Topological Sort Using Kahn's Algorithm BFS in DSA Java - Time & Space Complexity

Time Complexity: Topological Sort Using Kahn's Algorithm BFS

O(V + E)

Understanding Time Complexity

We want to understand how the time needed to perform topological sort using Kahn's Algorithm grows as the graph size increases.

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

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public List<Integer> kahnTopologicalSort(int V, List<List<Integer>> adj) {
    int[] indegree = new int[V];
    for (int i = 0; i < V; i++) {
        for (int node : adj.get(i)) {
            indegree[node]++;
        }
    }
    Queue<Integer> q = new LinkedList<>();
    for (int i = 0; i < V; i++) {
        if (indegree[i] == 0) q.add(i);
    }
    List<Integer> topo = new ArrayList<>();
    while (!q.isEmpty()) {
        int u = q.poll();
        topo.add(u);
        for (int v : adj.get(u)) {
            indegree[v]--;
            if (indegree[v] == 0) q.add(v);
        }
    }
    return topo;
}
    

This code performs a topological sort on a directed graph using Kahn's BFS algorithm.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Traversing all edges and nodes in the graph.
• How many times: Each edge is visited once when counting indegrees and once when processing nodes from the queue.

How Execution Grows With Input

The work grows with the number of nodes and edges. For example:

Input Size (V = nodes, E = edges)	Approx. Operations
V=10, E=15	About 25 operations (counting edges and nodes)
V=100, E=200	About 300 operations
V=1000, E=5000	About 6000 operations

Pattern observation: The operations grow roughly proportional to the sum of nodes and edges.

Final Time Complexity

Time Complexity: O(V + E)

This means the time needed grows linearly with the total number of nodes plus edges in the graph.

Common Mistake

[X] Wrong: "The algorithm takes O(V²) time because of nested loops."

[OK] Correct: The inner loops do not run V times for each node; they run once per edge overall, so total work is proportional to edges, not nodes squared.

Interview Connect

Understanding this time complexity helps you explain why Kahn's Algorithm is efficient for scheduling and dependency problems in real projects.

Self-Check

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

Practice

1. What is the main purpose of Kahn's Algorithm in graph theory?

easy

A. To find a topological ordering of a Directed Acyclic Graph (DAG)

B. To detect cycles in an undirected graph

C. To find the shortest path between two nodes

D. To calculate the minimum spanning tree

Solution

1. Step 1: Understand Kahn's Algorithm goal

   Kahn's Algorithm is designed to produce a linear ordering of vertices in a Directed Acyclic Graph (DAG) such that for every directed edge u -> v, u comes before v in the ordering.

2. Step 2: Compare with other graph algorithms

   Other options describe different algorithms: cycle detection, shortest path, and minimum spanning tree are unrelated to Kahn's Algorithm.

3. Final Answer:

   To find a topological ordering of a Directed Acyclic Graph (DAG) -> Option A

4. Quick Check:

   Topological sort = Kahn's Algorithm purpose [OK]

Hint: Kahn's Algorithm always finds order in DAGs [OK]

Common Mistakes:

• Confusing topological sort with shortest path
• Thinking it detects cycles in undirected graphs
• Mixing with minimum spanning tree algorithms

2. Which Java data structure is typically used to store nodes with zero incoming edges in Kahn's Algorithm?

easy

A. Queue

B. HashSet

C. Stack

D. PriorityQueue

Solution

1. Step 1: Identify the data structure for zero in-degree nodes

   Kahn's Algorithm uses a queue to process nodes with zero incoming edges in FIFO order to maintain correct topological order.

2. Step 2: Eliminate other options

   Stack is LIFO, which can change order; HashSet does not maintain order; PriorityQueue is not required for basic Kahn's Algorithm.

3. Final Answer:

   Queue -> Option A

4. Quick Check:

   Zero in-degree nodes stored in queue [OK]

Hint: Use queue for BFS order in Kahn's Algorithm [OK]

Common Mistakes:

• Using stack instead of queue
• Choosing unordered sets instead of ordered structures
• Confusing with priority queue usage

3. What is the output of the following Java code snippet using Kahn's Algorithm on this graph?

Graph edges: 5->0, 5->2, 4->0, 4->1, 2->3, 3->1

// After running Kahn's Algorithm, print the topological order.

medium

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

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

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

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

Solution

1. Step 1: Calculate in-degree and start nodes

   Nodes 4 and 5 have zero in-degree. Start queue with [4,5] (assuming ascending order).

2. Step 2: Process nodes in queue

   Process 4: output [4], reduce 0 to 1, 1 to 1. Process 5: output [4,5], reduce 0 to 0 (add 0), 2 to 0 (add 2); queue=[0,2]. Process 0: [4,5,0], no neighbors. Process 2: [4,5,0,2], reduce 3 to 0 (add 3). Process 3: [4,5,0,2,3], reduce 1 to 0 (add 1). Process 1: [4,5,0,2,3,1].

3. Final Answer:

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

4. Quick Check:

   Topological order = [4,5,0,2,3,1] [OK]

Hint: Start with zero in-degree nodes, process neighbors, repeat [OK]

Common Mistakes:

• Mixing order of nodes with zero in-degree
• Not updating in-degree after removing edges
• Adding nodes to queue too early or late

4. Identify the bug in this Java snippet implementing Kahn's Algorithm:

while (!queue.isEmpty()) {
  int node = queue.poll();
  result.add(node);
  for (int neighbor : graph.get(node)) {
    inDegree[neighbor]--;
    if (inDegree[node] == 0) {
      queue.add(neighbor);
    }
}

medium

A. Neighbors should not be iterated in this loop

B. Queue should be a stack for correct order

C. Result list should be cleared inside the loop

D. Incorrect check: should check inDegree[neighbor] == 0, not inDegree[node]

Solution

1. Step 1: Analyze the in-degree update and condition

   After decrementing inDegree[neighbor], the code incorrectly checks if inDegree[node] == 0 instead of inDegree[neighbor] == 0.

2. Step 2: Correct the condition to add neighbors with zero in-degree

   The correct condition is if (inDegree[neighbor] == 0) to add that neighbor to the queue for processing.

3. Final Answer:

   Incorrect check: should check inDegree[neighbor] == 0, not inDegree[node] -> Option D

4. Quick Check:

   Check zero in-degree on neighbor, not current node [OK]

Hint: Check in-degree of neighbor after decrement [OK]

Common Mistakes:

• Checking in-degree of wrong node
• Using stack instead of queue
• Clearing result inside loop mistakenly

5. Given a graph with edges: 1->2, 1->3, 3->4, 2->4, 4->5, and 5->6, which of the following is a valid topological order using Kahn's Algorithm?

hard

A. [1, 2, 3, 5, 4, 6]

B. [1, 3, 2, 4, 5, 6]

C. [6, 5, 4, 3, 2, 1]

D. [1, 2, 3, 4, 6, 5]

Solution

1. Step 1: Identify nodes with zero in-degree and dependencies

   Node 1 has zero in-degree. Nodes 2 and 3 depend on 1; 4 depends on 2 and 3; 5 depends on 4; 6 depends on 5.

2. Step 2: Validate each option for correct order

   [1, 3, 2, 4, 5, 6] respects dependencies: 1 before 2 and 3, 3 and 2 before 4, 4 before 5, 5 before 6. [1, 2, 3, 4, 6, 5] places 6 before 5, invalid. [6, 5, 4, 3, 2, 1] reverses order, invalid. [1, 2, 3, 5, 4, 6] places 5 before 4, invalid.

3. Final Answer:

   [1, 3, 2, 4, 5, 6] -> Option B

4. Quick Check:

   Dependencies respected in order D [OK]

Hint: Parents always appear before children in topological order [OK]

Common Mistakes:

• Placing dependent nodes before their parents
• Reversing the order of edges
• Ignoring multiple valid topological orders