BFS Breadth First Search on Graph in DSA Typescript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the queue with the start node.

DSA Typescript

const queue: number[] = [[1]];

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Agraph

B0

CstartNode

Dvisited

Attempts:

3 left

💡 Hint

Common Mistakes

Adding the whole graph or visited set instead of the start node.

Starting with an empty queue.

We start BFS by adding the start node to the queue to begin traversal.

2fill in blank

medium

Complete the code to mark the current node as visited.

DSA Typescript

visited[[1]] = true;

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Agraph

Bqueue

CstartNode

Dnode

Attempts:

3 left

💡 Hint

Common Mistakes

Marking the queue or graph as visited instead of the node.

Not marking nodes as visited, causing infinite loops.

We mark the current node as visited to avoid revisiting it.

3fill in blank

hard

Fix the error in the loop that processes neighbors of the current node.

DSA Typescript

for (const neighbor of graph[[1]]) {

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Anode

Bqueue

Cvisited

DstartNode

Attempts:

3 left

💡 Hint

Common Mistakes

Using queue or visited as index causes errors or wrong neighbors.

Using startNode instead of current node inside the loop.

We iterate over neighbors of the current node, so we use 'node' as the index.

4fill in blank

hard

Fill both blanks to check if a neighbor is unvisited and add it to the queue.

DSA Typescript

if (!visited[[1]]) {
  queue.[2](neighbor);
}

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Aneighbor

Bpush

Cpop

Dnode

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop instead of push adds/removes wrong elements.

Checking visited status of wrong variable.

We check if the neighbor is not visited and then add it to the queue using push.

5fill in blank

hard

Fill all three blanks to complete the BFS function that returns the order of visited nodes.

DSA Typescript

function bfs(graph: number[][], startNode: number): number[] {
  const visited: boolean[] = new Array(graph.length).fill(false);
  const queue: number[] = [startNode];
  const order: number[] = [];

  while (queue.length > 0) {
    const [1] = queue.shift()!;
    if (!visited[[2]]) {
      visited[[3]] = true;
      order.push(node);

      for (const neighbor of graph[node]) {
        if (!visited[neighbor]) {
          queue.push(neighbor);
        }
      }
    }
  }
  return order;
}

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Anode

BstartNode

Dqueue

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variable names inconsistently causes errors.

Not marking nodes as visited before processing neighbors.

We assign the shifted node to 'node', check and mark visited[node], and use 'node' consistently.