Topological Sort Using DFS in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to start DFS from each unvisited node.

DSA Python

for node in graph:
    if node not in visited:
        [1](node)

Drag options to blanks, or click blank then click option'

Abfs

Bdfs

Ctopo_sort

Dvisit

Attempts:

3 left

💡 Hint

Common Mistakes

Using BFS instead of DFS.

Calling a function that doesn't exist.

We use dfs to explore nodes deeply for topological sort.

2fill in blank

medium

Complete the code to mark the current node as visited.

DSA Python

def dfs(node):
    visited.add([1])
    for neighbor in graph[node]:
        if neighbor not in visited:
            dfs(neighbor)

Drag options to blanks, or click blank then click option'

Anode

Bneighbor

Cgraph

Dvisited

Attempts:

3 left

💡 Hint

Common Mistakes

Marking the neighbor instead of the current node.

Trying to add the whole graph.

We mark the current node as visited before exploring neighbors.

3fill in blank

hard

Fix the error in adding nodes to the result list after visiting all neighbors.

DSA Python

def dfs(node):
    visited.add(node)
    for neighbor in graph[node]:
        if neighbor not in visited:
            dfs(neighbor)
    result.[1](node)

Drag options to blanks, or click blank then click option'

Aappendleft

Binsert

Cappend

Dextend

Attempts:

3 left

💡 Hint

Common Mistakes

Using extend which expects an iterable.

Using insert without index.

We use append to add the node after visiting neighbors for correct order.

4fill in blank

hard

Fill both blanks to reverse the result list and print the topological order.

DSA Python

result.[1]()
print(' -> '.join(str(node) for node in result[2]))

Drag options to blanks, or click blank then click option'

Areverse

Bsort

C[::-1]

Dpop

Attempts:

3 left

💡 Hint

Common Mistakes

Using sort instead of reverse.

Using pop which removes elements.

We reverse the result list in place and then print it in reversed order using slicing.

5fill in blank

hard

Fill all three blanks to create a dictionary comprehension filtering nodes with no outgoing edges.

DSA Python

no_outgoing = {node: graph[node] for node in graph if len(graph[node]) [1] [2]
print(no_outgoing[3])

Drag options to blanks, or click blank then click option'

A==

B0

C.keys()

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using greater than instead of equality.

Printing the whole dictionary instead of keys.

We check if the length of adjacency list is zero and print keys of the filtered dictionary.

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 all dependencies come before the node itself

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 nodes reachable from a node are visited before the node itself.

2. Step 2: Relate DFS order to dependencies

   In Topological Sort, this means all dependencies (nodes with edges to current node) are processed before the node, preserving order.

3. Final Answer:

   To visit nodes in a way that all dependencies come before the node itself -> Option A

4. Quick Check:

   DFS ensures dependency order in Topological Sort [OK]

Hint: DFS visits dependencies before the node itself [OK]

Common Mistakes:

• Confusing DFS with BFS for topological order
• Thinking DFS finds shortest paths here
• Mixing cycle detection with topological sorting

2. Which of the following is the correct way to mark a node as visited in a DFS-based topological sort in Python?

easy

A. visited.add(node)

B. visited[node] == True

C. visited[node] = True

D. visited.append(node)

Solution

1. Step 1: Understand visited data structure type

   Usually, visited is a list or dictionary indexed by node, so marking visited means assigning True to that index.

2. Step 2: Identify correct syntax for assignment

   Using visited[node] = True correctly marks the node as visited. Using '==' is a comparison, not assignment.

3. Final Answer:

   visited[node] = True -> Option C

4. Quick Check:

   Assign True to visited[node] to mark visited [OK]

Hint: Use assignment (=), not comparison (==), to mark visited [OK]

Common Mistakes:

• Using '==' instead of '=' for assignment
• Using visited.append(node) when visited is a dict/list
• Using visited.add(node) when visited is not a set

3. What is the output of the following code snippet for topological sort using DFS?

graph = {0: [1, 2], 1: [2], 2: []}
visited = [False, False, False]
stack = []
def dfs(node):
    visited[node] = True
    for neighbor in graph[node]:
        if not visited[neighbor]:
            dfs(neighbor)
    stack.append(node)

for v in range(3):
    if not visited[v]:
        dfs(v)
print(stack[::-1])

medium

A. [0, 1, 2]

B. [2, 1, 0]

C. [1, 0, 2]

D. [2, 0, 1]

Solution

1. Step 1: Trace DFS calls and stack appends

   Start with node 0: visit 0, then neighbors 1 and 2. DFS(1) visits 1, then 2. DFS(2) visits 2, appends 2. Then append 1, then append 0.

2. Step 2: Reverse stack to get topological order

   Stack after DFS: [2, 1, 0]. Reversing gives [0, 1, 2].

3. Final Answer:

   [0, 1, 2] -> Option A

4. Quick Check:

   Stack reversed gives correct topological order [OK]

Hint: Reverse stack after DFS to get topological order [OK]

Common Mistakes:

• Not reversing the stack before printing
• Appending nodes before recursive DFS calls
• Confusing order of neighbors visited

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

def dfs(node):
    visited[node] = True
    for neighbor in graph[node]:
        if visited[neighbor]:
            dfs(neighbor)
    stack.append(node)

medium

A. DFS should not append nodes to stack

B. Stack should be cleared before DFS

C. Visited should be a set, not a list

D. The condition should be 'if not visited[neighbor]' before calling dfs

Solution

1. Step 1: Analyze the if condition in DFS

   The code calls dfs(neighbor) only if visited[neighbor] is True, which means neighbor was already visited. This causes repeated visits or infinite recursion.

2. Step 2: Correct the condition to avoid revisiting

   The condition should be 'if not visited[neighbor]' to visit unvisited neighbors only.

3. Final Answer:

   The condition should be 'if not visited[neighbor]' before calling dfs -> Option D

4. Quick Check:

   Visit neighbors only if not visited [OK]

Hint: Call dfs only on unvisited neighbors [OK]

Common Mistakes:

• Calling dfs on already visited nodes
• Not marking nodes visited before recursion
• Appending nodes to stack too early

5. Given a directed acyclic graph with edges: 5->2, 5->0, 4->0, 4->1, 2->3, 3->1, which of the following is a valid topological order using DFS-based topological sort?

hard

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

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

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

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

Solution

1. Step 1: Understand dependencies from edges

   Edges mean 5 before 2 and 0, 4 before 0 and 1, 2 before 3, 3 before 1. So 5 and 4 come before 2,3,1,0.

2. Step 2: Check each option for order validity

   [5, 4, 2, 3, 1, 0] respects all edges: 5 before 2 and 0, 4 before 0 and 1, 2 before 3, 3 before 1. Others violate these orders.

3. Final Answer:

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

4. Quick Check:

   All edges respected in order [OK]

Hint: Check all edges respected in order for valid topological sort [OK]

Common Mistakes:

• Ignoring edge directions in order
• Assuming any order of nodes is valid
• Not verifying all dependencies before nodes