DFS Depth First Search on Graph in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare the DFS function parameter for the graph adjacency list.

DSA C

void dfs(int node, int visited[], vector<vector<int>> [1], int n) {

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Alist

Bgraph

Cadj

Dedges

Attempts:

3 left

💡 Hint

Common Mistakes

Using a variable name that is not defined or inconsistent with the rest of the code.

The adjacency list is commonly named 'adj' to represent graph connections.

2fill in blank

medium

Complete the code to mark the current node as visited inside DFS.

DSA C

visited[node] = [1];

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A0

B1

C-1

Dnode

Attempts:

3 left

💡 Hint

Common Mistakes

Marking visited[node] as 0 which means unvisited.

Marking visited[node] as 1 means the node is visited.

3fill in blank

hard

Fix the error in the DFS recursive call to visit adjacent nodes.

DSA C

for(int i = 0; i < adj[node].size(); i++) {
    int adjNode = adj[node][i];
    if(!visited[[1]]) {
        dfs(adjNode, visited, adj, n);
    }
}

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AadjNode

Bnode

Ci

Dvisited

Attempts:

3 left

💡 Hint

Common Mistakes

Checking visited[node] instead of visited[adjNode].

Using the loop index 'i' as the node index.

We check if the adjacent node 'adjNode' is visited before calling DFS on it.

4fill in blank

hard

Fill both blanks to correctly initialize and call DFS for all nodes in the graph.

DSA C

int visited[n];
for(int i = 0; i < n; i++) {
    visited[i] = [1];
}
for(int i = 0; i < n; i++) {
    if(visited[i] == [2]) {
        dfs(i, visited, adj, n);
    }
}

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A0

B1

C-1

Dn

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing visited array with 1 instead of 0.

Checking for visited[i] == 1 instead of 0.

Initialize all nodes as unvisited (0) and call DFS on nodes that are unvisited (0).

5fill in blank

hard

Fill all three blanks to complete the DFS function that prints nodes in DFS order.

DSA C

void dfs(int node, int visited[], vector<vector<int>> [1], int n) {
    visited[node] = [2];
    printf("%d ", [3]);
    for(int i = 0; i < adj[node].size(); i++) {
        int adjNode = adj[node][i];
        if(!visited[adjNode]) {
            dfs(adjNode, visited, adj, n);
        }
    }
}

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Aadj

B1

Cnode

Dvisited

Attempts:

3 left

💡 Hint

Common Mistakes

Printing visited[node] instead of node.

Using wrong variable name for adjacency list.

The adjacency list is 'adj', visited[node] is set to 1, and we print the current node.

Practice

1. What is the main idea behind Depth First Search (DFS) on a graph?

easy

A. Visit all nodes at the current depth before moving deeper

B. Visit nodes randomly without any order

C. Explore as far as possible along each branch before backtracking

D. Only visit nodes with no outgoing edges

Solution

1. Step 1: Understand DFS traversal

   DFS explores a graph by going deep into one branch before backtracking to explore others.

2. Step 2: Compare with other traversal methods

   BFS visits nodes level by level, but DFS goes deep first.

3. Final Answer:

   Explore as far as possible along each branch before backtracking -> Option C

4. Quick Check:

   DFS = deep exploration first [OK]

Hint: DFS means go deep first, then backtrack [OK]

Common Mistakes:

• Confusing DFS with BFS
• Thinking DFS visits nodes level-wise
• Assuming DFS visits nodes randomly

2. Which of the following is the correct way to start a DFS function in C for a graph represented by adjacency list adj and visited array visited?

easy

A. void dfs(int node) { visited[node] = 1; for(int i=0; i

B. void dfs(int node) { visited[node] = true; for(int i=0; i

C. void dfs(int node) { visited[node] = 0; for(int i=0; i

D. void dfs(int node) { visited[node] = 1; for(int i=0; i

Solution

1. Step 1: Mark current node as visited

   Set visited[node] = 1 (or true) to avoid revisiting.

2. Step 2: Traverse neighbors only if not visited

   Check if neighbor is unvisited before recursive DFS call to prevent infinite loops.

3. Final Answer:

   void dfs(int node) { visited[node] = 1; for(int i=0; i -> Option D

4. Quick Check:

   Mark visited and check before recursion [OK]

Hint: Always check visited before recursive DFS call [OK]

Common Mistakes:

• Not checking if neighbor is visited
• Using wrong visited value (0 instead of 1)
• Using .size() instead of .length in C arrays

3. Consider the following graph adjacency list:
0: [1, 2]
1: [2]
2: [0, 3]
3: [3]
Starting DFS from node 2, what is the order of nodes visited?

medium

A. [2, 3, 0, 1]

B. [2, 0, 1, 3]

C. [2, 0, 3, 1]

D. [2, 1, 0, 3]

Solution

1. Step 1: Start DFS at node 2

   Visit 2, mark visited. Neighbors: 0 and 3.

2. Step 2: Visit neighbor 0 first

   From 2, go to 0. Visit 0, neighbors: 1 and 2 (2 already visited).

3. Step 3: Visit neighbor 1 from 0

   Visit 1, neighbor: 2 (visited). Backtrack.

4. Step 4: Back to 2, visit neighbor 3

   Visit 3, neighbor: 3 (already visited). End.

5. Final Answer:

   [2, 0, 1, 3] -> Option B

6. Quick Check:

   DFS order from 2 = [2,0,1,3] [OK]

Hint: Follow neighbors depth-first, mark visited to avoid repeats [OK]

Common Mistakes:

• Visiting nodes multiple times
• Visiting neighbors in wrong order
• Ignoring visited checks causing loops

4. The following DFS code has a bug. What is the error?

void dfs(int node) {
  visited[node] = 1;
  for(int i=0; i<adj[node].length; i++) {
    dfs(adj[node][i]);
  }
}

medium

A. Missing check if neighbor is already visited before recursive call

B. Visited array is not initialized

C. Incorrect loop condition syntax

D. Visited[node] should be set after recursion

Solution

1. Step 1: Analyze DFS recursion calls

   The code calls dfs on all neighbors without checking if they are visited.

2. Step 2: Identify risk of infinite recursion

   Without visited check, DFS may revisit nodes endlessly causing stack overflow.

3. Final Answer:

   Missing check if neighbor is already visited before recursive call -> Option A

4. Quick Check:

   Always check visited before recursion [OK]

Hint: Check visited before recursive DFS call to avoid loops [OK]

Common Mistakes:

• Assuming visited check is optional
• Setting visited after recursion
• Ignoring infinite recursion risk

5. Given a directed graph with possible cycles, how can DFS be modified to detect if a cycle exists?

hard

A. Use a recursion stack array to track nodes in the current path and check if a node is revisited in this stack

B. Only mark nodes visited after all neighbors are processed

C. Use BFS instead of DFS to detect cycles

D. Ignore visited array and rely on node values

Solution

1. Step 1: Understand cycle detection in DFS

   Detect cycles by tracking nodes currently in recursion (call) stack.

2. Step 2: Use recursion stack array

   Maintain an array marking nodes in current DFS path; if a neighbor is in this stack, a cycle exists.

3. Step 3: Differentiate from visited array

   Visited marks all visited nodes; recursion stack tracks current path only.

4. Final Answer:

   Use a recursion stack array to track nodes in the current path and check if a node is revisited in this stack -> Option A

5. Quick Check:

   Recursion stack detects cycles in DFS [OK]

Hint: Track recursion stack to find cycles during DFS [OK]

Common Mistakes:

• Confusing visited with recursion stack
• Using BFS for cycle detection in directed graphs
• Not tracking current path nodes