Connected Components Using BFS in DSA Typescript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the visited set before BFS.

DSA Typescript

const visited = new Set[1]();

Drag options to blanks, or click blank then click option'

A<string>

B()

C[]

D{}

Attempts:

3 left

💡 Hint

Common Mistakes

Using brackets [] instead of parentheses () to create a Set.

Trying to initialize Set with curly braces {} which is an object literal.

We create a new Set by calling new Set() with parentheses.

2fill in blank

medium

Complete the code to add the starting node to the queue in BFS.

DSA Typescript

queue.push[1](startNode);

Drag options to blanks, or click blank then click option'

A.add

B.enqueue

C.push

D()

Attempts:

3 left

💡 Hint

Common Mistakes

Using .add which is for Sets, not arrays.

Writing queue.push without parentheses, which is a function reference, not a call.

We call the push method with parentheses and the node as argument: push(startNode).

3fill in blank

hard

Fix the error in the BFS loop condition to process nodes while queue is not empty.

DSA Typescript

while (queue[1] 0) {

Drag options to blanks, or click blank then click option'

A>

B==

C!==

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using == 0 which stops immediately.

Using < 0 which is never true for length.

We check if queue length is greater than 0 to continue processing nodes.

4fill in blank

hard

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

DSA Typescript

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

Drag options to blanks, or click blank then click option'

A.push

B.pop

D.shift

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop or shift to add elements instead of push.

Checking queue with wrong method instead of includes.

We check if neighbor is not in queue using includes, then add it with push.

5fill in blank

hard

Fill all three blanks to complete the BFS function that finds connected components.

DSA Typescript

function bfs(graph: Map<string, string[]>): string[][] {
  const visited = new Set<string>();
  const components: string[][] = [];
  for (const node of graph.keys()) {
    if (!visited.has(node)) {
      const queue = [node];
      const component: string[] = [];
      while (queue.length [1] 0) {
        const current = queue[2]();
        if (!visited.has(current)) {
          visited.add(current);
          component.push(current);
          for (const neighbor of graph.get(current) || []) {
            if (!visited.has(neighbor)) {
              queue.push(neighbor);
            }
          }
        }
      }
      components.push(component);
    }
  }
  return components;
}

Drag options to blanks, or click blank then click option'

A>

Bshift

Cpop

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop() removes last element, which is DFS behavior.

Using < 0 or == 0 in loop condition stops loop incorrectly.

We loop while queue length > 0, remove the first element with shift. pop is a distractor here.