DFS Depth First Search on Graph in DSA C# - Time & Space Complexity

Time Complexity: DFS Depth First Search on Graph

O(n + m)

Understanding Time Complexity

We want to understand how the time needed to explore a graph grows as the graph gets bigger using DFS.

How does the number of steps change when the graph has more nodes and edges?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

void DFS(int node, bool[] visited, List<int>[] graph) {
    visited[node] = true;
    foreach (int neighbor in graph[node]) {
        if (!visited[neighbor]) {
            DFS(neighbor, visited, graph);
        }
    }
}
    

This code visits all nodes reachable from the starting node using depth-first search on an adjacency list graph.

Identify Repeating Operations

• Primary operation: Visiting each node and exploring all its neighbors.
• How many times: Each node is visited once, and each edge is checked once in total.

How Execution Grows With Input

As the graph grows, the steps grow with the number of nodes plus the number of edges.

Input Size (n nodes, m edges)	Approx. Operations
10 nodes, 15 edges	About 25 steps (10 + 15)
100 nodes, 200 edges	About 300 steps (100 + 200)
1000 nodes, 5000 edges	About 6000 steps (1000 + 5000)

Pattern observation: The work grows roughly by adding nodes and edges together, so bigger graphs take proportionally more time.

Final Time Complexity

Time Complexity: O(n + m)

This means the time grows linearly with the total number of nodes and edges in the graph.

Common Mistake

[X] Wrong: "DFS always takes O(n²) time because it uses recursion and loops."

[OK] Correct: DFS visits each node and edge only once, so it depends on the actual graph size, not just the number of nodes squared.

Interview Connect

Understanding DFS time complexity helps you explain how graph traversal scales, a key skill in many coding problems.

Self-Check

"What if the graph was represented by an adjacency matrix instead of adjacency lists? How would the time complexity change?"

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