Shortest Path in Unweighted Graph Using BFS in DSA Javascript - Time & Space Complexity

Time Complexity: Shortest Path in Unweighted Graph Using BFS

O(n + m)

Understanding Time Complexity

We want to know how the time needed to find the shortest path in an unweighted graph grows as the graph gets bigger.

Specifically, how does the search process scale with more nodes and edges?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function shortestPathBFS(graph, start) {
  const queue = [start];
  const distances = { [start]: 0 };
  while (queue.length > 0) {
    const node = queue.shift();
    for (const neighbor of graph[node]) {
      if (distances[neighbor] === undefined) {
        distances[neighbor] = distances[node] + 1;
        queue.push(neighbor);
      }
    }
  }
  return distances;
}
    

This code finds the shortest distance from the start node to all other nodes using a breadth-first search.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Visiting each node and checking its neighbors in the graph.
• How many times: Each node is visited once, and each edge is checked once during the search.

How Execution Grows With Input

As the number of nodes and edges grows, the work grows roughly in proportion to their total count.

Input Size (n nodes, m edges)	Approx. Operations
10 nodes, 15 edges	About 25 checks (nodes + edges)
100 nodes, 200 edges	About 300 checks
1000 nodes, 5000 edges	About 6000 checks

Pattern observation: The total operations grow roughly as the sum of nodes and edges, increasing linearly as the graph grows.

Final Time Complexity

Time Complexity: O(n + m)

This means the time to find shortest paths grows linearly with the number of nodes plus edges in the graph.

Common Mistake

[X] Wrong: "The BFS time depends only on the number of nodes, so it is O(n)."

[OK] Correct: BFS also checks all edges connected to nodes, so edges affect the time too, making it O(n + m).

Interview Connect

Understanding BFS time complexity helps you explain how graph searches scale, a key skill in many coding challenges and real-world problems.

Self-Check

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

Practice

1. What does BFS (Breadth-First Search) help find in an unweighted graph?

easy

A. The longest path between two nodes

B. The node with the highest degree

C. The total number of nodes in the graph

D. The shortest path from a start node to all other nodes

Solution

1. Step 1: Understand BFS traversal

   BFS explores nodes level by level starting from the source node, visiting all neighbors before moving deeper.

2. Step 2: BFS finds shortest paths in unweighted graphs

   Because BFS visits nodes in order of increasing distance, the first time it reaches a node is via the shortest path.

3. Final Answer:

   The shortest path from a start node to all other nodes -> Option D

4. Quick Check:

   BFS = shortest path in unweighted graph [OK]

Hint: BFS visits nodes by distance layers to find shortest paths [OK]

Common Mistakes:

• Confusing BFS with DFS for shortest path
• Thinking BFS finds longest path
• Assuming BFS counts total nodes only

2. Which JavaScript data structure is best to implement the BFS queue for shortest path?

easy

A. Map to store node distances

B. Array with push and shift methods

C. Set to store unique nodes

D. Object with key-value pairs

Solution

1. Step 1: Identify BFS queue operations

   BFS requires adding nodes at the end and removing from the front, which is a queue behavior.

2. Step 2: Use Array methods for queue

   JavaScript arrays support push (add end) and shift (remove front), perfect for BFS queue.

3. Final Answer:

   Array with push and shift methods -> Option B

4. Quick Check:

   Queue = Array push/shift [OK]

Hint: Use array push/shift for BFS queue in JavaScript [OK]

Common Mistakes:

• Using Set as queue (no order guaranteed)
• Using Map for queue operations
• Using Object which has no order

3. What is the output distance array after running BFS from node 0 on this graph?

const graph = {
  0: [1, 2],
  1: [0, 3],
  2: [0, 3],
  3: [1, 2, 4],
  4: [3]
};
// BFS from node 0

medium

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

B. [0, 1, 2, 3, 4]

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

D. [0, 2, 2, 3, 4]

Solution

1. Step 1: Initialize distances and start BFS

   Distance to start node 0 is 0. Neighbors 1 and 2 get distance 1.

2. Step 2: Visit neighbors of 1 and 2

   Node 3 is neighbor of 1 and 2, so distance is 2. Node 4 is neighbor of 3, distance 3.

3. Final Answer:

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

4. Quick Check:

   Distances = [0,1,1,2,3] [OK]

Hint: Distance increments by 1 each BFS layer [OK]

Common Mistakes:

• Assigning wrong distances to neighbors
• Mixing up node indices
• Forgetting to update distance only once

4. Identify the bug in this BFS shortest path code snippet:

function bfs(graph, start) {
  const queue = [];
  const distances = {};
  queue.push(start);
  distances[start] = 0;
  while (queue.length > 0) {
    const node = queue.pop();
    for (const neighbor of graph[node]) {
      if (!(neighbor in distances)) {
        distances[neighbor] = distances[node] + 1;
        queue.push(neighbor);
      }
    }
  }
  return distances;
}

medium

A. Using pop() instead of shift() causes wrong BFS order

B. Distances should be initialized with -1

C. Neighbors should be checked before pushing start node

D. Queue should be a Set to avoid duplicates

Solution

1. Step 1: Check queue operations in BFS

   BFS requires FIFO order, so nodes must be removed from front using shift(), not pop() which removes from end.

2. Step 2: Effect of using pop()

   Using pop() makes it LIFO, turning BFS into DFS order, breaking shortest path logic.

3. Final Answer:

   Using pop() instead of shift() causes wrong BFS order -> Option A

4. Quick Check:

   Queue removal must be shift() for BFS [OK]

Hint: Use shift() to dequeue in BFS, not pop() [OK]

Common Mistakes:

• Using pop() instead of shift() for queue
• Initializing distances incorrectly
• Using Set for queue which loses order

5. Given a graph with nodes 0 to 5 and edges: 0-1, 0-2, 1-3, 2-3, 3-4, 4-5, what is the shortest path from node 0 to node 5 using BFS?

hard

A. [0, 1, 3, 4, 5]

B. [0, 2, 3, 4, 5]

C. [0, 1, 2, 3, 4, 5]

D. [0, 2, 4, 5]

Solution

1. Step 1: Explore neighbors from node 0

   Neighbors are 1 and 2, both distance 1.

2. Step 2: From neighbors, explore node 3

   Node 3 is neighbor of both 1 and 2, distance 2.

3. Step 3: From node 3, go to 4 then 5

   Node 4 distance 3, node 5 distance 4. The path via 1 is [0,1,3,4,5].

4. Final Answer:

   [0, 1, 3, 4, 5] -> Option A

5. Quick Check:

   Shortest path = [0,1,3,4,5] [OK]

Hint: Trace BFS layers from start to target [OK]

Common Mistakes:

• Choosing longer paths ignoring BFS order
• Skipping nodes in path
• Confusing edges and neighbors