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

Practice - 5 Tasks

1fill in blank

easy

Complete the code to start DFS from the given node.

DSA Javascript

function dfs(graph, start) {
  const visited = new Set();
  function explore(node) {
    if (visited.has(node)) return;
    visited.add(node);
    for (const neighbor of graph[node]) {
      explore(neighbor);
    }
  }
  explore([1]);
  return visited;
}

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

Avisited

Bgraph

Cnode

Dstart

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'graph' instead of 'start' as the initial node.

Using 'node' which is undefined in this scope.

Using 'visited' which is a Set, not a node.

The DFS starts from the given 'start' node, so we call explore(start).

2fill in blank

medium

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

DSA Javascript

function explore(node) {
  if ([1]) return;
  visited.add(node);
  for (const neighbor of graph[node]) {
    explore(neighbor);
  }
}

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

Agraph.has(node)

Bvisited.add(node)

Cvisited.has(node)

Dnode in visited

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'visited.add(node)' which adds instead of checks.

Using 'node in visited' which does not work for Set.

Using 'graph.has(node)' which checks graph keys, not visited nodes.

We check if the node is already in the visited set using visited.has(node).

3fill in blank

hard

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

DSA Javascript

for (const neighbor of [1]) {
  explore(neighbor);
}

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

Avisited

Bgraph[node]

Cgraph

Dnode

Attempts:

3 left

💡 Hint

Common Mistakes

Looping over 'graph' which is the whole graph object.

Looping over 'visited' which is a Set of visited nodes.

Looping over 'node' which is a single node, not iterable.

We need to loop over the neighbors of the current node, which are in graph[node].

4fill in blank

hard

Fill both blanks to create a DFS function that returns the visited nodes set.

DSA Javascript

function dfs(graph, [1]) {
  const visited = new Set();
  function explore(node) {
    if (visited.has(node)) return;
    visited.add(node);
    for (const neighbor of graph[node]) {
      explore(neighbor);
    }
  }
  explore([2]);
  return visited;
}

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

Astart

Bnode

Cgraph

Dvisited

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variable names for parameter and call.

Using 'graph' or 'visited' instead of the start node.

The parameter is 'start' and we call explore(start) to begin DFS.

5fill in blank

hard

Fill all three blanks to create a DFS that collects nodes in order visited in an array.

DSA Javascript

function dfs(graph, [1]) {
  const visited = new Set();
  const order = [];
  function explore(node) {
    if (visited.has(node)) return;
    visited.add(node);
    order.push([2]);
    for (const neighbor of graph[node]) {
      explore(neighbor);
    }
  }
  explore([3]);
  return order;
}

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

Astart

Bnode

Dvisited

Attempts:

3 left

💡 Hint

Common Mistakes

Pushing 'start' instead of 'node' to order.

Calling explore with 'node' which is undefined outside.

Using 'visited' instead of node or start.

Parameter is 'start', we push 'node' to order, and call explore(start).

Practice

1. What does DFS (Depth First Search) do when traversing a graph?

easy

A. It only visits nodes connected directly to the start node.

B. It visits nodes level by level from the start node.

C. It explores as far as possible along each branch before backtracking.

D. It randomly visits nodes without any order.

Solution

1. Step 1: Understand DFS traversal

   DFS explores a path deeply before moving to another path.

2. Step 2: Compare with other traversals

   BFS visits nodes level by level, but DFS goes deep first.

3. Final Answer:

   It explores as far as possible along each branch before backtracking. -> Option C

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 nodes randomly
• Assuming DFS only visits direct neighbors

2. Which of the following is the correct way to start a DFS function in JavaScript for a graph represented as adjacency list graph?

easy

A. function dfs(node, visited = new Set()) {

B. function dfs(node, visited = []) {

C. function dfs(node, visited = {}) {

D. function dfs(node, visited = 0) {

Solution

1. Step 1: Identify suitable data structure for visited nodes

   A Set is ideal to track visited nodes efficiently.

2. Step 2: Check function signature correctness

   Using visited = new Set() initializes visited nodes properly.

3. Final Answer:

   function dfs(node, visited = new Set()) { -> Option A

4. Quick Check:

   Use Set for visited nodes [OK]

Hint: Use Set to track visited nodes in DFS [OK]

Common Mistakes:

• Using array or object incorrectly for visited
• Initializing visited as a number
• Not initializing visited parameter

3. Given the graph adjacency list:

const graph = {
  1: [2, 3],
  2: [4],
  3: [],
  4: []
};

function dfs(node, visited = new Set()) {
  visited.add(node);
  for (const neighbor of graph[node]) {
    if (!visited.has(neighbor)) {
      dfs(neighbor, visited);
    }
  }
  return Array.from(visited);
}

console.log(dfs(1));

What is the output?

medium

A. [1, 3, 2, 4]

B. [1, 4, 2, 3]

C. [1, 2, 3, 4]

D. [1, 2, 4, 3]

Solution

1. Step 1: Start DFS from node 1

   Visit 1, then neighbors 2 and 3 in order.

2. Step 2: Visit neighbors recursively

   From 2, visit 4. From 3, no neighbors. Order: 1, 2, 4, 3.

3. Final Answer:

   [1, 2, 4, 3] -> Option D

4. Quick Check:

   DFS order = 1,2,4,3 [OK]

Hint: Follow neighbors deeply before next sibling [OK]

Common Mistakes:

• Visiting neighbors in wrong order
• Confusing BFS order with DFS
• Not converting Set to array properly

4. Find the error in this DFS code snippet for graph traversal:

const graph = {
  1: [2, 3],
  2: [4],
  3: [],
  4: []
};

function dfs(node, visited = new Set()) {
  visited.add(node);
  for (const neighbor of graph[node]) {
    if (visited.has(neighbor)) {
      dfs(neighbor, visited);
    }
  }
  return visited;
}

console.log(Array.from(dfs(1)));

medium

A. The condition should be if (!visited.has(neighbor)) instead of if (visited.has(neighbor))

B. The visited set should be an array instead

C. The graph object keys should be strings, not numbers

D. The function should return an array, not a set

Solution

1. Step 1: Check the neighbor visit condition

   It incorrectly calls dfs if neighbor is already visited, causing infinite recursion.

2. Step 2: Correct the condition

   Change to if (!visited.has(neighbor)) to visit unvisited neighbors only.

3. Final Answer:

   The condition should be if (!visited.has(neighbor)) instead of if (visited.has(neighbor)) -> Option A

4. Quick Check:

   Visit only unvisited neighbors [OK]

Hint: Call DFS only if neighbor not visited [OK]

Common Mistakes:

• Calling DFS on already visited nodes
• Using wrong data structure for visited
• Returning wrong type from DFS

5. You want to find if there is a path from node 1 to node 5 in this graph:

const graph = {
  1: [2, 3],
  2: [4],
  3: [5],
  4: [],
  5: []
};

function dfs(node, target, visited = new Set()) {
  if (node === target) return true;
  visited.add(node);
  for (const neighbor of graph[node]) {
    if (!visited.has(neighbor)) {
      if (dfs(neighbor, target, visited)) return true;
    }
  }
  return false;
}

console.log(dfs(1, 5));

What will be the output and why?

hard

A. false, because the visited set is not updated correctly

B. true, because DFS finds a path 1 -> 3 -> 5

C. true, because DFS visits all nodes regardless of target

D. false, because node 5 is not directly connected to 1

Solution

1. Step 1: Check DFS path search logic

   DFS returns true immediately when target node is found.

2. Step 2: Trace path from 1 to 5

   Path exists: 1 -> 3 -> 5, so DFS returns true.

3. Final Answer:

   true, because DFS finds a path 1 -> 3 -> 5 -> Option B

4. Quick Check:

   DFS finds path = true [OK]

Hint: DFS returns true when target found [OK]

Common Mistakes:

• Assuming direct connection needed
• Ignoring early return on target found
• Not updating visited set properly