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DSA C++programming~10 mins

BFS Breadth First Search on Graph in DSA C++ - Interactive Practice

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Practice - 5 Tasks
Answer the questions below
1fill in blank
easy

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

DSA C++
std::queue<int> q;
q.[1](start);
Drag options to blanks, or click blank then click option'
Apush
Bpop
Cempty
Dfront
Attempts:
3 left
💡 Hint
Common Mistakes
Using pop instead of push to add elements.
Using front which only accesses the first element.
Using empty which checks if the queue is empty.
2fill in blank
medium

Complete the code to mark a node as visited in BFS.

DSA C++
std::vector<bool> visited(n, false);
visited[[1]] = true;
Drag options to blanks, or click blank then click option'
Aqueue
Badj
Cstart
Dnode
Attempts:
3 left
💡 Hint
Common Mistakes
Marking the queue or adjacency list as visited.
Using a variable not representing a node index.
3fill in blank
hard

Fix the error in the BFS loop to get the current node from the queue.

DSA C++
while (!q.empty()) {
    int current = q.[1]();
    q.pop();
    // process current
}
Drag options to blanks, or click blank then click option'
Afront
Bpop
Cpush
Dempty
Attempts:
3 left
💡 Hint
Common Mistakes
Using pop() to get the element, which returns void.
Using push() or empty() incorrectly.
4fill in blank
hard

Fill both blanks to check if a neighbor is unvisited and then mark it visited.

DSA C++
for (int neighbor : adj[current]) {
    if (!visited[[1]]) {
        visited[[2]] = true;
        q.push(neighbor);
    }
}
Drag options to blanks, or click blank then click option'
Aneighbor
Bcurrent
Cstart
Dadj
Attempts:
3 left
💡 Hint
Common Mistakes
Using current or start instead of neighbor.
Using adjacency list variable instead of node index.
5fill in blank
hard

Fill all three blanks to complete BFS that prints nodes in order.

DSA C++
std::queue<int> q;
std::vector<bool> visited(n, false);
q.[1](start);
visited[[2]] = true;

while (!q.empty()) {
    int current = q.[3]();
    q.pop();
    std::cout << current << " ";
    for (int neighbor : adj[current]) {
        if (!visited[neighbor]) {
            visited[neighbor] = true;
            q.push(neighbor);
        }
    }
}
Drag options to blanks, or click blank then click option'
Apush
Bstart
Cfront
Dpop
Attempts:
3 left
💡 Hint
Common Mistakes
Using pop() to get current node.
Not marking start node visited.
Using wrong variable names.

Practice

(1/5)
1. What does BFS (Breadth First Search) primarily use to keep track of nodes to visit next in a graph?
easy
A. A queue data structure
B. A stack data structure
C. A priority queue
D. A hash map

Solution

  1. Step 1: Understand BFS traversal method

    BFS explores nodes level by level, visiting all neighbors before moving deeper.
  2. Step 2: Identify the data structure used

    To maintain the order of visiting nodes level-wise, BFS uses a queue which follows FIFO (First In First Out).
  3. Final Answer:

    A queue data structure -> Option A
  4. Quick Check:

    BFS uses queue = B [OK]
Hint: BFS uses FIFO order, so it needs a queue [OK]
Common Mistakes:
  • Confusing BFS with DFS which uses a stack
  • Thinking BFS uses a priority queue
  • Assuming BFS uses a hash map for order
2. Which of the following is the correct way to declare a queue for BFS traversal of integers in C++?
easy
A. vector<int> q;
B. queue<int> q;
C. stack<int> q;
D. list<int> q;

Solution

  1. Step 1: Identify the correct STL container for queue

    In C++, the standard queue container is declared as queue<type>.
  2. Step 2: Match the declaration for integer queue

    queue<int> q; correctly declares a queue of integers.
  3. Final Answer:

    queue<int> q; -> Option B
  4. Quick Check:

    Correct queue syntax = queue<int> q; [OK]
Hint: Use queue<int> for BFS integer queue in C++ [OK]
Common Mistakes:
  • Using stack instead of queue
  • Using vector or list which are not queues
  • Missing template angle brackets
3. Consider the following graph represented as adjacency list:
0: 1, 2
1: 0, 3
2: 0, 3
3: 1, 2
If BFS starts from node 0, what is the order of nodes visited?
medium
A. [0, 1, 3, 2]
B. [0, 2, 1, 3]
C. [0, 3, 1, 2]
D. [0, 1, 2, 3]

Solution

  1. Step 1: Start BFS from node 0

    Visit node 0 first, enqueue neighbors 1 and 2.
  2. Step 2: Visit neighbors in queue order

    Dequeue 1, enqueue its neighbor 3 (0 already visited). Then dequeue 2, enqueue 3 but already visited.
  3. Step 3: Visit node 3

    Finally dequeue 3.
  4. Final Answer:

    [0, 1, 2, 3] -> Option D
  5. Quick Check:

    BFS order = [0, 1, 2, 3] [OK]
Hint: Visit neighbors in queue order, avoid revisits [OK]
Common Mistakes:
  • Visiting neighbors in wrong order
  • Revisiting nodes already visited
  • Confusing BFS with DFS order
4. The following BFS code snippet has a bug. What is the error?
void bfs(int start, vector& adj) {
  vector<bool> visited(adj.size(), false);
  queue<int> q;
  q.push(start);
  while (!q.empty()) {
    int node = q.front();
    q.pop();
    cout << node << " ";
    for (int neighbor : adj[node]) {
      if (!visited[neighbor]) {
        q.push(neighbor);
      }
    }
  }
}
medium
A. Neighbors are not iterated correctly
B. Queue is not initialized properly
C. Visited array is never updated to true for visited nodes
D. Start node is not pushed into the queue

Solution

  1. Step 1: Check visited array usage

    Visited array is created but nodes are never marked visited after visiting.
  2. Step 2: Understand impact on BFS

    Without marking visited true, nodes can be enqueued multiple times causing infinite loops or repeated visits.
  3. Final Answer:

    Visited array is never updated to true for visited nodes -> Option C
  4. Quick Check:

    Must mark visited[node] = true after popping [OK]
Hint: Always mark nodes visited after dequeuing [OK]
Common Mistakes:
  • Forgetting to mark visited nodes
  • Pushing start node multiple times
  • Incorrect loop conditions
5. Given a graph with nodes 0 to 4 and edges: 0-1, 0-2, 1-3, 2-3, 3-4. Using BFS starting from node 0, which node is visited at the 3rd position in the BFS order?
hard
A. 2
B. 4
C. 1
D. 3

Solution

  1. Step 1: List neighbors and BFS order

    Start at 0, enqueue neighbors 1 and 2. BFS order so far: [0]
  2. Step 2: Visit nodes in queue order

    Dequeue 1 (2nd visited), enqueue 3. Dequeue 2 (3rd visited), enqueue 3 but already enqueued.
  3. Step 3: Confirm 3rd visited node

    The third node visited is 2.
  4. Final Answer:

    2 -> Option A
  5. Quick Check:

    Third visited node = 2 [OK]
Hint: Track queue order carefully to find nth visited node [OK]
Common Mistakes:
  • Mixing order of neighbors
  • Counting start node incorrectly
  • Ignoring repeated nodes in queue
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10 questions~5 mineasy hard