BFS Breadth First Search on Graph in DSA Javascript - 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 grows as the graph gets bigger using BFS.

How does BFS's work time change when the graph has more nodes and edges?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function bfs(graph, start) {
  const visited = new Set();
  const queue = [start];
  visited.add(start);

  while (queue.length > 0) {
    const node = queue.shift();
    for (const neighbor of graph[node]) {
      if (!visited.has(neighbor)) {
        visited.add(neighbor);
        queue.push(neighbor);
      }
    }
  }
}
    

This code explores all nodes reachable from the start node by visiting neighbors level by level.

Identify Repeating Operations

• 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 visited once, and each edge is checked once during neighbor visits.

How Execution Grows With Input

As the graph grows, BFS visits every node and edge once.

Input Size (n nodes, m edges)	Approx. Operations
10 nodes, 15 edges	About 10 node visits + 15 edge checks = 25 operations
100 nodes, 200 edges	About 100 node visits + 200 edge checks = 300 operations
1000 nodes, 3000 edges	About 1000 node visits + 3000 edge checks = 4000 operations

Pattern observation: The total work grows roughly in proportion to the number of nodes plus edges.

Final Time Complexity

Time Complexity: O(n + m)

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

Common Mistake

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

[OK] Correct: The inner loop only runs over neighbors, so total work over all nodes is proportional to edges, not nodes squared.

Interview Connect

Knowing BFS time complexity helps you explain how graph searches scale 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 stack to explore nodes deeply first

B. A hash map to store visited nodes only

C. A queue to explore nodes level by level

D. A priority queue to explore nodes by weight

Solution

1. Step 1: Understand BFS traversal method

   BFS explores nodes level by level, so it needs to remember nodes in the order they are discovered.

2. Step 2: Identify the data structure for order

   A queue follows First-In-First-Out (FIFO), perfect for level order traversal in BFS.

3. Final Answer:

   A queue to explore nodes level by level -> Option C

4. Quick Check:

   BFS uses queue = A [OK]

Hint: BFS uses queue, DFS uses stack [OK]

Common Mistakes:

• Confusing BFS with DFS and choosing stack
• Thinking BFS uses priority queue
• Ignoring the order of node visits

2. Which of the following is the correct syntax to enqueue a node `v` into a queue named `queue` in JavaScript?

easy

A. queue.push(v);

B. queue.enqueue(v);

C. queue.add(v);

D. queue.insert(v);

Solution

1. Step 1: Recall JavaScript array methods for queue

   JavaScript arrays use `push()` to add elements at the end, simulating enqueue.

2. Step 2: Verify other options

   Methods like `enqueue()`, `add()`, or `insert()` are not standard array methods in JavaScript.

3. Final Answer:

   queue.push(v); -> Option A

4. Quick Check:

   Enqueue in JS array = push() [OK]

Hint: Use push() to add at queue's end in JS [OK]

Common Mistakes:

• Using non-existent enqueue() method
• Confusing push() with pop()
• Trying to use add() or insert() which don't exist

3. Consider the following graph represented as adjacency list:

const graph = {
  1: [2, 3],
  2: [4],
  3: [4],
  4: []
};

function bfs(start) {
  const queue = [start];
  const visited = new Set([start]);
  const order = [];

  while(queue.length > 0) {
    const node = queue.shift();
    order.push(node);
    for(const neighbor of graph[node]) {
      if(!visited.has(neighbor)) {
        visited.add(neighbor);
        queue.push(neighbor);
      }
    }
  }
  return order;
}

console.log(bfs(1));

What will be the output printed?

medium

A. [1, 3, 2, 4]

B. [1, 2, 3, 4]

C. [1, 4, 2, 3]

D. [4, 2, 3, 1]

Solution

1. Step 1: Start BFS from node 1

   Queue = [1], visited = {1}, order = []. Dequeue 1, order = [1]. Enqueue neighbors 2 and 3.

2. Step 2: Visit nodes in queue order

   Queue = [2,3]. Dequeue 2, order = [1,2]. Enqueue neighbor 4. Queue = [3,4]. Dequeue 3, order = [1,2,3]. Neighbor 4 already visited, skip. Dequeue 4, order = [1,2,3,4].

3. Final Answer:

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

4. Quick Check:

   BFS order = [1, 2, 3, 4] [OK]

Hint: BFS visits neighbors in queue order [OK]

Common Mistakes:

• Mixing order of neighbors
• Not marking visited before enqueue
• Using stack behavior instead of queue

4. The following BFS code has a bug. What is the error?

function bfs(start, graph) {
  const queue = [start];
  const visited = new Set();
  const order = [];

  while(queue.length > 0) {
    const node = queue.shift();
    order.push(node);
    for(const neighbor of graph[node]) {
      if(!visited.has(neighbor)) {
        queue.push(neighbor);
        visited.add(neighbor);
      }
    }
  }
  return order;
}

medium

A. Visited nodes are not marked before enqueueing, causing duplicates

B. Visited set is never updated, causing infinite loop

C. Visited nodes are marked after enqueueing, causing duplicates

D. Queue is not initialized properly

Solution

1. Step 1: Check order of visited marking

   Visited nodes are marked after enqueueing neighbors, which can cause duplicates if neighbors appear multiple times.

2. Step 2: Correct approach

   Mark nodes as visited before enqueueing to avoid adding duplicates to the queue.

3. Final Answer:

   Visited nodes are not marked before enqueueing, causing duplicates -> Option A

4. Quick Check:

   Mark visited before enqueue = no duplicates [OK]

Hint: Mark visited before enqueue to avoid duplicates [OK]

Common Mistakes:

• Marking visited after enqueue
• Not marking visited at all
• Initializing queue incorrectly

5. Given a graph with nodes and edges, how can BFS be adapted to find the shortest path length from a start node to a target node? Choose the best approach.

hard

A. Use a stack to explore nodes and update shortest distance dynamically

B. Use DFS and record path length when target is found

C. Use BFS but do not track visited nodes to find all paths

D. Use BFS with a queue and track distance for each node when first visited

Solution

1. Step 1: Understand BFS property for shortest path

   BFS explores nodes level by level, so the first time we reach the target node is the shortest path.

2. Step 2: Track distance during BFS

   Maintain a distance map or array, updating distance when a node is first visited to record shortest path length.

3. Final Answer:

   Use BFS with a queue and track distance for each node when first visited -> Option D

4. Quick Check:

   BFS + distance tracking = shortest path [OK]

Hint: Track distance on first visit in BFS for shortest path [OK]

Common Mistakes:

• Using DFS which may not find shortest path
• Not tracking visited nodes causing infinite loops
• Using stack instead of queue for BFS