Connected Components Using BFS in DSA C - Build from Scratch

Connected Components Using BFS

📖 Scenario: You are working with a simple network of computers represented as an undirected graph. Each computer is a node, and connections between them are edges. You want to find how many separate groups of connected computers exist in the network.

🎯 Goal: Build a program that uses Breadth-First Search (BFS) to find and count the number of connected components in an undirected graph.

📋 What You'll Learn

Create an adjacency matrix to represent the graph with 6 nodes.

Create a visited array to track visited nodes.

Implement BFS to traverse connected nodes.

Count and print the number of connected components.

💡 Why This Matters

🌍 Real World

Finding connected components helps identify isolated groups in networks like social media, computer networks, or road maps.

💼 Career

Understanding graph traversal and connected components is essential for roles in software development, data analysis, and network engineering.

1

Create the Graph Adjacency Matrix

Create a 6x6 integer adjacency matrix called graph with these exact connections (undirected): edges between nodes 0-1, 0-2, 1-2, 3-4. Use 1 for connected and 0 for no connection. Initialize all other entries to 0.

DSA C

# Your code here

Hint

Use a 2D array with 6 rows and 6 columns. Put 1 where nodes are connected and 0 otherwise.

2

Create the Visited Array

Create an integer array called visited of size 6 and initialize all elements to 0 to track if a node has been visited.

DSA C

int graph[6][6] = {
    {0, 1, 1, 0, 0, 0},
    {1, 0, 1, 0, 0, 0},
    {1, 1, 0, 0, 0, 0},
    {0, 0, 0, 0, 1, 0},
    {0, 0, 0, 1, 0, 0},
    {0, 0, 0, 0, 0, 0}
};

# Your code here

Hint

Use an integer array with 6 zeros to mark all nodes as not visited.

3

Implement BFS Function

Write a function called bfs that takes an integer start, the graph adjacency matrix, and the visited array. Use a queue to visit all nodes connected to start. Mark nodes as visited when they are dequeued.

DSA C

int graph[6][6] = {
    {0, 1, 1, 0, 0, 0},
    {1, 0, 1, 0, 0, 0},
    {1, 1, 0, 0, 0, 0},
    {0, 0, 0, 0, 1, 0},
    {0, 0, 0, 1, 0, 0},
    {0, 0, 0, 0, 0, 0}
};

int visited[6] = {0, 0, 0, 0, 0, 0};

// Your code here

Hint

Use an array as a queue with front and rear indices. Mark nodes visited when added to the queue.

4

Count and Print Connected Components

Write a main function that uses bfs to count how many connected components are in the graph. For each node from 0 to 5, if it is not visited, call bfs on it and increment a counter. Then print the count with printf("%d\n", count);.

DSA C

#include 
#include 

int graph[6][6] = {
    {0, 1, 1, 0, 0, 0},
    {1, 0, 1, 0, 0, 0},
    {1, 1, 0, 0, 0, 0},
    {0, 0, 0, 0, 1, 0},
    {0, 0, 0, 1, 0, 0},
    {0, 0, 0, 0, 0, 0}
};

int visited[6] = {0, 0, 0, 0, 0, 0};

void bfs(int start) {
    int queue[6];
    int front = 0, rear = 0;
    queue[rear++] = start;
    visited[start] = 1;

    while (front < rear) {
        int current = queue[front++];
        for (int i = 0; i < 6; i++) {
            if (graph[current][i] == 1 && visited[i] == 0) {
                queue[rear++] = i;
                visited[i] = 1;
            }
        }
    }
}

// Your code here

Hint

Loop through all nodes, call bfs if not visited, and count how many times bfs is called. Print the count.

Practice

1. What does the BFS algorithm help find in an undirected graph when used to find connected components?

easy

A. All nodes reachable from a starting node forming one connected component

B. The shortest path between two nodes only

C. The number of edges in the graph

D. The degree of each node

Solution

1. Step 1: Understand BFS traversal in undirected graphs

   BFS explores all nodes reachable from a given start node by visiting neighbors level by level.

2. Step 2: Connect BFS to connected components

   By running BFS from an unvisited node, we find all nodes in that node's connected component.

3. Final Answer:

   All nodes reachable from a starting node forming one connected component -> Option A

4. Quick Check:

   BFS finds connected components by exploring reachable nodes [OK]

Hint: BFS visits all nodes in one connected component [OK]

Common Mistakes:

• Thinking BFS only finds shortest paths
• Confusing connected components with edge counts
• Assuming BFS finds node degrees

2. Which of the following is the correct way to mark a node as visited in BFS when using an array named visited and current node u?

easy

A. visited[u] = 0;

B. visited[u] = true;

C. visited[u] == true;

D. visited[u] = null;

Solution

1. Step 1: Understand marking visited nodes

   To mark a node visited, assign true to visited[u], indicating it is visited.

2. Step 2: Check syntax correctness

   Using visited[u] = true; correctly assigns the boolean true. Using == is a comparison, not assignment.

3. Final Answer:

   visited[u] = true; -> Option B

4. Quick Check:

   Assign true to mark visited[u] [OK]

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

Common Mistakes:

• Using == instead of = for assignment
• Assigning 0 instead of true to visited
• Assigning null which is invalid in C

3. Given the following graph adjacency list and BFS code snippet, what is the output number of connected components?

int n = 5;
int adj[5][5] = {
  {0,1,0,0,0},
  {1,0,0,0,0},
  {0,0,0,1,1},
  {0,0,1,0,0},
  {0,0,1,0,0}
};
bool visited[5] = {false};

int count = 0;
for (int i = 0; i < n; i++) {
  if (!visited[i]) {
    bfs(i, adj, visited, n);
    count++;
  }
}
printf("%d\n", count);

medium

A. 2

B. 1

C. 3

D. 4

Solution

1. Step 1: Analyze adjacency matrix for connected components

   Nodes 0 and 1 are connected; nodes 2, 3, 4 form another connected group.

2. Step 2: Count connected components using BFS

   BFS from node 0 visits nodes 0 and 1; BFS from node 2 visits nodes 2, 3, 4. Total components = 2.

3. Final Answer:

   2 -> Option A

4. Quick Check:

   Two groups found by BFS = 2 [OK]

Hint: Count BFS runs on unvisited nodes for components [OK]

Common Mistakes:

• Counting edges instead of components
• Misreading adjacency matrix connections
• Forgetting to mark visited nodes

4. Identify the bug in this BFS function for counting connected components:

void bfs(int start, int adj[][5], bool visited[], int n) {
  queue<int> q;
  q.push(start);
  visited[start] = true;
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (int v = 0; v < n; v++) {
      if (adj[u][v] == 1 && visited[v] == false) {
        q.push(v);
      }
    }
  }
}

medium

A. Loop condition should be while(q.size() > 1)

B. Queue is not initialized properly

C. Start node is not marked visited

D. Missing visited[v] = true; inside the loop

Solution

1. Step 1: Check BFS neighbor processing

   When a neighbor v is found unvisited, it must be marked visited immediately to avoid re-adding.

2. Step 2: Identify missing visited mark

   The code pushes v to queue but does not mark visited[v] = true; causing repeated visits.

3. Final Answer:

   Missing visited[v] = true; inside the loop -> Option D

4. Quick Check:

   Mark visited when enqueueing neighbors [OK]

Hint: Mark visited when pushing to queue, not after popping [OK]

Common Mistakes:

• Marking visited only after popping from queue
• Incorrect queue initialization
• Wrong loop condition for BFS

5. You have a graph with 7 nodes and edges: (0-1), (1-2), (3-4), (5-6). Using BFS to find connected components, what is the minimum number of BFS calls needed to find all components?

hard

A. 1

B. 2

C. 3

D. 4

Solution

1. Step 1: Identify connected components from edges

   Edges connect nodes into groups: {0,1,2}, {3,4}, {5,6}. Since nodes are numbered 0 to 6, there are no other nodes.

2. Step 2: Count connected components

   Groups are: {0,1,2}, {3,4}, {5,6} -> 3 components total.

3. Step 3: BFS calls equal connected components

   Each BFS call finds one component, so minimum BFS calls = 3.

4. Final Answer:

   3 -> Option C

5. Quick Check:

   Number of BFS calls = number of connected components [OK]

Hint: Count connected groups; BFS calls = groups [OK]

Common Mistakes:

• Counting edges instead of components
• Miscounting isolated nodes
• Assuming BFS calls equal nodes