DFS Depth First Search on Graph in DSA Typescript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to start DFS from the given node.

DSA Typescript

function dfs(graph: Record<string, string[]>, node: string, visited: Set<string>) {
  if (visited.has(node)) return;
  visited.add(node);
  console.log(node);
  for (const neighbor of graph[node]) {
    dfs(graph, [1], visited);
  }
}

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

Anode

Bneighbor

Cvisited

Dgraph

Attempts:

3 left

💡 Hint

Common Mistakes

Passing the current node again instead of the neighbor.

Passing the visited set incorrectly.

Passing the whole graph instead of a node.

The DFS function should call itself recursively on each neighbor of the current node.

2fill in blank

medium

Complete the code to check if a node has been visited before processing it.

DSA Typescript

function dfs(graph: Record<string, string[]>, node: string, visited: Set<string>) {
  if ([1]) return;
  visited.add(node);
  console.log(node);
  for (const neighbor of graph[node]) {
    dfs(graph, neighbor, visited);
  }
}

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

Avisited.contains(node)

Bnode in visited

Cvisited.has(node)

Dvisited.includes(node)

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'in' operator which doesn't work for Set.

Using contains or includes which are not Set methods.

The Set method to check if an element exists is has().

3fill in blank

hard

Fix the error in the DFS function to avoid infinite recursion.

DSA Typescript

function dfs(graph: Record<string, string[]>, node: string, visited: Set<string>) {
  if (visited.has(node)) return;
  visited.add(node);
  console.log(node);
  for (const neighbor of graph[node]) {
    dfs(graph, neighbor, [1]);
  }
}

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

Avisited

Bnew Set()

Cnew Set([node])

Dundefined

Attempts:

3 left

💡 Hint

Common Mistakes

Passing a new Set() causes infinite recursion.

Passing undefined causes runtime errors.

The same visited set must be passed to keep track of visited nodes and prevent infinite loops.

4fill in blank

hard

Fill both blanks to create a DFS function that returns all visited nodes as an array.

DSA Typescript

function dfs(graph: Record<string, string[]>, node: string, visited: Set<string>, result: string[]) {
  if ([1]) return;
  visited.add(node);
  result.push(node);
  for (const neighbor of graph[node]) {
    dfs(graph, neighbor, visited, [2]);
  }
  return result;
}

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

Avisited.has(node)

Bresult

Cvisited

Dnode

Attempts:

3 left

💡 Hint

Common Mistakes

Passing visited instead of result to recursive calls.

Not checking if node is visited before processing.

Check if node is visited with visited.has(node) and pass the result array to recursive calls to collect nodes.

5fill in blank

hard

Fill all three blanks to implement DFS that returns nodes in order visited, avoiding cycles.

DSA Typescript

function dfs(graph: Record<string, string[]>, node: string, visited: Set<string> = new Set(), result: string[] = []) {
  if ([1]) return result;
  visited.add(node);
  result.push(node);
  for (const neighbor of graph[node]) {
    dfs(graph, [2], visited, [3]);
  }
  return result;
}

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

Avisited.has(node)

Bneighbor

Cresult

Dnode

Attempts:

3 left

💡 Hint

Common Mistakes

Using node instead of neighbor in recursive call.

Not passing the result array causing incomplete results.

Check if node is visited with visited.has(node), recurse on neighbor, and pass the result array to collect visited nodes.

Practice

1. What is the main idea behind Depth First Search (DFS) on a graph?

easy

A. Explore as far as possible along each branch before backtracking

B. Visit all neighbors of a node before moving deeper

C. Visit nodes in order of their distance from the start node

D. Randomly visit nodes until all are visited

Solution

1. Step 1: Understand DFS traversal strategy

   DFS explores nodes by going deep into one branch before backtracking to explore other branches.

2. Step 2: Compare with other traversal methods

   BFS visits neighbors first, but DFS goes deep first, matching Explore as far as possible along each branch before backtracking.

3. Final Answer:

   Explore as far as possible along each branch before backtracking -> Option A

4. Quick Check:

   DFS = deep exploration first [OK]

Hint: DFS goes deep first, BFS goes wide first [OK]

Common Mistakes:

• Confusing DFS with BFS
• Thinking DFS visits neighbors first
• Assuming DFS visits nodes by distance

2. Which of the following is the correct TypeScript function signature for a DFS function on a graph represented as an adjacency list?

easy

A. function dfs(graph: Map, start: number): void

B. function dfs(graph: number[], start: number): void

C. function dfs(graph: number[][], start: number): void

D. function dfs(graph: string[][], start: string): void

Solution

1. Step 1: Identify graph representation type

   An adjacency list for a graph with numbered nodes is usually number[][], where each index holds neighbors.

2. Step 2: Check parameter types

   Start node is a number, so function signature with graph: number[][] and start: number is correct.

3. Final Answer:

   function dfs(graph: number[][], start: number): void -> Option C

4. Quick Check:

   Adjacency list = number[][], start node = number [OK]

Hint: Adjacency list uses array of arrays for neighbors [OK]

Common Mistakes:

• Using number[] instead of number[][] for adjacency list
• Using Map with incorrect value type
• Using string types for numeric graphs

3. What is the output of the following DFS traversal starting from node 0?

const graph = [[1, 2], [2], [0, 3], [3]];
const visited: boolean[] = [];
const result: number[] = [];
function dfs(node: number) {
  if (visited[node]) return;
  visited[node] = true;
  result.push(node);
  for (const neighbor of graph[node]) {
    dfs(neighbor);
  }
}
dfs(0);
console.log(result);

medium

A. [0, 2, 3, 1]

B. [0, 2, 1, 3]

C. [0, 1, 2]

D. [0, 1, 2, 3]

Solution

1. Step 1: Start DFS at node 0

   Visit 0, mark visited, add to result: [0]. Neighbors: 1, 2.

2. Step 2: Visit neighbor 1

   Visit 1, mark visited, add: [0,1]. Neighbor: 2.

3. Step 3: Visit neighbor 2 from 1

   Visit 2, mark visited, add: [0,1,2]. Neighbors: 0 (visited), 3.

4. Step 4: Visit neighbor 3 from 2

   Visit 3, mark visited, add: [0,1,2,3]. Neighbor: 3 (already visited).

5. Final Answer:

   [0, 2, 1, 3] -> Option B

6. Quick Check:

   DFS order = [0,2,1,3] [OK]

Hint: Follow neighbors recursively, mark visited to avoid repeats [OK]

Common Mistakes:

• Not marking nodes visited before recursion
• Visiting neighbors in wrong order
• Ignoring already visited nodes

4. Identify the bug in this DFS implementation and choose the fix:

function dfs(node: number, graph: number[][], visited: boolean[]) {
  visited[node] = true;
  for (const neighbor of graph[node]) {
    if (!visited[neighbor]) {
      dfs(neighbor, graph, visited);
    }
  }
}

const graph = [[1], [2], [0]];
const visited = new Array(graph.length).fill(false);
dfs(0, graph, visited);
console.log(visited);

medium

A. Mark visited[node] = true after recursive calls

B. Initialize visited array with true instead of false

C. Add a base case to return if node is already visited at start

D. No bug, code works correctly

Solution

1. Step 1: Check visited marking timing

   Visited[node] is marked true before recursion, preventing revisits.

2. Step 2: Verify recursion and base case

   Recursion only happens if neighbor not visited, so no infinite loops.

3. Final Answer:

   No bug, code works correctly -> Option D

4. Quick Check:

   Visited marked before recursion = correct DFS [OK]

Hint: Mark visited before recursion to avoid cycles [OK]

Common Mistakes:

• Marking visited after recursion causing infinite loops
• Initializing visited array incorrectly
• Missing base case check

5. Given a graph with possible cycles, how can you modify DFS to detect if a cycle exists starting from a given node? Choose the best approach.

hard

A. Use a recursion stack array to track nodes in the current path and detect back edges

B. Only mark nodes visited after all neighbors are visited

C. Use BFS instead of DFS to detect cycles

D. Ignore visited nodes and continue traversal

Solution

1. Step 1: Understand cycle detection in DFS

   Detecting cycles requires knowing if a node is in the current recursion path (stack).

2. Step 2: Use recursion stack tracking

   Maintain a separate array to track nodes currently in recursion; if a neighbor is in this stack, a cycle exists.

3. Step 3: Evaluate other options

   Marking visited after neighbors or using BFS won't reliably detect cycles; ignoring visited nodes causes infinite loops.

4. Final Answer:

   Use a recursion stack array to track nodes in the current path and detect back edges -> Option A

5. Quick Check:

   Cycle detection = recursion stack tracking [OK]

Hint: Track recursion path to find back edges indicating cycles [OK]

Common Mistakes:

• Not tracking recursion stack causing missed cycles
• Using BFS which doesn't track recursion path
• Marking visited too late or ignoring visited nodes