Shortest Path in Unweighted Graph Using BFS in DSA Typescript - Build from Scratch

Shortest Path in Unweighted Graph Using BFS

📖 Scenario: Imagine you are navigating a city with intersections connected by roads. You want to find the shortest number of roads to travel from one intersection to another.

🎯 Goal: You will build a program that finds the shortest path length between two intersections in an unweighted graph using Breadth-First Search (BFS).

📋 What You'll Learn

Create a graph as an adjacency list with exact nodes and edges

Set up start and end nodes for the path search

Implement BFS to find the shortest path length

Print the shortest path length or -1 if no path exists

💡 Why This Matters

🌍 Real World

Finding shortest routes in maps, social networks, or network routing where all connections have equal cost.

💼 Career

Understanding BFS and shortest path algorithms is essential for software engineers working on navigation, recommendation systems, and network analysis.

1

Create the graph as an adjacency list

Create a variable called graph as a Map<string, string[]> with these exact entries: 'A' connected to ['B', 'C'], 'B' connected to ['A', 'D', 'E'], 'C' connected to ['A', 'F'], 'D' connected to ['B'], 'E' connected to ['B', 'F'], and 'F' connected to ['C', 'E'].

DSA Typescript

// Your code here

Hint

Use new Map() with an array of key-value pairs where keys are node names and values are arrays of connected nodes.

2

Set start and end nodes

Create two variables: start set to 'A' and end set to 'F'.

DSA Typescript

const graph: Map = new Map([
  ['A', ['B', 'C']],
  ['B', ['A', 'D', 'E']],
  ['C', ['A', 'F']],
  ['D', ['B']],
  ['E', ['B', 'F']],
  ['F', ['C', 'E']]
]);
// Your code here

Hint

Just create two string variables with the exact names and values.

3

Implement BFS to find shortest path length

Write a function called shortestPathLength that takes graph, start, and end as parameters and returns the shortest path length as a number. Use a queue for BFS, a Set called visited to track visited nodes, and a Map called distance to store distances from start. Return -1 if end is not reachable.

DSA Typescript

const graph: Map = new Map([
  ['A', ['B', 'C']],
  ['B', ['A', 'D', 'E']],
  ['C', ['A', 'F']],
  ['D', ['B']],
  ['E', ['B', 'F']],
  ['F', ['C', 'E']]
]);

const start: string = 'A';
const end: string = 'F';

// Your code here

Hint

Use a queue to explore nodes level by level. Track visited nodes to avoid repeats. Store distance from start for each node. Return distance when you reach the end.

4

Print the shortest path length

Call shortestPathLength(graph, start, end) and print the result using console.log.

DSA Typescript

const graph: Map = new Map([
  ['A', ['B', 'C']],
  ['B', ['A', 'D', 'E']],
  ['C', ['A', 'F']],
  ['D', ['B']],
  ['E', ['B', 'F']],
  ['F', ['C', 'E']]
]);

const start: string = 'A';
const end: string = 'F';

function shortestPathLength(graph: Map, start: string, end: string): number {
  const queue: string[] = [];
  const visited: Set = new Set();
  const distance: Map = new Map();

  queue.push(start);
  visited.add(start);
  distance.set(start, 0);

  while (queue.length > 0) {
    const current = queue.shift()!;
    if (current === end) {
      return distance.get(current)!;
    }
    const neighbors = graph.get(current) || [];
    for (const neighbor of neighbors) {
      if (!visited.has(neighbor)) {
        visited.add(neighbor);
        distance.set(neighbor, (distance.get(current) || 0) + 1);
        queue.push(neighbor);
      }
    }
  }

  return -1;
}

// Your code here

Hint

Call the function with the graph, start, and end nodes, then print the returned number.

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