Shortest Path in Unweighted Graph Using BFS in DSA Typescript - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of BFS shortest path distances

What is the output of the following TypeScript code that finds shortest path distances from node 0 in an unweighted graph?

DSA Typescript

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

const graph = [
  [1, 2],
  [0, 3],
  [0, 3],
  [1, 2, 4],
  [3]
];
console.log(bfsShortestPath(graph, 0));

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

B [0, 1, 1, 1, 2]

C [0, 2, 2, 3, 4]

D [0, 1, 1, 2, 3]

💡 Hint

Think about how BFS explores neighbors level by level starting from the start node.

The BFS starts at node 0 with distance 0. Nodes 1 and 2 are directly connected to 0, so distance 1. Node 3 is connected to 1 and 2, so distance 2. Node 4 is connected to 3, so distance 3.

🧠 Conceptual

intermediate

Understanding BFS queue behavior

In BFS for shortest path in an unweighted graph, what is the main reason for using a queue data structure?

A To explore nodes in the order they are discovered, ensuring shortest path layers are processed first

B To store nodes in sorted order by their values

C To keep track of visited nodes using a stack-like behavior

D To randomly select the next node to explore

💡 Hint

Think about how BFS explores nodes level by level.

A queue ensures nodes are processed in the order they are discovered, which corresponds to increasing distance from the start node. This guarantees shortest path discovery in unweighted graphs.

🔧 Debug

advanced

Identify the bug in BFS shortest path code

What error will this TypeScript BFS shortest path code produce when run, and why?

DSA Typescript

function bfsShortestPath(graph: number[][], start: number): number[] {
  const distances = Array(graph.length).fill(-1);
  const queue: number[] = [];
  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;
}

A TypeError because distances array is not initialized properly

B SyntaxError due to missing semicolon after queue.push(start)

C The code runs but returns incorrect distances because it uses pop() instead of shift() causing DFS behavior

D Runtime error because graph[node] is undefined

💡 Hint

Check how the queue is used to remove elements.

Using pop() removes the last element, making the traversal depth-first instead of breadth-first, so distances are incorrect for shortest path.

🚀 Application

advanced

Shortest path reconstruction from BFS distances

Given BFS distances from start node 0, how can you reconstruct the actual shortest path to node 4 in the graph below? Graph adjacency list: 0: [1, 2] 1: [0, 3] 2: [0, 3] 3: [1, 2, 4] 4: [3] Distances from 0: [0, 1, 1, 2, 3]

A Start from node 4 and move to any neighbor with distance one less until reaching 0, collecting nodes in reverse

B Start from node 0 and move to neighbors with larger distance until reaching 4

C Randomly pick neighbors until node 4 is reached

D Use a stack to store nodes visited during BFS and pop until node 4 is found

💡 Hint

Shortest path can be traced backward using distances.

By moving from the target node to neighbors with distance one less, you trace the shortest path back to the start node.

🧠 Conceptual

expert

Why BFS guarantees shortest path in unweighted graphs

Why does BFS always find the shortest path in an unweighted graph?

A Because BFS uses a stack to explore the deepest nodes first

B Because BFS explores nodes in order of increasing distance from the start node using a queue

C Because BFS sorts neighbors by their node values before visiting

D Because BFS uses recursion to explore all paths simultaneously

💡 Hint

Think about the order nodes are visited in BFS.

BFS visits nodes in layers based on their distance from the start node, so the first time a node is visited is via the shortest path.