Topological Sort Using Kahn's Algorithm BFS in DSA C

Mental Model

We order tasks so that each task comes after all tasks it depends on. We do this by repeatedly picking tasks with no remaining dependencies.

Analogy: Imagine you have to bake a cake, but you must first prepare the batter and preheat the oven. You can only do a step when all its prerequisites are done. Kahn's algorithm finds the order to do all steps without breaking any rules.

Graph:
0 -> 1 -> 3
↓
2

In-degree array:
[0]=0, [1]=1, [2]=1, [3]=2

Queue initially holds nodes with in-degree 0:
[0]

Dry Run Walkthrough

Input: Graph edges: 0->1, 0->2, 1->3, 2->3; nodes = 4

Goal: Find a linear order of nodes so that every edge u->v means u comes before v

Step 1: Calculate in-degree for each node

In-degree array: [0]=0, [1]=1, [2]=1, [3]=2

Why: We need to know which nodes have no dependencies to start with

Step 2: Add nodes with in-degree 0 to queue

Queue: [0]

Why: Only nodes with no dependencies can be processed first

Step 3: Remove node 0 from queue and add to result

Result: 0
Queue: []

Why: Process node 0 as it has no dependencies

Step 4: Decrease in-degree of neighbors 1 and 2 by 1

In-degree array: [1]=0, [2]=0, [3]=2

Why: Removing 0 means its neighbors have one less dependency

Step 5: Add nodes 1 and 2 with in-degree 0 to queue

Queue: [1, 2]

Why: Now nodes 1 and 2 have no dependencies and can be processed

Step 6: Remove node 1 from queue and add to result

Result: 0 -> 1
Queue: [2]

Why: Process node 1 next

Step 7: Decrease in-degree of neighbor 3 by 1

In-degree array: [3]=1

Why: Removing 1 reduces dependencies of 3

Step 8: Remove node 2 from queue and add to result

Result: 0 -> 1 -> 2
Queue: []

Why: Process node 2 next

Step 9: Decrease in-degree of neighbor 3 by 1

In-degree array: [3]=0

Why: Removing 2 reduces dependencies of 3 to zero

Step 10: Add node 3 to queue

Queue: [3]

Why: Node 3 now has no dependencies

Step 11: Remove node 3 from queue and add to result

Result: 0 -> 1 -> 2 -> 3
Queue: []

Why: Process last node 3

Result:

Final topological order: 0 -> 1 -> 2 -> 3

Annotated Code

DSA C

#include <stdio.h>
#include <stdlib.h>

#define MAX_NODES 100

// Queue structure for BFS
typedef struct {
    int items[MAX_NODES];
    int front, rear;
} Queue;

void initQueue(Queue* q) {
    q->front = 0;
    q->rear = -1;
}

int isEmpty(Queue* q) {
    return q->rear < q->front;
}

void enqueue(Queue* q, int val) {
    q->items[++q->rear] = val;
}

int dequeue(Queue* q) {
    return q->items[q->front++];
}

// Graph adjacency list
typedef struct Node {
    int vertex;
    struct Node* next;
} Node;

Node* createNode(int v) {
    Node* newNode = (Node*)malloc(sizeof(Node));
    newNode->vertex = v;
    newNode->next = NULL;
    return newNode;
}

void addEdge(Node* adj[], int src, int dest) {
    Node* newNode = createNode(dest);
    newNode->next = adj[src];
    adj[src] = newNode;
}

void kahnTopologicalSort(Node* adj[], int n) {
    int inDegree[MAX_NODES] = {0};
    int i;

    // Calculate in-degree of each node
    for (i = 0; i < n; i++) {
        Node* temp = adj[i];
        while (temp) {
            inDegree[temp->vertex]++;
            temp = temp->next;
        }
    }

    Queue q;
    initQueue(&q);

    // Enqueue nodes with in-degree 0
    for (i = 0; i < n; i++) {
        if (inDegree[i] == 0) {
            enqueue(&q, i);
        }
    }

    int count = 0;
    int topOrder[MAX_NODES];

    while (!isEmpty(&q)) {
        int u = dequeue(&q); // Remove node with no dependencies
        topOrder[count++] = u;

        // Decrease in-degree of neighbors
        Node* temp = adj[u];
        while (temp) {
            inDegree[temp->vertex]--;
            if (inDegree[temp->vertex] == 0) {
                enqueue(&q, temp->vertex);
            }
            temp = temp->next;
        }
    }

    // Check if there was a cycle
    if (count != n) {
        printf("Graph has a cycle, topological sort not possible\n");
        return;
    }

    // Print topological order
    for (i = 0; i < count; i++) {
        printf("%d", topOrder[i]);
        if (i != count - 1) printf(" -> ");
    }
    printf("\n");
}

int main() {
    int n = 4;
    Node* adj[MAX_NODES] = {NULL};

    addEdge(adj, 0, 1);
    addEdge(adj, 0, 2);
    addEdge(adj, 1, 3);
    addEdge(adj, 2, 3);

    kahnTopologicalSort(adj, n);

    return 0;
}

for (i = 0; i < n; i++) { Node* temp = adj[i]; while (temp) { inDegree[temp->vertex]++; temp = temp->next; } }

Calculate in-degree of each node by counting incoming edges

for (i = 0; i < n; i++) { if (inDegree[i] == 0) { enqueue(&q, i); } }

Add all nodes with zero in-degree to the queue to start processing

while (!isEmpty(&q)) { int u = dequeue(&q); topOrder[count++] = u;

Remove node with no dependencies and add to topological order

Node* temp = adj[u]; while (temp) { inDegree[temp->vertex]--; if (inDegree[temp->vertex] == 0) { enqueue(&q, temp->vertex); } temp = temp->next; }

Decrease in-degree of neighbors and enqueue if they become zero

if (count != n) { printf("Graph has a cycle, topological sort not possible\n"); return; }

Detect cycle if not all nodes are processed

Output

0 -> 1 -> 2 -> 3

Complexity Analysis

Time: O(V + E) because we visit each node and edge once during in-degree calculation and BFS

Space: O(V + E) for adjacency list and in-degree array

vs Alternative: Compared to DFS-based topological sort, Kahn's algorithm uses BFS and explicit in-degree tracking, which can be easier to understand and detect cycles early

Edge Cases

Graph with cycle

Algorithm detects cycle because not all nodes get processed; prints error message

DSA C

if (count != n) {
    printf("Graph has a cycle, topological sort not possible\n");
    return;
}

Graph with single node and no edges

Node has in-degree 0, gets processed immediately, output is that single node

DSA C

for (i = 0; i < n; i++) {
    if (inDegree[i] == 0) {
        enqueue(&q, i);
    }
}

Graph with multiple disconnected components

All nodes with zero in-degree from all components get processed in order, producing a valid topological order

DSA C

for (i = 0; i < n; i++) {
    if (inDegree[i] == 0) {
        enqueue(&q, i);
    }
}

When to Use This Pattern

When you need to order tasks with dependencies and want to detect cycles, use Kahn's algorithm BFS for topological sorting because it processes nodes with zero dependencies first.

Common Mistakes

Mistake: Not updating in-degree of neighbors after removing a node
Fix: Decrease in-degree of all neighbors and enqueue those whose in-degree becomes zero

Mistake: Not checking for cycles by comparing processed node count to total nodes
Fix: Add a check after BFS to detect if a cycle exists by verifying if all nodes were processed

Summary

Finds an order of nodes so all dependencies come before dependent nodes using BFS and in-degree counts.

Use when you have a directed acyclic graph and want to find a valid linear ordering of nodes.

The key insight is to repeatedly process nodes with zero in-degree and update neighbors' in-degree accordingly.