Connected Components Using BFS in DSA C - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is a connected component in an undirected graph?

A connected component is a group of nodes where each node is reachable from any other node in the same group by following edges.

beginner

How does BFS help in finding connected components?

BFS explores all nodes reachable from a starting node. By running BFS from unvisited nodes, we find all nodes in that connected component.

beginner

What data structure is commonly used to implement BFS?

A queue is used to keep track of nodes to visit next in BFS.

beginner

In BFS for connected components, why do we mark nodes as visited?

Marking nodes as visited prevents revisiting the same node, avoiding infinite loops and repeated work.

intermediate

What is the time complexity of finding connected components using BFS in a graph with V vertices and E edges?

The time complexity is O(V + E) because each vertex and edge is visited at most once during BFS.

What is the first step when using BFS to find connected components?

A Mark all nodes as visited

B Start BFS from an unvisited node

C Add all nodes to the queue

D Remove edges from the graph

We start BFS from an unvisited node to explore its connected component.

Which data structure is essential for BFS traversal?

A Heap

B Stack

C Queue

D Set

A queue is used to process nodes in the order they are discovered in BFS.

How do you know when you have found all connected components in a graph?

A When BFS runs once

B When the queue is empty

C When all edges are removed

D When all nodes are visited

All connected components are found when every node in the graph has been visited.

What happens if you do not mark nodes as visited in BFS?

A You may visit nodes multiple times causing infinite loops

B BFS will run faster

C The graph will change

D Edges will be ignored

Without marking visited nodes, BFS can revisit nodes repeatedly, causing infinite loops.

What is the main goal of BFS in connected components?

A To explore all nodes in one connected component

B To sort nodes by value

C To find the shortest path between two nodes

D To remove cycles

BFS explores all nodes reachable from a starting node, identifying one connected component.

Explain step-by-step how BFS is used to find all connected components in an undirected graph.

Think about visiting nodes and marking them to avoid repeats.

Describe why marking nodes as visited is important in BFS when finding connected components.

Consider what happens if you visit the same node multiple times.

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