive dfs
• Not resetting recStack[node] to false
• Confusing visited and recStack roles

5. Given a directed graph represented as adjacency list:

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

Which nodes are part of at least one cycle?

hard

A. [1, 2, 3]

B. [0, 1, 2, 3]

C. [1, 3]

D. [0, 4]

Solution

1. Step 1: Identify cycles in graph

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

2. Step 2: Check other nodes

   Node 0 points to 1 but is not in cycle; node 4 has no outgoing edges.

3. Final Answer:

   [1, 2, 3] -> Option A

4. Quick Check:

   Cycle nodes = [1, 2, 3] [OK]

Hint: Cycle nodes form closed loop edges [OK]

Common Mistakes:

• Including nodes that only lead to cycle but not in it
• Missing nodes inside the cycle
• Confusing cycle presence with reachability