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

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Time Complexity: Cycle Detection in Directed Graph
O(n + e)
Understanding Time Complexity

We want to understand how long it takes to check if a directed graph has a cycle.

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

Scenario Under Consideration

Analyze the time complexity of the following code snippet.


def dfs(node):
    visited[node] = True
    recursion_stack[node] = True
    for neighbor in graph[node]:
        if not visited[neighbor]:
            if dfs(neighbor):
                return True
        elif recursion_stack[neighbor]:
            return True
    recursion_stack[node] = False
    return False

for node in range(num_nodes):
    if not visited[node]:
        if dfs(node):
            print("Cycle detected")
            break

This code uses depth-first search to detect cycles by tracking nodes in the current path.

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 DFS.
How Execution Grows With Input

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

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

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

Final Time Complexity

Time Complexity: O(n + e)

This means the time to detect a cycle grows linearly with the number of nodes plus edges.

Common Mistake

[X] Wrong: "Cycle detection takes O(n²) because of nested loops."

[OK] Correct: DFS visits each node and edge only once, so it does not repeatedly check all pairs.

Interview Connect

Understanding this time complexity helps you explain efficient graph traversal in interviews confidently.

Self-Check

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

Practice

(1/5)
1. What does cycle detection in a directed graph help us find?
easy
A. The total number of nodes in the graph
B. The shortest path between two nodes
C. The maximum degree of any node
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 starting from a node and following directed edges, you can return to the same node.
  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 = path returning to start node [OK]
Hint: Cycle means path loops back to start node following edges [OK]
Common Mistakes:
  • Confusing cycle with shortest path
  • Ignoring edge directions
  • Thinking cycle means any repeated node without direction
2. Which Python data structure is commonly used to keep track of nodes currently in the recursion stack during cycle detection in a directed graph?
easy
A. Set
B. List
C. Dictionary
D. Queue

Solution

  1. Step 1: Identify recursion stack tracking

    During DFS, nodes currently being explored are tracked to detect cycles.
  2. Step 2: Choose data structure for fast membership check

    Set allows O(1) check if a node is in recursion stack.
  3. Final Answer:

    Set -> Option A
  4. Quick Check:

    Recursion stack uses set for fast lookup [OK]
Hint: Use set for fast check if node is in recursion stack [OK]
Common Mistakes:
  • Using list which is slower for membership check
  • Using queue which is FIFO, not suitable here
  • Using dictionary without keys for membership
3. What will be the output of the following code snippet?
graph = {0: [1], 1: [2], 2: [0]}
visited = set()
rec_stack = set()
def dfs(node):
    if node in rec_stack:
        return True
    if node in visited:
        return False
    visited.add(node)
    rec_stack.add(node)
    for neighbor in graph.get(node, []):
        if dfs(neighbor):
            return True
    rec_stack.remove(node)
    return False

print(dfs(0))
medium
A. True
B. False
C. None
D. Error

Solution

  1. Step 1: Trace DFS calls starting from node 0

    Node 0 calls dfs(1), which calls dfs(2), which calls dfs(0) again.
  2. Step 2: Detect cycle when node 0 is in recursion stack

    When dfs(0) is called second time, node 0 is in rec_stack, so returns True.
  3. Final Answer:

    True -> Option A
  4. Quick Check:

    Cycle detected = True [OK]
Hint: Cycle if node revisited in recursion stack returns True [OK]
Common Mistakes:
  • Ignoring recursion stack check
  • Confusing visited with recursion stack
  • Assuming no cycle returns True
4. Find the bug in this cycle detection code snippet for a directed graph:
def dfs(node, graph, visited, rec_stack):
    visited.add(node)
    rec_stack.add(node)
    for neighbor in graph[node]:
        if neighbor not in visited:
            dfs(neighbor, graph, visited, rec_stack)
        elif neighbor in rec_stack:
            return True
    rec_stack.remove(node)
    return False
medium
A. Should use list instead of set for rec_stack
B. Missing return value when dfs(neighbor) is called recursively
C. Visited set should be cleared after recursion
D. Graph keys should be integers only

Solution

  1. Step 1: Check recursive call return usage

    The recursive call dfs(neighbor, ...) is made but its return value is ignored.
  2. Step 2: Fix by returning True if recursive call finds cycle

    Should return True immediately if dfs(neighbor, ...) returns True.
  3. Final Answer:

    Missing return value when dfs(neighbor) is called recursively -> Option B
  4. Quick Check:

    Return True on recursive cycle detection [OK]
Hint: Return True immediately when recursive call detects cycle [OK]
Common Mistakes:
  • Ignoring recursive call return value
  • Using wrong data structure for rec_stack
  • Clearing visited incorrectly
5. Given a directed graph represented as adjacency list:
graph = {
  'A': ['B'],
  'B': ['C', 'D'],
  'C': ['E'],
  'D': ['F'],
  'E': ['B'],
  'F': []
}
Which nodes are part of at least one cycle?
hard
A. A, D, F
B. A, B, C, E
C. B, C, E
D. No nodes form a cycle

Solution

  1. Step 1: Identify cycles by tracing edges

    Path B -> C -> E -> B forms a cycle involving B, C, E.
  2. Step 2: Check other nodes

    A leads to B but is not in cycle; D and F have no cycles.
  3. Final Answer:

    B, C, E -> Option C
  4. Quick Check:

    Cycle nodes = B, C, E [OK]
Hint: Trace edges to find nodes revisited in cycle [OK]
Common Mistakes:
  • Including nodes only leading to cycle but not in it
  • Ignoring indirect cycles
  • Assuming all connected nodes form cycles
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