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

Practice - 5 Tasks

1fill in blank

easy

Complete the code to start DFS from the given node.

DSA C#

void DFS(int node) {
    visited[node] = true;
    Console.WriteLine(node);
    foreach (var neighbor in graph[node]) {
        if (!visited[[1]]) {
            DFS(neighbor);
        }
    }
}

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

Anode

Bvisited

Cneighbor

Dgraph

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'node' instead of 'neighbor' in the visited check.

Checking 'visited' array directly without index.

We check if the neighbor node has been visited before calling DFS on it.

2fill in blank

medium

Complete the code to initialize the visited array for DFS.

DSA C#

int n = graph.Count;
bool[] visited = new bool[[1]];

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

An + 1

Bn - 1

Cgraph.Length - 1

Dn

Attempts:

3 left

💡 Hint

Common Mistakes

Using n + 1 which causes extra unused space.

Using n - 1 which causes out of range errors.

The visited array size should match the number of nodes, which is 'n'.

3fill in blank

hard

Fix the error in the DFS recursive call to avoid infinite recursion.

DSA C#

void DFS(int node) {
    visited[node] = true;
    Console.WriteLine(node);
    foreach (var neighbor in graph[node]) {
        if (!visited[neighbor]) {
            DFS([1]);
        }
    }
}

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

Aneighbor

Bnode

Cvisited[node]

Dgraph[node]

Attempts:

3 left

💡 Hint

Common Mistakes

Calling DFS(node) inside the loop instead of DFS(neighbor).

Passing visited[node] which is a boolean, not a node index.

We must call DFS on the neighbor node, not the current node, to avoid infinite recursion.

4fill in blank

hard

Fill both blanks to create a DFS function that prints nodes and marks them visited.

DSA C#

void DFS(int [1]) {
    visited[[2]] = true;
    Console.WriteLine([2]);
    foreach (var neighbor in graph[[2]]) {
        if (!visited[neighbor]) {
            DFS(neighbor);
        }
    }
}

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

Anode

Bneighbor

Cvisited

Dgraph

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variables for parameter and visited index.

Using 'neighbor' instead of 'node' in visited array.

The parameter and the visited index must both be 'node' to correctly mark and print the current node.

5fill in blank

hard

Fill all three blanks to create a dictionary comprehension that maps each node to its degree (number of neighbors).

DSA C#

var degrees = new Dictionary<int, int>();
foreach (var [1] in graph.Keys) {
    degrees[[2]] = graph[[3]].Count;
}

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

Anode

Dneighbor

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'neighbor' instead of 'node' as loop variable.

Mixing different variable names for keys.

We iterate over each node key, then assign degrees[node] = graph[node].Count to get the number of neighbors.

Practice

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

easy

A. Visit nodes in order of their distance from the start node

B. Visit all neighbors of a node before moving deeper

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

D. Randomly visit nodes until all are visited

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 neighbors first, 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 neighbors first
• Assuming DFS visits nodes by distance

2. Which of the following is the correct way to start a DFS traversal in C# using recursion on a graph represented by adjacency list Dictionary<int, List<int>> graph?

easy

A. void DFS(int node, HashSet<int> visited) { if (visited.Contains(node)) return; visited.Add(node); foreach (var neighbor in graph[node]) DFS(neighbor, visited); }

B. void DFS(int node) { visited.Add(node); foreach (var neighbor in graph[node]) DFS(neighbor); }

C. void DFS(int node, List<int> visited) { if (visited.Contains(node)) return; visited.Add(node); foreach (var neighbor in graph[node]) DFS(neighbor, visited); }

D. void DFS(int node, HashSet<int> visited) { visited.Add(node); foreach (var neighbor in graph) DFS(neighbor, visited); }

Solution

1. Step 1: Check parameters and visited tracking

   void DFS(int node, HashSet<int> visited) { if (visited.Contains(node)) return; visited.Add(node); foreach (var neighbor in graph[node]) DFS(neighbor, visited); } uses HashSet to track visited nodes and prevents revisiting.

2. Step 2: Verify iteration over neighbors

   void DFS(int node, HashSet<int> visited) { if (visited.Contains(node)) return; visited.Add(node); foreach (var neighbor in graph[node]) DFS(neighbor, visited); } correctly iterates over neighbors of the current node from the graph dictionary.

3. Final Answer:

   void DFS(int node, HashSet<int> visited) { if (visited.Contains(node)) return; visited.Add(node); foreach (var neighbor in graph[node]) DFS(neighbor, visited); } -> Option A

4. Quick Check:

   Correct recursion with visited check [OK]

Hint: Use HashSet to track visited nodes in DFS recursion [OK]

Common Mistakes:

• Not checking if node is visited before recursion
• Using List instead of HashSet for visited (slow lookup)
• Iterating over graph keys instead of neighbors

3. Given the graph adjacency list:

var graph = new Dictionary {
  {1, new List<int>{2, 3}},
  {2, new List<int>{4}},
  {3, new List<int>{4}},
  {4, new List<int>()}
};

List<int> visited = new List<int>();
DFS(1, visited); // DFS as defined in option C above
Console.WriteLine(string.Join(", ", visited));

What is the output?

medium

A. 1, 2, 3, 4

B. 1, 2, 4, 3

C. 1, 3, 4, 2

D. 1, 4, 2, 3

Solution

1. Step 1: Start DFS from node 1

   Visit 1, mark visited. Neighbors: 2, 3.

2. Step 2: Visit neighbor 2 first

   Visit 2, then its neighbor 4. 4 has no neighbors, backtrack.

3. Step 3: Backtrack to 1 and visit neighbor 3

   Visit 3, then neighbor 4 (already visited, skip).

4. Step 4: Order of visited nodes

   Visited nodes in order: 1, 2, 4, 3.

5. Final Answer:

   1, 2, 4, 3 -> Option B

6. Quick Check:

   DFS order = 1,2,4,3 [OK]

Hint: DFS visits neighbors in order, backtracks after deepest node [OK]

Common Mistakes:

• Assuming BFS order instead of DFS
• Visiting node 3 before node 2
• Revisiting already visited nodes

4. Consider this DFS code snippet in C#:

void DFS(int node, HashSet<int> visited) {
  visited.Add(node);
  foreach (var neighbor in graph.Keys) {
    if (!visited.Contains(neighbor)) {
      DFS(neighbor, visited);
    }
  }
}

What is the main problem with this DFS implementation?

medium

A. It iterates over all graph nodes instead of neighbors of the current node

B. It does not add the current node to visited

C. It uses List instead of HashSet for visited nodes

D. It does not check if the node is null before recursion

Solution

1. Step 1: Analyze the foreach loop

   The loop iterates over all keys in the graph, not neighbors of the current node.

2. Step 2: Understand DFS neighbor traversal

   DFS should explore neighbors of the current node, not all nodes in the graph.

3. Final Answer:

   It iterates over all graph nodes instead of neighbors of the current node -> Option A

4. Quick Check:

   DFS must visit neighbors, not all nodes [OK]

Hint: Loop over neighbors, not all graph keys [OK]

Common Mistakes:

• Looping over graph.Keys instead of graph[node]
• Forgetting to add node to visited
• Not checking visited before recursion

5. You want to find if there is a path from node 1 to node 5 in this directed graph:

var graph = new Dictionary {
  {1, new List<int>{2, 3}},
  {2, new List<int>{4}},
  {3, new List<int>{4, 5}},
  {4, new List<int>{}},
  {5, new List<int>{}}
};

bool DFS(int current, int target, HashSet<int> visited) {
  if (current == target) return true;
  visited.Add(current);
  foreach (var neighbor in graph[current]) {
    if (!visited.Contains(neighbor)) {
      if (DFS(neighbor, target, visited)) return true;
    }
  }
  return false;
}

var visited = new HashSet<int>();
bool result = DFS(1, 5, visited);
Console.WriteLine(result);

What will be printed?

hard

A. Runtime exception

B. False

C. Compilation error

D. True

Solution

1. Step 1: Start DFS from node 1 looking for 5

   Visit 1, neighbors 2 and 3.

2. Step 2: Explore neighbor 2

   From 2, visit 4. 4 has no neighbors, backtrack.

3. Step 3: Explore neighbor 3

   From 3, neighbors are 4 and 5. 4 visited, then 5 found matches target.

4. Step 4: DFS returns true when target found

   DFS returns true and prints True.

5. Final Answer:

   True -> Option D

6. Quick Check:

   Path exists from 1 to 5 [OK]

Hint: DFS returns true immediately when target found [OK]

Common Mistakes:

• Assuming no path because 5 is not direct neighbor
• Not returning true when target found
• Forgetting to mark visited nodes