Topological Sort Using DFS in DSA Go - Time & Space Complexity

Time Complexity: Topological Sort Using DFS

O(n + e)

Understanding Time Complexity

We want to understand how the time needed to perform topological sort using DFS changes as the graph size grows.

Specifically, how the number of nodes and edges affects the work done.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

func topoSortDFS(graph map[int][]int, node int, visited map[int]bool, stack *[]int) {
    visited[node] = true
    for _, neighbor := range graph[node] {
        if !visited[neighbor] {
            topoSortDFS(graph, neighbor, visited, stack)
        }
    }
    *stack = append(*stack, node)
}

func TopologicalSort(graph map[int][]int) []int {
    visited := make(map[int]bool)
    stack := []int{}
    for node := range graph {
        if !visited[node] {
            topoSortDFS(graph, node, visited, &stack)
        }
    }
    // reverse stack for correct order
    for i, j := 0, len(stack)-1; i < j; i, j = i+1, j-1 {
        stack[i], stack[j] = stack[j], stack[i]
    }
    return stack
}
    

This code performs a topological sort on a directed graph using depth-first search (DFS).

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

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

How Execution Grows With Input

As the number of nodes (n) and edges (e) increase, the DFS explores all nodes and edges once.

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

Pattern observation: The work grows 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 total number of nodes and edges in the graph.

Common Mistake

[X] Wrong: "Topological sort takes O(n²) time because of nested loops."

[OK] Correct: The DFS visits each node and edge only once, so it does not repeatedly check all pairs. The graph structure guides the visits efficiently.

Interview Connect

Understanding this time complexity helps you explain how graph algorithms scale and why DFS is efficient for ordering tasks with dependencies.

Self-Check

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

Practice

1. What is the main purpose of using DFS in Topological Sort on a directed graph?

easy

A. To visit nodes in a way that ensures dependencies come before dependents

B. To find the shortest path between two nodes

C. To detect cycles in an undirected graph

D. To count the number of connected components

Solution

1. Step 1: Understand DFS role in Topological Sort

   DFS visits nodes deeply, ensuring all dependencies of a node are visited before the node itself.

2. Step 2: Relate DFS order to dependencies

   By pushing nodes to a stack after visiting all neighbors, we get an order where dependencies come before dependents.

3. Final Answer:

   To visit nodes in a way that ensures dependencies come before dependents -> Option A

4. Quick Check:

   DFS ensures dependency order = A [OK]

Hint: DFS post-order gives correct dependency order [OK]

Common Mistakes:

• Confusing DFS with BFS for topological sort
• Thinking topological sort finds shortest paths
• Assuming DFS detects cycles in undirected graphs here

2. Which of the following Go code snippets correctly declares a recursive DFS function for topological sort?

easy

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

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

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

D. func dfs(node int, graph map[int][]int, visited map[int]bool, stack *[]int) { if !visited[node] { visited[node] = true for _, neighbor := range graph[node] { dfs(neighbor, graph, visited, stack) } *stack = append(*stack, node) } }

Solution

1. Step 1: Check visited condition and marking

   func dfs(node int, graph map[int][]int, visited map[int]bool, stack *[]int) { if !visited[node] { visited[node] = true for _, neighbor := range graph[node] { dfs(neighbor, graph, visited, stack) } *stack = append(*stack, node) } } correctly checks if node is not visited before processing and marks it visited inside the condition.

2. Step 2: Confirm stack append and recursion

   It recursively calls dfs on neighbors and appends node to stack after visiting neighbors, using pointer to modify stack.

3. Final Answer:

   Option D code snippet -> Option D

4. Quick Check:

   Correct DFS recursion and stack append = D [OK]

Hint: Use pointer for stack to append in recursive DFS [OK]

Common Mistakes:

• Passing stack by value causing no changes
• Marking visited after recursion causing repeated visits
• Appending node before visiting neighbors

3. What is the output order of nodes after performing topological sort using DFS on this graph?

graph := map[int][]int{
  5: {2, 0},
  4: {0, 1},
  2: {3},
  3: {1},
  0: {},
  1: {},
}

Assuming DFS visits nodes in ascending order starting from 5, then 4.

medium

A. [5, 4, 2, 3, 1, 0]

B. [4, 5, 2, 3, 1, 0]

C. [4, 5, 0, 1, 2, 3]

D. [5, 4, 3, 2, 1, 0]

Solution

1. Step 1: Perform DFS starting from node 5

   DFS from 5 (neighbors ascending: 0 then 2): DFS(0) append 0; then DFS(2) -> DFS(3) -> DFS(1) append 1; append 3; append 2; append 5. Then DFS from 4 (0,1 visited), append 4.

2. Step 2: Collect nodes in post-order (append after visiting neighbors)

   Post-order stack push order: 0, 1, 3, 2, 5, 4. Reversing gives [4, 5, 2, 3, 1, 0].

3. Step 3: Consider DFS starting order and visited nodes

   Since DFS visits 5 first, then 4, the final topological order is the reverse of the post-order stack: [4, 5, 2, 3, 1, 0]. However, the question states DFS visits nodes in ascending order starting from 5, then 4. The actual post-order stack is [0,1,3,2,5,4], so reversed order is [4,5,2,3,1,0].

4. Final Answer:

   [4, 5, 2, 3, 1, 0] -> Option B

5. Quick Check:

   Post-order reverse = [4,5,2,3,1,0] [OK]

Hint: Reverse DFS post-order for topological sort output [OK]

Common Mistakes:

• Not reversing the stack after DFS
• Visiting nodes in wrong order
• Ignoring already visited nodes

4. Identify the bug in this DFS-based topological sort code snippet:

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

medium

A. Visited is marked after recursion, causing repeated visits

B. Stack is appended before visiting neighbors

C. Visited map is not initialized

D. Function does not return any value

Solution

1. Step 1: Check visited marking timing

   Visited is marked after recursive calls, so neighbors may revisit the same node multiple times.

2. Step 2: Correct visited marking position

   Visited should be marked before recursion to avoid repeated visits and infinite loops.

3. Final Answer:

   Visited is marked after recursion, causing repeated visits -> Option A

4. Quick Check:

   Mark visited before recursion = B [OK]

Hint: Mark visited before recursion to avoid repeats [OK]

Common Mistakes:

• Marking visited too late causing infinite recursion
• Appending stack before recursion
• Ignoring initialization of visited map

5. Given a directed graph with edges:

1 -> 2, 1 -> 3, 3 -> 4, 2 -> 4, 4 -> 5

If you want to find a valid order to complete tasks represented by nodes, which topological sort order is correct?

hard

A. [3, 1, 2, 4, 5]

B. [5, 4, 3, 2, 1]

C. [1, 2, 3, 4, 5]

D. [2, 1, 3, 4, 5]

Solution

1. Step 1: Analyze dependencies from edges

   Node 1 must come before 2 and 3; 2 and 3 before 4; 4 before 5.

2. Step 2: Find order respecting dependencies

   Order [1, 2, 3, 4, 5] respects all edges: 1 before 2 and 3, 2 and 3 before 4, 4 before 5.

3. Final Answer:

   [1, 2, 3, 4, 5] -> Option C

4. Quick Check:

   Dependencies respected = C [OK]

Hint: Start with nodes having no incoming edges [OK]

Common Mistakes:

• Reversing order ignoring dependencies
• Placing dependent before dependency
• Choosing order with cycles