BFS Breadth First Search on Graph in DSA C++ - Time & Space Complexity

Time Complexity: BFS Breadth First Search on Graph

O(n + m)

Understanding Time Complexity

We want to understand how the time needed to explore a graph using BFS changes as the graph grows.

How does BFS scale when the graph has more nodes and edges?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

void bfs(int start, const vector<vector<int>>& graph) {
    int n = graph.size();
    vector<bool> visited(n, false);
    queue<int> q;
    q.push(start);
    visited[start] = true;

    while (!q.empty()) {
        int node = q.front();
        q.pop();
        for (int neighbor : graph[node]) {
            if (!visited[neighbor]) {
                visited[neighbor] = true;
                q.push(neighbor);
            }
        }
    }
}
    

This code explores all nodes reachable from the start node using BFS, visiting each node and its neighbors once.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop runs as long as there are nodes to visit, and inside it, the for loop visits all neighbors of the current node.
• How many times: Each node is enqueued and dequeued once, and each edge is checked once when exploring neighbors.

How Execution Grows With Input

As the graph grows, BFS visits every node and edge once, so the work grows with the total nodes plus edges.

Input Size (n nodes, m edges)	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 total work grows roughly in a straight line with the sum of nodes and edges.

Final Time Complexity

Time Complexity: O(n + m)

This means BFS takes time proportional to the number of nodes plus the number of edges in the graph.

Common Mistake

[X] Wrong: "BFS always takes O(n²) time because it uses nested loops."

[OK] Correct: BFS visits each node and edge only once, so it does not repeatedly check all pairs. The loops do not multiply to n squared.

Interview Connect

Understanding BFS time complexity helps you explain how graph traversal scales and why BFS is efficient for many problems.

Self-Check

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

Practice

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