Connected Components Using BFS in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to add the starting node to the queue in BFS.

DSA Python

queue = []
visited = set()
start_node = 0
queue.[1](start_node)
visited.add(start_node)

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

Aappend

Bpop

Cremove

Dinsert

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop instead of append will remove an element instead of adding.

Using remove will cause an error if the element is not in the list.

We use append() to add the start_node to the end of the queue for BFS.

2fill in blank

medium

Complete the code to dequeue the first element from the queue in BFS.

DSA Python

current = queue.[1](0)

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

Aappend

Bpop

Cinsert

Dremove

Attempts:

3 left

💡 Hint

Common Mistakes

Using append will add an element instead of removing.

Using remove without argument causes error.

pop(0) removes and returns the first element from the queue, which is needed for BFS.

3fill in blank

hard

Fix the error in the BFS loop to check if a neighbor is not visited before adding.

DSA Python

for neighbor in graph[current]:
    if neighbor not in [1]:
        queue.append(neighbor)
        visited.add(neighbor)

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

Agraph

Bqueue

Cvisited

Dcurrent

Attempts:

3 left

💡 Hint

Common Mistakes

Checking in queue instead of visited can cause duplicates.

Checking in graph or current is incorrect.

We check if neighbor is not in visited set before adding to queue to avoid revisiting nodes.

4fill in blank

hard

Fill both blanks to complete the function that finds connected components using BFS.

DSA Python

def connected_components(graph):
    visited = set()
    components = []
    for node in graph:
        if node not in [1]:
            component = []
            queue = [node]
            visited.add(node)
            while queue:
                current = queue.pop(0)
                component.append(current)
                for neighbor in graph[current]:
                    if neighbor not in [2]:
                        queue.append(neighbor)
                        visited.add(neighbor)
            components.append(component)
    return components

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

Avisited

Bqueue

Cgraph

Dcomponents

Attempts:

3 left

💡 Hint

Common Mistakes

Using queue or graph instead of visited causes logic errors.

Using components in condition is incorrect.

We check if node and neighbor are not in visited set to avoid revisiting nodes.

5fill in blank

hard

Fill all three blanks to create a dictionary comprehension that maps each node to its component index.

DSA Python

def map_nodes_to_components(components):
    return {node: [1] for idx, comp in enumerate(components) for node in [2] if node not in [3]

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

Aidx

Bcomp

Cvisited

Dcomponents

Attempts:

3 left

💡 Hint

Common Mistakes

Using components instead of comp causes errors.

Not checking visited causes duplicate keys.

We assign each node the index idx of its component comp, skipping nodes already in visited to avoid duplicates.

Practice

1. What does the BFS algorithm help find in an undirected graph when used for connected components?

easy

A. All nodes reachable from a starting node forming one connected component

B. The shortest path between two nodes only

C. The maximum degree of any node in the graph

D. The total number of edges in the graph

Solution

1. Step 1: Understand BFS in graph context

   BFS explores all nodes reachable from a given start node by visiting neighbors level by level.

2. Step 2: Connect BFS to connected components

   Using BFS from an unvisited node finds all nodes in that node's connected component.

3. Final Answer:

   All nodes reachable from a starting node forming one connected component -> Option A

4. Quick Check:

   BFS finds connected components by exploring reachable nodes [OK]

Hint: BFS explores all nodes connected to start node [OK]

Common Mistakes:

• Confusing BFS with shortest path only
• Thinking BFS counts edges
• Mixing BFS with node degree calculation

2. Which of the following is the correct way to initialize a queue for BFS in Python?

easy

A. queue = []

B. queue = dict()

C. queue = set()

D. queue = collections.deque()

Solution

1. Step 1: Identify BFS queue requirements

   BFS requires a queue data structure to process nodes in FIFO order.

2. Step 2: Choose correct Python queue type

   collections.deque() is efficient for queue operations (append and popleft).

3. Final Answer:

   queue = collections.deque() -> Option D

4. Quick Check:

   Use deque for BFS queue in Python [OK]

Hint: Use deque for efficient FIFO queue in BFS [OK]

Common Mistakes:

• Using list which is inefficient for popping front
• Using set which is unordered
• Using dict which is key-value store

3. What is the output of the following code snippet that finds connected components using BFS?

graph = {0: [1], 1: [0, 2], 2: [1], 3: []}
visited = set()
components = []

for node in graph:
    if node not in visited:
        queue = collections.deque([node])
        comp = []
        while queue:
            current = queue.popleft()
            if current not in visited:
                visited.add(current)
                comp.append(current)
                queue.extend(n for n in graph[current] if n not in visited)
        components.append(sorted(comp))
print(components)

medium

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

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

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

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

Solution

1. Step 1: Analyze graph structure

   Nodes 0,1,2 are connected; node 3 is isolated.

2. Step 2: Trace BFS for each component

   Starting at 0, BFS visits 0,1,2; then 3 alone forms second component.

3. Final Answer:

   [[0, 1, 2], [3]] -> Option C

4. Quick Check:

   Connected components = [[0,1,2],[3]] [OK]

Hint: Group nodes reachable from each unvisited node [OK]

Common Mistakes:

• Not sorting component lists
• Mixing isolated node with others
• Forgetting to mark visited nodes

4. Identify the bug in this BFS connected components code snippet:

def bfs_components(graph):
    visited = set()
    components = []
    for node in graph:
        if node not in visited:
            queue = [node]
            comp = []
            while queue:
                current = queue.pop()
                if current not in visited:
                    visited.add(current)
                    comp.append(current)
                    for neighbor in graph[current]:
                        if neighbor not in visited:
                            queue.append(neighbor)
            components.append(comp)
    return components

medium

A. Using list pop() removes from end, not front, breaking BFS order

B. Visited set is not updated correctly

C. Neighbors are not checked for visited status

D. Components list is not returned

Solution

1. Step 1: Check queue operations

   Using list pop() removes last element, making it DFS, not BFS.

2. Step 2: Understand BFS queue behavior

   BFS requires FIFO, so pop(0) or deque.popleft() is needed.

3. Final Answer:

   Using list pop() removes from end, not front, breaking BFS order -> Option A

4. Quick Check:

   pop() removes last, use popleft() for BFS [OK]

Hint: Use popleft() or pop(0) for BFS queue [OK]

Common Mistakes:

• Using pop() instead of popleft()
• Not marking visited before enqueue
• Appending neighbors without check

5. Given a graph represented as adjacency list:

graph = {0: [1, 2], 1: [0, 3], 2: [0], 3: [1], 4: [5], 5: [4], 6: []}

Write a BFS-based approach to find the number of connected components and return the size of the largest component.

hard

A. Number of components = 4, Largest size = 2

B. Number of components = 3, Largest size = 4

C. Number of components = 2, Largest size = 4

D. Number of components = 3, Largest size = 2

Solution

1. Step 1: Identify connected components

   Components are: {0,1,2,3}, {4,5}, {6} totaling 3 components.

2. Step 2: Calculate sizes

   Sizes are 4, 2, and 1 respectively; largest is 4 nodes.

3. Step 3: Re-check sizes carefully

   Node 6 is isolated (size 1), so largest component size is 4.

4. Final Answer:

   Number of components = 3, Largest size = 4 -> Option B

5. Quick Check:

   3 components, largest has 4 nodes [OK]

Hint: Count components and track max size during BFS [OK]

Common Mistakes:

• Counting isolated nodes incorrectly
• Mixing component sizes
• Forgetting to update largest size