Topological Sort Using Kahn's Algorithm BFS in DSA C - 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.

void kahnTopologicalSort(int V, vector<int> adj[]) {
    vector<int> indegree(V, 0);
    for (int i = 0; i < V; i++) {
        for (int node : adj[i]) {
            indegree[node]++;
        }
    }
    queue<int> q;
    for (int i = 0; i < V; i++) {
        if (indegree[i] == 0) q.push(i);
    }
    while (!q.empty()) {
        int u = q.front(); q.pop();
        // process u
        for (int node : adj[u]) {
            indegree[node]--;
            if (indegree[node] == 0) q.push(node);
        }
    }
}
    

This code performs a topological sort on a graph with V nodes 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 roughly with the number of nodes plus the number of edges.

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

Pattern observation: The operations increase linearly with the sum of nodes and edges.

Final Time Complexity

Time Complexity: O(V + E)

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

Common Mistake

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

[OK] Correct: The inner loops do not run for all pairs; they run once per edge, so total work depends on edges, not all pairs of nodes.

Interview Connect

Understanding this time complexity helps you explain how graph algorithms scale and why Kahn's method is efficient for topological sorting.

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 calculate the minimum spanning tree

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

C. To detect cycles in an undirected graph

D. To find the shortest path between two nodes

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 like shortest path, cycle detection, and minimum spanning tree solve different problems unrelated to topological sorting.

3. Final Answer:

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

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 it up with minimum spanning tree algorithms

2. Which data structure is primarily used in Kahn's Algorithm to keep track of nodes with zero incoming edges?

easy

A. Priority Queue

B. Stack

C. Queue

D. Hash Map

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 a first-in-first-out manner, ensuring correct topological order.

2. Step 2: Why not other data structures?

   A stack would reverse the order, a priority queue is not necessary for basic Kahn's Algorithm, and a hash map stores key-value pairs, not order.

3. Final Answer:

   Queue -> Option C

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
• Confusing priority queue with normal queue
• Thinking hash map manages order

3. Consider the directed graph with edges: 5->2, 5->0, 4->0, 4->1, 2->3, 3->1. What is the topological order produced by Kahn's Algorithm?

medium

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

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

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

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

Solution

1. Step 1: Calculate in-degree of each node

   Nodes: 0(in-degree 2),1(2),2(1),3(1),4(0),5(0). Start with nodes 4 and 5 in queue (pushed in ascending order: queue [4,5]).

2. Step 2: Process nodes in queue and update in-degree

   Pop 4: output [4], reduce 0->1, 1->1. Queue: [5]
   Pop 5: output [4,5], reduce 2->0 (push 2), 0->0 (push 0). Queue: [2,0]
   Pop 2: output [4,5,2], reduce 3->0 (push 3). Queue: [0,3]
   Pop 0: output [4,5,2,0], no neighbors. Queue: [3]
   Pop 3: output [4,5,2,0,3], reduce 1->0 (push 1). Queue: [1]
   Pop 1: output [4,5,2,0,3,1]. All nodes processed.

3. Final Answer:

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

4. Quick Check:

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

Hint: Start with zero in-degree nodes, update neighbors' in-degree [OK]

Common Mistakes:

• Not updating in-degree after removing nodes
• Wrong initial zero in-degree nodes
• Mixing order of nodes in queue

4. Given the following code snippet for Kahn's Algorithm, what is the error?

int indegree[V] = {0};
for (int i = 0; i < V; i++) {
  for (int j = 0; j < adj[i].size(); j++) {
    indegree[adj[i][j]]++;
  }
}
queue<int> q;
for (int i = 0; i < V; i++) {
  if (indegree[i] == 0) q.push(i);
}
while (!q.empty()) {
  int u = q.front(); q.pop();
  cout << u << " ";
  for (int v : adj[u]) {
    indegree[v]--;
    if (indegree[v] == 0) q.push(v);
  }
}

medium

A. No error, the code is correct

B. The adjacency list is not declared properly

C. The indegree array is not initialized correctly

D. The loop to push zero indegree nodes is missing

Solution

1. Step 1: Check indegree initialization

   Indegree array is initialized to zero for all vertices correctly using {0} syntax.

2. Step 2: Verify adjacency list and loops

   Loops correctly calculate indegree, push zero indegree nodes, and process queue with BFS logic.

3. Final Answer:

   No error, the code is correct -> Option A

4. Quick Check:

   Code matches Kahn's Algorithm steps [OK]

Hint: Check initialization and loops carefully for correctness [OK]

Common Mistakes:

• Forgetting to initialize indegree array
• Incorrect adjacency list declaration
• Missing zero indegree node insertion

5. You have a graph with edges: 1->2, 1->3, 3->4, 2->4, 4->5, and 5->3. What will Kahn's Algorithm detect when run on this graph?

hard

A. A valid topological order exists

B. The algorithm will run infinitely without stopping

C. The graph is disconnected but topological order exists

D. The graph contains a cycle, so no topological order

Solution

1. Step 1: Identify cycle in the graph

   The edges 3->4, 4->5, and 5->3 form a cycle (3->4->5->3), so the graph is not a DAG.

2. Step 2: Understand Kahn's Algorithm behavior on cycles

   Kahn's Algorithm fails to process all nodes if a cycle exists because some nodes never reach zero in-degree, indicating no valid topological order.

3. Final Answer:

   The graph contains a cycle, so no topological order -> Option D

4. Quick Check:

   Cycle detected = no topological order [OK]

Hint: Cycle means no topological order in Kahn's Algorithm [OK]

Common Mistakes:

• Assuming topological order exists despite cycle
• Thinking disconnected graph always causes failure
• Believing algorithm runs infinitely on cycle