Cycle Detection in Directed Graph in DSA Go - 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.

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

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

func hasCycle(graph map[int][]int) bool {
    visited := make(map[int]bool)
    recStack := make(map[int]bool)

    var dfs func(int) bool
    dfs = func(node int) bool {
        if recStack[node] {
            return true
        }
        if visited[node] {
            return false
        }
        visited[node] = true
        recStack[node] = true

        for _, neighbor := range graph[node] {
            if dfs(neighbor) {
                return true
            }
        }
        recStack[node] = false
        return false
    }

    for node := range graph {
        if dfs(node) {
            return true
        }
    }
    return false
}
    

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

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

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

How Execution Grows With Input

As the graph grows, the DFS explores more nodes and edges.

Input Size (n nodes, e edges)	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 operations grow roughly in proportion to the number of nodes plus edges.

Final Time Complexity

Time Complexity: O(n + e)

This means the time to detect a cycle grows linearly with the number of nodes and 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, so DFS does not repeat work on the same node, keeping the time linear.

Interview Connect

Understanding this time complexity helps you explain how efficient your cycle detection is and shows you can analyze graph algorithms clearly.

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 the direction of edges

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 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 A

4. Quick Check:

   Cycle detection = finding directed loops [OK]

Hint: Cycle means a loop following edge directions [OK]

Common Mistakes:

• Confusing cycle with shortest path
• Ignoring edge directions
• Thinking cycle means any repeated node

2. Which of the following Go code snippets correctly declares a graph as an adjacency list for cycle detection?

easy

A. graph := make(map[int][]int)

B. graph := []int{}

C. graph := map[int]int{}

D. graph := [][]int{}

Solution

1. Step 1: Understand adjacency list representation

   An adjacency list for a directed graph maps each node to a list of nodes it points to.

2. Step 2: Check Go syntax for adjacency list

   Using map[int][]int creates a map from int to slice of ints, perfect for adjacency list.

3. Final Answer:

   graph := make(map[int][]int) -> Option A

4. Quick Check:

   Adjacency list = map[int][]int [OK]

Hint: Adjacency list uses map of int to slice of int [OK]

Common Mistakes:

• Using slice of int only (no mapping)
• Using map[int]int which stores single int per node
• Using [][]int without map keys

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

package main
import "fmt"

func dfs(node int, graph map[int][]int, visited, recStack map[int]bool) bool {
    if recStack[node] {
        return true
    }
    if visited[node] {
        return false
    }
    visited[node] = true
    recStack[node] = true
    for _, neighbor := range graph[node] {
        if dfs(neighbor, graph, visited, recStack) {
            return true
        }
    }
    recStack[node] = false
    return false
}

func main() {
    graph := map[int][]int{
        1: {2},
        2: {3},
        3: {1},
    }
    visited := make(map[int]bool)
    recStack := make(map[int]bool)
    fmt.Println(dfs(1, graph, visited, recStack))
}

medium

A. false

B. true

C. Compilation error

D. Runtime panic

Solution

1. Step 1: Analyze graph structure

   Graph edges: 1->2, 2->3, 3->1 form a cycle.

2. Step 2: Trace DFS cycle detection

   DFS from node 1 visits 2, then 3, then back to 1 which is in recursion stack, so returns true.

3. Final Answer:

   true -> Option B

4. Quick Check:

   Cycle detected = true [OK]

Hint: Cycle if node revisited in recursion stack [OK]

Common Mistakes:

• Ignoring recursion stack for cycle detection
• Assuming visited alone detects cycle
• Confusing false with no cycle

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

func dfs(node int, graph map[int][]int, visited map[int]bool) bool {
    if visited[node] {
        return true
    }
    visited[node] = true
    for _, neighbor := range graph[node] {
        if dfs(neighbor, graph, visited) {
            return true
        }
    }
    return false
}

medium

A. Return type should be int, not bool

B. Visited map should be a slice, not a map

C. Missing recursion stack to track nodes in current path

D. Graph should be a slice of slices, not a map

Solution

1. Step 1: Understand cycle detection logic

   Cycle detection in directed graph needs to track nodes in current recursion path (recursion stack).

2. Step 2: Identify missing recursion stack

   This code only uses visited to mark visited nodes, which cannot detect cycles correctly in directed graphs.

3. Final Answer:

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

4. Quick Check:

   Recursion stack needed for directed cycle detection [OK]

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

Common Mistakes:

• Using only visited to detect cycles
• Confusing undirected and directed cycle detection
• Ignoring recursion stack leads to false negatives

5. Given a directed graph represented as adjacency list:

graph := map[int][]int{
    1: {2, 3},
    2: {4},
    3: {4, 5},
    4: {6},
    5: {6},
    6: {3},
}

Which nodes are part of a cycle in this graph?

hard

A. [1, 3, 5]

B. [1, 2, 5]

C. [2, 4, 5]

D. [3, 4, 6]

Solution

1. Step 1: Analyze edges for cycles

   Edges: 3->4, 4->6, 6->3 form a cycle involving nodes 3, 4, 6.

2. Step 2: Confirm cycle nodes

   Nodes 1, 2, 5 lead into cycle but are not part of cycle themselves.

3. Final Answer:

   [3, 4, 6] -> Option D

4. Quick Check:

   Cycle nodes = [3,4,6] [OK]

Hint: Trace back edges to find cycle nodes [OK]

Common Mistakes:

• Including nodes that only lead to cycle
• Missing nodes that form the cycle
• Confusing cycle membership with reachability