BFS Breadth First Search on Graph in DSA Typescript - 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 change when the graph has more nodes and edges?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function bfs(graph: Map<number, number[]>, start: number): number[] {
  const visited = new Set<number>();
  const queue: number[] = [];
  const result: number[] = [];

  queue.push(start);
  visited.add(start);

  while (queue.length > 0) {
    const node = queue.shift()!;
    result.push(node);

    for (const neighbor of graph.get(node) || []) {
      if (!visited.has(neighbor)) {
        visited.add(neighbor);
        queue.push(neighbor);
      }
    }
  }

  return result;
}
    

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

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop runs as long as there are nodes in the queue, processing each node once.
• How many times: Each node is visited once, and each edge is checked once when exploring neighbors.

How Execution Grows With Input

As the graph grows, BFS visits every node and checks every edge connected to those nodes.

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, 5000 edges	About 1000 node visits + 5000 edge checks = 6000 operations

Pattern observation: The 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 number of nodes plus 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; it scales with nodes plus edges, not their square.

Interview Connect

Understanding BFS time complexity helps you explain how graph searches work efficiently, a key skill for many coding challenges and real-world 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 priority queue to select nodes by weight

B. A stack to process nodes in depth order

C. A hash map to store visited nodes only

D. A queue to process nodes level by level

Solution

1. Step 1: Understand BFS traversal method

   BFS explores nodes level by level, so it needs a structure that processes nodes in the order they are discovered.

2. Step 2: Identify the data structure used

   A queue follows First-In-First-Out (FIFO) order, which matches BFS's level order traversal.

3. Final Answer:

   A queue to process nodes level by level -> Option D

4. Quick Check:

   BFS uses a queue = A [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 only uses a visited set without a queue

2. Which of the following is the correct TypeScript syntax to initialize a queue for BFS using an array?

easy

A. const queue = new Queue<number>();

B. const queue: number[] = [];

C. let queue = {};

D. var queue = new ArrayQueue();

Solution

1. Step 1: Recognize TypeScript array initialization

   To create a queue using an array, declare an array with type annotation and initialize it empty.

2. Step 2: Check each option

   const queue: number[] = []; correctly declares an empty number array. const queue = new Queue<number>(); uses a class Queue which is not built-in. let queue = {}; initializes an object, not a queue. var queue = new ArrayQueue(); uses a non-existent ArrayQueue.

3. Final Answer:

   const queue: number[] = []; -> Option B

4. Quick Check:

   Array queue syntax = A [OK]

Hint: Use typed empty array for queue in TypeScript [OK]

Common Mistakes:

• Using undefined classes like Queue without import
• Initializing queue as an object {} instead of array
• Using var instead of const or let (less preferred)

3. Given the graph adjacency list below and BFS starting from node 0, what is the order of nodes visited?

const graph = {
  0: [1, 2],
  1: [3],
  2: [3],
  3: []
};

medium

A. [0, 1, 2, 3]

B. [0, 2, 1, 3]

C. [0, 3, 1, 2]

D. [1, 2, 0, 3]

Solution

1. Step 1: Start BFS from node 0

   Visit node 0 first, then enqueue its neighbors 1 and 2.

2. Step 2: Visit neighbors in queue order

   Dequeue 1, visit it, enqueue its neighbor 3. Then dequeue 2, visit it, enqueue 3 again but already visited. Finally, dequeue 3.

3. Final Answer:

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

4. Quick Check:

   BFS order = [0, 1, 2, 3] [OK]

Hint: Visit nodes level by level, enqueue neighbors in order [OK]

Common Mistakes:

• Visiting neighbors in wrong order
• Visiting node 3 twice
• Starting BFS from wrong node

4. Consider this BFS code snippet in TypeScript. What is the error?

function bfs(graph: Record, start: number) {
  const queue: number[] = [start];
  const visited = new Set<number>();
  while (queue.length > 0) {
    const node = queue.shift();
    if (!visited.has(node)) {
      visited.add(node);
      for (const neighbor of graph[node]) {
        queue.push(neighbor);
      }
    }
  }
  return Array.from(visited);
}

medium

A. queue.shift() can return undefined, causing errors

B. visited set is not updated correctly

C. Neighbors are not added to queue

D. Function does not return anything

Solution

1. Step 1: Analyze queue.shift() usage

   queue.shift() returns number | undefined. If queue is empty, node can be undefined, causing errors when accessing graph[node].

2. Step 2: Identify missing check for undefined

   Code should check if node is undefined before using it. Without this, runtime error can occur.

3. Final Answer:

   queue.shift() can return undefined, causing errors -> Option A

4. Quick Check:

   Check for undefined after shift = B [OK]

Hint: Always check queue.shift() result for undefined [OK]

Common Mistakes:

• Ignoring possible undefined from shift()
• Assuming visited set is wrong when it is correct
• Thinking neighbors are not added when they are

5. You want to find the shortest path length from node 0 to node 4 in this graph using BFS:

const graph = {
  0: [1, 2],
  1: [3],
  2: [3, 4],
  3: [4],
  4: []
};

Which BFS modification helps find the shortest path length correctly?

hard

A. Ignore visited set to explore all paths

B. Use a stack instead of a queue

C. Track distance for each node when first visited

D. Visit nodes multiple times to update distance

Solution

1. Step 1: Understand BFS for shortest path

   BFS finds shortest paths in unweighted graphs by exploring neighbors level by level.

2. Step 2: Track distance when visiting nodes

   Maintain a distance map or array to record the number of edges from start node to each node when first visited.

3. Final Answer:

   Track distance for each node when first visited -> Option C

4. Quick Check:

   Distance tracking in BFS = D [OK]

Hint: Store distance on first visit to each node [OK]

Common Mistakes:

• Using stack breaks BFS order
• Visiting nodes multiple times wastes time
• Ignoring visited set causes infinite loops