Topological Sort Using DFS in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare the adjacency list for the graph.

DSA C

int adj[[1]][100]; // adjacency list for graph
int visited[100];

Drag options to blanks, or click blank then click option'

A10

B50

C100

D200

Attempts:

3 left

💡 Hint

Common Mistakes

Using a size larger than the number of nodes wastes memory.

Using a smaller size causes out-of-bounds errors.

The graph size is set to 10 nodes, so the adjacency list should be declared with size 10.

2fill in blank

medium

Complete the DFS function header to accept the current node.

DSA C

void dfs(int [1]) {
    visited[node] = 1;
    // DFS logic
}

Drag options to blanks, or click blank then click option'

Astart

Bnode

Cindex

Dcurrent

Attempts:

3 left

💡 Hint

Common Mistakes

Using a parameter name that doesn't match the variable used inside the function.

The parameter name 'node' clearly represents the current node in DFS traversal.

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][0]; i++) {
    int adjNode = adj[node][i + 1];
    if (!visited[adjNode]) {
        dfs([1]);
    }
}

Drag options to blanks, or click blank then click option'

AadjNode

Bi

Cnode

Dvisited

Attempts:

3 left

💡 Hint

Common Mistakes

Calling dfs on the loop index 'i' instead of the adjacent node.

Calling dfs on the current node again causing infinite recursion.

The DFS should be called on the adjacent node, which is 'adjNode'.

4fill in blank

hard

Fill both blanks to push the node onto the stack after visiting all neighbors.

DSA C

void dfs(int node) {
    visited[node] = 1;
    for (int i = 0; i < adj[node][0]; i++) {
        int adjNode = adj[node][i + 1];
        if (!visited[adjNode]) {
            dfs(adjNode);
        }
    }
    stack[[1]] = node;
    [2]++;
}

Drag options to blanks, or click blank then click option'

Atop

Bcount

Cindex

Dsize

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variables for index and increment causing errors.

Forgetting to increment the stack pointer.

The variable 'top' is used as the stack pointer to store nodes and increment after pushing.

5fill in blank

hard

Fill all three blanks to print the topological order from the stack.

DSA C

for (int i = [1] - 1; i >= 0; i--) {
    printf("%d ", stack[[2]]);
}
printf("\n");

// Reset stack pointer
[3] = 0;

Drag options to blanks, or click blank then click option'

Atop

Bi

Dcount

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong variable names causing compilation errors.

Not resetting the stack pointer after printing.

We start from 'top - 1' down to 0, print stack[i], and reset 'top' to 0 after printing.

Practice

1. What is the main purpose of using DFS in Topological Sort on a Directed Acyclic Graph (DAG)?

easy

A. To visit nodes in a way that all dependencies come before the node

B. To find the shortest path between two nodes

C. To detect cycles in an undirected graph

D. To count the number of connected components

Solution

1. Step 1: Understand DFS role in Topological Sort

   DFS visits nodes deeply, ensuring all nodes reachable from a node are visited before the node itself.

2. Step 2: Relate DFS to dependency order

   This ensures nodes appear after all their dependencies, which is the goal of topological sorting.

3. Final Answer:

   To visit nodes in a way that all dependencies come before the node -> Option A

4. Quick Check:

   DFS orders nodes respecting dependencies [OK]

Hint: DFS ensures dependencies come before nodes in topological order [OK]

Common Mistakes:

• Confusing topological sort with shortest path
• Thinking DFS detects cycles in undirected graphs here
• Mixing connected components with topological order

2. Which of the following is the correct way to mark a node as visited in a DFS-based topological sort in C?

easy

A. visited(node) = true;

B. visited[node] = 1;

C. visited[node] == true;

D. visited[node] = false;

Solution

1. Step 1: Understand C syntax for arrays

   In C, arrays use square brackets to access elements, so visited[node] is correct.

2. Step 2: Assigning visited status

   Assigning 1 marks the node as visited; '==' is comparison, not assignment; 'false' would mark unvisited.

3. Final Answer:

   visited[node] = 1; -> Option B

4. Quick Check:

   Use visited[node] = 1 to mark visited [OK]

Hint: Use visited[node] = 1 to mark visited in C arrays [OK]

Common Mistakes:

• Using parentheses instead of brackets for arrays
• Using '==' instead of '=' for assignment
• Assigning false instead of true/1 to mark visited

3. What is the output order of nodes after performing DFS-based topological sort on this graph?

Edges:
0 -> 1
0 -> 2
1 -> 3
2 -> 3

Assume nodes are pushed to stack after visiting all neighbors and stack is printed top to bottom.

medium

A. [0, 2, 1, 3]

B. [3, 2, 1, 0]

C. [0, 1, 2, 3]

D. [3, 1, 2, 0]

Solution

1. Step 1: Perform DFS starting from node 0

   Visit 0, then neighbors 1 then 2 (edge order). From 1 to 3, from 2 to 3 (visited).

2. Step 2: Push nodes to stack after visiting neighbors

   Push 3, then 1, then 3 already visited, then 2, then 0. Stack top to bottom: [3, 1, 2, 0].

3. Final Answer:

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

4. Quick Check:

   Stack top to bottom after DFS = [3, 1, 2, 0] [OK]

Hint: Push nodes after neighbors; output is stack top to bottom [OK]

Common Mistakes:

• Pushing nodes before visiting neighbors
• Confusing order of neighbors visited
• Printing stack bottom to top instead of top to bottom

4. Identify the error in this DFS-based topological sort code snippet in C:

void dfs(int node) {
  visited[node] = 1;
  for (int i = 0; i < adj[node].size(); i++) {
    int nbr = adj[node][i];
    if (!visited[nbr])
      dfs(nbr);
  }
  stack.push(node);
}

medium

A. Stack push should be before recursive calls

B. Missing base case for cycle detection

C. Using adj[node].size() is invalid in C

D. visited array should be initialized to 2 instead of 1

Solution

1. Step 1: Check syntax for adjacency list size in C

   adj[node].size() is C++ syntax; C does not support .size() method on arrays.

2. Step 2: Identify correct way to get adjacency list size in C

   In C, adjacency lists are usually arrays with separate size variables; .size() causes compile error.

3. Final Answer:

   Using adj[node].size() is invalid in C -> Option C

4. Quick Check:

   adj[node].size() is C++ only, not C [OK]

Hint: Use array length variables in C, not .size() method [OK]

Common Mistakes:

• Confusing C and C++ syntax for containers
• Ignoring cycle detection in topological sort
• Pushing nodes before recursive calls

5. Given a graph with edges:

5 -> 2
5 -> 0
4 -> 0
4 -> 1
2 -> 3
3 -> 1

Which of the following is a valid topological order using DFS-based topological sort?

hard

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

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

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

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

Solution

1. Step 1: Analyze prerequisites from edges

   (u -> v means u before v): 5 before 2 and 0; 4 before 0 and 1; 2 before 3; 3 before 1.

2. Step 2: Check each option for prerequisite order

   [4, 5, 2, 3, 1, 0]: 4 before 0 and 1, 5 before 2 and 0, 2 before 3, 3 before 1, all correct.

3. Final Answer:

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

4. Quick Check:

   All edges respected in order D [OK]

Hint: Check each edge u->v: u appears before v in order [OK]

Common Mistakes:

• Ignoring edge directions in order
• Placing dependent nodes before their prerequisites
• Assuming any order with all nodes is valid