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 taken by Depth First Search (DFS) changes as the graph size grows.

Specifically, how does DFS's work increase when the graph has more nodes and edges?

Scenario Under Consideration

Analyze the time complexity of the following DFS code on a graph.

void dfs(int node, vector<vector<int>>& graph, vector<bool>& visited) {
    visited[node] = true;
    for (int neighbor : graph[node]) {
        if (!visited[neighbor]) {
            dfs(neighbor, graph, visited);
        }
    }
}
    

This code visits each node and explores all its neighbors recursively.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The for-loop over neighbors inside the dfs function.
• How many times: Each edge is checked once, and each node is visited once.

How Execution Grows With Input

As the graph grows, DFS visits every node and edge once.

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

Pattern observation: The work grows roughly with the sum of nodes and edges.

Final Time Complexity

Time Complexity: O(n + m)

This means DFS takes time proportional to the number of nodes plus the number of edges in the graph.

Common Mistake

[X] Wrong: "DFS always takes O(n²) time because it uses recursion inside a loop."

[OK] Correct: DFS visits each node and edge only once, so it does not repeat work unnecessarily. The recursion just helps explore neighbors efficiently.

Interview Connect

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

Self-Check

"What if the graph is represented as 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. Explore as far as possible along each branch before backtracking

B. Visit all neighbors of a node before moving deeper

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

D. Randomly visit nodes without any order

Solution

1. Step 1: Understand DFS traversal

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

2. Step 2: Compare with other traversal methods

   BFS visits neighbors first, but DFS goes deep first, matching Explore as far as possible along each branch before backtracking.

3. Final Answer:

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

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 C++ syntax to start a DFS from node 0 using recursion?

easy

A. DFS(visited, graph, 0);

B. dfs(0, visited, graph);

C. dfs(graph, 0, visited);

D. dfs(visited, 0, graph);

Solution

1. Step 1: Check typical DFS function signature

   Usually, DFS is called as dfs(node, visited, graph) where node is current node.

2. Step 2: Match the correct order

   dfs(0, visited, graph); matches the common order: dfs(0, visited, graph).

3. Final Answer:

   dfs(0, visited, graph); -> Option B

4. Quick Check:

   Call dfs with node first [OK]

Hint: DFS call starts with node, then visited, then graph [OK]

Common Mistakes:

• Swapping argument order
• Using uppercase DFS instead of dfs
• Passing graph before node

3. What will be the output order of nodes visited by DFS starting from node 0 in the graph below?

Graph adjacency list:
0: [1, 2]
1: [3]
2: []
3: []

Assuming DFS visits neighbors in ascending order.

medium

A. [0, 3, 1, 2]

B. [0, 2, 1, 3]

C. [0, 1, 3, 2]

D. [0, 1, 2, 3]

Solution

1. Step 1: Start DFS at node 0

   Visit 0, then explore neighbors in ascending order: 1, then 2.

2. Step 2: Explore neighbor 1

   From 1, visit 3 (only neighbor), then backtrack.

3. Step 3: Explore neighbor 2

   Node 2 has no neighbors, so visit and backtrack.

4. Final Answer:

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

5. Quick Check:

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

Hint: Visit neighbors in order, go deep first [OK]

Common Mistakes:

• Visiting node 2 before node 1
• Not backtracking properly
• Confusing BFS order with DFS

4. Given the following DFS code snippet, what is the error?

void dfs(int node, vector<bool> &visited, vector &graph) {
  visited[node] = true;
  for (int i = 0; i < graph[node].size(); i++) {
    int neighbor = graph[node][i];
    if (visited[neighbor] = false) {
      dfs(neighbor, visited, graph);
    }
  }
}

medium

A. The function should return an int instead of void

B. The loop should use 'i <= graph[node].size()' instead of '<'

C. The visited vector should be passed by value, not reference

D. The condition 'if (visited[neighbor] = false)' uses assignment instead of comparison

Solution

1. Step 1: Check the if condition inside the loop

   The condition uses '=' which assigns false instead of comparing.

2. Step 2: Understand the impact of assignment in condition

   This causes the condition to always be false, so DFS won't recurse properly.

3. Final Answer:

   The condition 'if (visited[neighbor] = false)' uses assignment instead of comparison -> Option D

4. Quick Check:

   Use '==' for comparison, not '=' [OK]

Hint: Use '==' to compare, not '=' [OK]

Common Mistakes:

• Using '=' instead of '==' in conditions
• Loop boundary off-by-one errors
• Passing visited by value causing no effect

5. You have a graph with 5 nodes (0 to 4) and edges: 0-1, 0-2, 1-3, 1-4. Using DFS starting at node 0, how can you modify the algorithm to count the number of nodes reachable from node 0?

hard

A. Add a counter variable incremented each time a node is visited during DFS

B. Count nodes after DFS by checking which nodes have no neighbors

C. Use BFS instead of DFS to count nodes reachable

D. Count nodes by summing the degrees of all nodes

Solution

1. Step 1: Understand counting reachable nodes

   Counting nodes visited during DFS means incrementing a counter when marking a node visited.

2. Step 2: Why other options fail

   Counting nodes with no neighbors or summing degrees won't give reachable count; BFS is alternative but not required.

3. Final Answer:

   Add a counter variable incremented each time a node is visited during DFS -> Option A

4. Quick Check:

   Count nodes when visited in DFS [OK]

Hint: Increment count inside DFS when visiting nodes [OK]

Common Mistakes:

• Counting nodes with no neighbors instead of visited
• Using BFS unnecessarily
• Summing degrees instead of counting visited