Topological Sort Using DFS in DSA Javascript - Time & Space Complexity

Time Complexity: Topological Sort Using DFS

O(n + e)

Understanding Time Complexity

We want to understand how the time needed to perform topological sort using DFS changes as the graph size grows.

Specifically, how the number of nodes and edges affects the work done.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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();
}
    

This code performs a topological sort on a directed graph using depth-first search (DFS).

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: DFS visits each node and explores all its neighbors.
• How many times: Each node is visited once, and each edge is checked once during DFS.

How Execution Grows With Input

As the number of nodes and edges increases, the total work grows roughly with the sum of nodes and edges.

Input Size (n nodes, e edges)	Approx. Operations
10 nodes, 15 edges	About 25 operations (10 + 15)
100 nodes, 200 edges	About 300 operations (100 + 200)
1000 nodes, 5000 edges	About 6000 operations (1000 + 5000)

Pattern observation: The work grows linearly with the number of nodes plus edges.

Final Time Complexity

Time Complexity: O(n + e)

This means the time needed grows in a straight line with the total number of nodes and edges in the graph.

Common Mistake

[X] Wrong: "DFS visits nodes multiple times, so time is O(n * e)."

[OK] Correct: Each node is marked visited and processed only once, so edges are checked only once, not repeatedly.

Interview Connect

Understanding this time complexity helps you explain how efficient graph algorithms work and shows you can analyze recursive traversals clearly.

Self-Check

"What if the graph is represented as an adjacency matrix instead of adjacency lists? How would the time complexity change?"

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