Why Shortest Path in Unweighted Graph Using BFS in DSA Typescript?

The Big Idea

What if you could always find the quickest way without guessing or getting lost?

The Scenario

Imagine you are in a city with many streets and intersections. You want to find the shortest way to walk from your home to a friend's house without knowing the map well.

If you try to guess the path by walking randomly or checking every street one by one, it will take a lot of time and effort.

The Problem

Manually checking every possible path is slow and confusing. You might walk in circles or take longer routes without realizing it.

It is easy to get lost or waste time exploring paths that do not lead to your friend's house quickly.

The Solution

Using the Breadth-First Search (BFS) method, you explore all nearby intersections first before moving further away.

This way, you find the shortest path step-by-step without missing any closer options.

BFS guarantees the shortest path in an unweighted graph because it checks all neighbors at the current distance before going deeper.

Before vs After

✗ Before

function findPath(graph, start, end) {
  // Try all paths manually, very complex and slow
  // No clear order of checking neighbors
}

✓ After

function bfsShortestPath(graph, start, end) {
  const queue = [start];
  const visited = new Set([start]);
  const distance = new Map();
  distance.set(start, 0);
  while (queue.length) {
    const node = queue.shift();
    if (node === end) return distance.get(node);
    for (const neighbor of graph[node]) {
      if (!visited.has(neighbor)) {
        visited.add(neighbor);
        distance.set(neighbor, distance.get(node) + 1);
        queue.push(neighbor);
      }
    }
  }
  return -1;
}

What It Enables

You can quickly find the shortest route in any network of connections without guessing or wasting time.

Real Life Example

Finding the quickest way to send a message through a network of friends or computers where each connection has the same cost.

Key Takeaways

Manual path search is slow and error-prone.

BFS explores neighbors level by level to find the shortest path.

This method works best for unweighted graphs where all edges have equal cost.

Practice

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

easy

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

B. The heaviest edge in the graph

C. The longest path between two nodes

D. The total number of edges in the graph

Solution

1. Step 1: Understand BFS traversal

   BFS explores nodes level by level starting from the source node.

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

   Because BFS visits nodes in order of increasing distance, it finds shortest paths.

3. Final Answer:

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

4. Quick Check:

   BFS = shortest path in unweighted graph [OK]

Hint: BFS explores neighbors level-wise to find shortest paths [OK]

Common Mistakes:

• Confusing BFS with DFS for shortest path
• Thinking BFS finds longest path
• Assuming BFS works for weighted graphs without modification

2. Which TypeScript data structure is best to use as a queue for BFS implementation?

easy

A. Array with push() and pop()

B. Array with push() and shift()

C. Set with add() and delete()

D. Map with set() and get()

Solution

1. Step 1: Identify queue operations needed for BFS

   BFS requires FIFO order: enqueue at end, dequeue from front.

2. Step 2: Choose Array methods matching FIFO

   push() adds to end, shift() removes from front, so Array with push() and shift() works as a queue.

3. Final Answer:

   Array with push() and shift() -> Option B

4. Quick Check:

   Queue = push + shift [OK]

Hint: Use push() to enqueue and shift() to dequeue in Array [OK]

Common Mistakes:

• Using pop() instead of shift() causes LIFO behavior
• Using Set or Map which don't maintain order
• Using Array methods incorrectly for queue

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

const graph = [[1,2],[0,3],[0,3],[1,2]];
// Nodes: 0,1,2,3
// Edges: 0-1, 0-2, 1-3, 2-3

medium

A. [0, 1, 1, 2]

B. [0, 2, 2, 3]

C. [1, 0, 1, 2]

D. [0, 1, 2, 1]

Solution

1. Step 1: Initialize distances and start BFS from node 0

   Distance to 0 is 0, others infinity initially.

2. Step 2: Visit neighbors level-wise

   Neighbors of 0 are 1 and 2, set distance 1 for both. Then neighbors of 1 and 2 include 3, set distance 2 for 3.

3. Final Answer:

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

4. Quick Check:

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

Hint: Distance increments by 1 per BFS level [OK]

Common Mistakes:

• Mixing up node indices and distances
• Assigning wrong distances to neighbors
• Confusing BFS with DFS order

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

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

medium

A. Neighbors are not checked for visited status

B. Distances array initialized with -1 instead of 0

C. Start node is not added to queue

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

Solution

1. Step 1: Check queue operations in BFS

   BFS requires FIFO queue, but pop() removes from end (LIFO).

2. Step 2: Identify impact of pop() usage

   Using pop() makes traversal DFS-like, breaking shortest path logic.

3. Final Answer:

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

4. Quick Check:

   Queue must be FIFO: push + shift [OK]

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

Common Mistakes:

• Confusing pop() with shift() for queue
• Ignoring visited check for neighbors
• Incorrect initialization of distances

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

hard

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

B. [0, 2, 3, 4]

C. [0, 1, 3, 4]

D. [0, 4]

Solution

1. Step 1: Explore neighbors from node 0

   Neighbors are 1 and 2, both distance 1.

2. Step 2: Explore neighbors of 1 and 2

   Both connect to 3 at distance 2.

3. Step 3: From 3, move to 4 at distance 3

   Shortest path is 0 -> 1 -> 3 -> 4 or 0 -> 2 -> 3 -> 4.

4. Step 4: Choose path found first by BFS

   BFS visits 1 before 2, so path is [0,1,3,4].

5. Final Answer:

   [0, 1, 3, 4] -> Option C

6. Quick Check:

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

Hint: BFS finds shortest path by earliest discovered route [OK]

Common Mistakes:

• Assuming direct edge from 0 to 4 exists
• Choosing longer path ignoring BFS order
• Confusing path with distance array