Topological Sort Using DFS in DSA Javascript - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Topological Sort on a Simple DAG

What is the output of the following JavaScript code performing topological sort using DFS on the given graph?

DSA Javascript

function topologicalSort(graph) {
  const visited = new Set();
  const stack = [];

  function dfs(node) {
    visited.add(node);
    for (const neighbor of graph[node]) {
      if (!visited.has(neighbor)) {
        dfs(neighbor);
      }
    }
    stack.push(node);
  }

  for (const node in graph) {
    if (!visited.has(node)) {
      dfs(node);
    }
  }

  return stack.reverse();
}

const graph = {
  5: [2, 0],
  4: [0, 1],
  2: [3],
  3: [1],
  0: [],
  1: []
};

console.log(topologicalSort(graph));

A [5, 4, 2, 3, 1, 0]

B [4, 5, 2, 3, 1, 0]

C [5, 4, 0, 1, 2, 3]

D [4, 5, 3, 2, 1, 0]

💡 Hint

Remember that DFS pushes nodes after visiting all neighbors, so the stack reversed gives the topological order.

The DFS visits nodes starting from 5 and 4. Node 5 leads to 2 and 0, and 2 leads to 3, which leads to 1. Node 4 leads to 0 and 1. The final reversed stack is [5, 4, 2, 3, 1, 0].

❓ Predict Output

intermediate

Detecting Output for a Graph with Multiple Independent Chains

What is the output of the topological sort using DFS on this graph?

DSA Javascript

function topologicalSort(graph) {
  const visited = new Set();
  const stack = [];

  function dfs(node) {
    visited.add(node);
    for (const neighbor of graph[node]) {
      if (!visited.has(neighbor)) {
        dfs(neighbor);
      }
    }
    stack.push(node);
  }

  for (const node in graph) {
    if (!visited.has(node)) {
      dfs(node);
    }
  }

  return stack.reverse();
}

const graph = {
  a: ['b'],
  b: ['c'],
  c: [],
  d: ['e'],
  e: []
};

console.log(topologicalSort(graph));

A [a, b, c, d, e]

B [d, e, a, b, c]

C [a, d, b, e, c]

D [d, a, e, b, c]

💡 Hint

The DFS visits nodes in the order of object keys, so 'a' then 'b' then 'c', then 'd' then 'e'.

DFS starts at 'a', visits 'b' and 'c', then pushes them in stack order. Then DFS starts at 'd', visits 'e', and pushes them. The reversed stack is [d, e, a, b, c].

🔧 Debug

advanced

Identify the Error in DFS Topological Sort Implementation

What error will this code produce when run, and why?

DSA Javascript

function topologicalSort(graph) {
  const visited = new Set();
  const stack = [];

  function dfs(node) {
    if (visited.has(node)) return;
    for (const neighbor of graph[node]) {
      dfs(neighbor);
    }
    visited.add(node);
    stack.push(node);
  }

  for (const node in graph) {
    dfs(node);
  }

  return stack.reverse();
}

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

console.log(topologicalSort(graph));

A Output: [3, 2, 1]

B Correct output: [1, 2, 3]

C Stack overflow due to infinite recursion

D TypeError: Cannot read property of undefined

💡 Hint

Check when nodes are marked visited in the DFS function.

The code calls dfs on neighbors before marking the current node visited, so it revisits nodes repeatedly causing infinite recursion and stack overflow.

❓ Predict Output

advanced

Topological Sort Output with Cycle Detection Missing

What will be the output or error of this topological sort code on a graph with a cycle?

DSA Javascript

function topologicalSort(graph) {
  const visited = new Set();
  const stack = [];

  function dfs(node) {
    visited.add(node);
    for (const neighbor of graph[node]) {
      if (!visited.has(neighbor)) {
        dfs(neighbor);
      }
    }
    stack.push(node);
  }

  for (const node in graph) {
    if (!visited.has(node)) {
      dfs(node);
    }
  }

  return stack.reverse();
}

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

console.log(topologicalSort(graph));

A Stack overflow due to infinite recursion

B [1, 2, 3]

C [3, 2, 1]

D Output with repeated nodes in stack

💡 Hint

The graph has a cycle but the code does not detect it.

The DFS revisits nodes in the cycle endlessly because visited is only marked after recursion, causing infinite recursion and stack overflow.

🧠 Conceptual

expert

Understanding the Role of the Stack in DFS Topological Sort

Why does the DFS-based topological sort algorithm push nodes onto a stack only after visiting all their neighbors?

A To ensure nodes appear before their dependencies in the final order

B To detect cycles in the graph during traversal

C To reduce the time complexity of the algorithm

D To ensure nodes appear after their dependencies in the final order

💡 Hint

Think about the order in which nodes must appear in a topological sort.

Pushing nodes after visiting all neighbors ensures that dependencies come before the node in the stack. Reversing the stack gives the correct topological order where each node appears after its dependencies.

Practice

1. What is the main purpose of using DFS in Topological Sort on a directed graph?

easy

A. To count the number of connected components

B. To find the shortest path between two nodes

C. To detect cycles in an undirected graph

D. To visit nodes in a way that ensures dependencies come before dependents

Solution

1. Step 1: Understand DFS role in Topological Sort

   DFS visits nodes deeply, ensuring all nodes reachable from a node are visited before it is processed.

2. Step 2: Relate DFS order to dependencies

   In Topological Sort, nodes are added after visiting all their dependencies, so dependencies come before dependents.

3. Final Answer:

   To visit nodes in a way that ensures dependencies come before dependents -> Option D

4. Quick Check:

   DFS ensures dependency order = A [OK]

Hint: DFS post-order gives correct dependency order [OK]

Common Mistakes:

• Confusing DFS with BFS for topological sorting
• Thinking DFS finds shortest paths
• Mixing cycle detection with topological order

2. Which of the following is the correct way to mark a node as visited in a DFS-based topological sort in JavaScript?

easy

A. visited = node;

B. visited.push(node);

C. visited[node] = true;

D. visited[node] == true;

Solution

1. Step 1: Understand visited marking in DFS

   Visited is usually an array or object where each node's visited status is stored as a boolean.

2. Step 2: Correct syntax for marking visited

   Assigning true to visited[node] marks that node visited; using '==' is comparison, not assignment.

3. Final Answer:

   visited[node] = true; -> Option C

4. Quick Check:

   Assign true to visited[node] = A [OK]

Hint: Use assignment (=), not comparison (==), to mark visited [OK]

Common Mistakes:

• Using '==' instead of '=' for assignment
• Using push on visited array instead of index assignment
• Overwriting visited variable instead of marking node

3. What will be the output of the following code snippet performing topological sort using DFS on the graph with edges: 5->2, 5->0, 4->0, 4->1, 2->3, 3->1?

const graph = {
  5: [2, 0],
  4: [0, 1],
  2: [3],
  3: [1],
  0: [],
  1: []
};

const visited = {};
const stack = [];

function dfs(node) {
  visited[node] = true;
  for (const neighbor of graph[node]) {
    if (!visited[neighbor]) dfs(neighbor);
  }
  stack.push(node);
}

for (const node in graph) {
  if (!visited[node]) dfs(node);
}

console.log(stack.reverse());

medium

A. [5, 4, 2, 3, 1, 0]

B. [4, 5, 0, 2, 3, 1]

C. [5, 4, 0, 2, 3, 1]

D. [4, 5, 2, 3, 1, 0]

Solution

1. Step 1: Perform DFS and track stack push order

   DFS visits nodes deeply, pushing nodes after visiting neighbors. For node 5: visits 2, then 3, then 1, then 0; for node 4: visits 0 and 1. Stack push order is [1,3,2,0,5,4] reversed to [4,5,0,2,3,1].

2. Step 2: Reverse stack to get topological order

   Reversing stack gives [4, 5, 0, 2, 3, 1], which respects dependencies.

3. Final Answer:

   [4, 5, 0, 2, 3, 1] -> Option B

4. Quick Check:

   DFS post-order reversed = [4,5,0,2,3,1] [OK]

Hint: Reverse DFS finishing order for topological sort output [OK]

Common Mistakes:

• Not reversing the stack before output
• Visiting nodes in wrong order due to object key iteration
• Confusing push order with final topological order

4. Identify the bug in this DFS-based topological sort code snippet:

function dfs(node, graph, visited, stack) {
  if (visited[node]) return;
  visited[node] = true;
  for (const neighbor of graph[node]) {
    dfs(neighbor, graph, visited, stack);
  }
  stack.push(node);
}

const graph = {0: [1], 1: [2], 2: []};
const visited = {};
const stack = [];
for (const node in graph) {
  dfs(node, graph, visited, stack);
}
console.log(stack.reverse());

medium

A. No bug; code works correctly

B. Visited check missing before recursive call

C. Stack should be cleared before DFS

D. Visited should be reset inside DFS function

Solution

1. Step 1: Analyze visited check and recursion

   The code checks if node is visited before recursion, preventing repeated visits.

2. Step 2: Confirm stack usage and output

   Nodes are pushed after visiting neighbors, and stack is reversed for correct order.

3. Final Answer:

   No bug; code works correctly -> Option A

4. Quick Check:

   No bug; code works correctly = C [OK]

Hint: Check visited before recursion to avoid repeats [OK]

Common Mistakes:

• Calling dfs without visited check causing infinite recursion
• Not reversing stack before output
• Resetting visited inside DFS causing wrong results

5. Given a directed graph with nodes and edges, how can you modify the DFS-based topological sort to detect if the graph has a cycle?

hard

A. Use an extra recursion stack to track nodes in the current DFS path and detect back edges

B. Count the number of nodes visited and compare with total nodes

C. Sort nodes by their degree before DFS

D. Run BFS instead of DFS to detect cycles

Solution

1. Step 1: Understand cycle detection in DFS

   Cycle exists if a node is visited again in the current DFS path (back edge).

2. Step 2: Use recursion stack to track current path

   Maintain a recursion stack array marking nodes currently in DFS call; if a neighbor is in recursion stack, cycle detected.

3. Final Answer:

   Use an extra recursion stack to track nodes in the current DFS path and detect back edges -> Option A

4. Quick Check:

   Recursion stack detects cycles = D [OK]

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

Common Mistakes:

• Only using visited array without recursion stack
• Trying to detect cycles by node count mismatch
• Using BFS without proper cycle detection logic