DFS Depth First Search on Graph in DSA Python - 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 the number of nodes and edges affects the work done.

Scenario Under Consideration

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

def dfs(graph, node, visited):
    visited.add(node)
    for neighbor in graph[node]:
        if neighbor not in visited:
            dfs(graph, neighbor, visited)

This code visits all nodes reachable from the starting node using recursion.

Identify Repeating Operations

Look for loops and recursive calls that repeat work.

• Primary operation: Visiting each node and exploring its neighbors.
• How many times: Each node is visited once, and each edge is checked 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 25 operations (10 nodes + 15 edges)
100 nodes, 200 edges	About 300 operations (100 + 200)
1000 nodes, 5000 edges	About 6000 operations (1000 + 5000)

Pattern observation: Operations grow roughly with the sum of nodes and edges.

Final Time Complexity

Time Complexity: O(n + m)

This means the time grows linearly with 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 does not multiply work but helps explore neighbors efficiently.

Interview Connect

Understanding DFS time complexity helps you explain how graph traversal scales and why it is efficient for many 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 does DFS (Depth First Search) do when exploring a graph?

easy

A. It explores as far as possible along each branch before backtracking.

B. It explores all neighbors at the current depth before going deeper.

C. It visits nodes randomly without any order.

D. It only visits nodes with the highest value first.

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 explores neighbors level by level, but DFS goes deep first.

3. Final Answer:

   It explores 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 nodes randomly
• Assuming DFS visits nodes by value

2. Which of the following is the correct Python syntax to start a DFS function on a graph represented as adjacency list graph from node start?

easy

A. def dfs(start, graph):

B. def dfs(graph, start):

C. def dfs(graph):

D. def dfs(start):

Solution

1. Step 1: Identify required parameters for DFS

   DFS needs the graph and the starting node to begin traversal.

2. Step 2: Check function signature correctness

   def dfs(graph, start): correctly defines a function with parameters (graph, start).

3. Final Answer:

   def dfs(graph, start): -> Option B

4. Quick Check:

   DFS needs graph and start node [OK]

Hint: DFS function needs graph and start node parameters [OK]

Common Mistakes:

• Swapping parameter order incorrectly
• Missing one of the parameters
• Using only one parameter when two are needed

3. What is the output of the following DFS traversal starting from node 0?

graph = {0: [1, 2], 1: [2], 2: [0, 3], 3: [3]}
visited = set()
def dfs(node):
    if node not in visited:
        print(node, end=' ')
        visited.add(node)
        for neighbor in graph[node]:
            dfs(neighbor)
dfs(0)

medium

A. 0 3 2 1

B. 0 2 3 1

C. 0 1 3 2

D. 0 1 2 3

Solution

1. Step 1: Start DFS from node 0

   Visit 0, mark visited. Neighbors: 1, 2.

2. Step 2: Visit neighbor 1

   Visit 1, mark visited. Neighbor: 2.

3. Step 3: Visit neighbor 2

   Visit 2, mark visited. Neighbors: 0 (visited), 3.

4. Step 4: Visit neighbor 3

   Visit 3, mark visited. Neighbor: 3 (already visited).

5. Final Answer:

   0 1 2 3 -> Option D

6. Quick Check:

   DFS order = 0 1 2 3 [OK]

Hint: Follow neighbors depth-first, avoid revisiting nodes [OK]

Common Mistakes:

• Visiting neighbors in wrong order
• Revisiting already visited nodes
• Confusing BFS order with DFS

4. Find the error in this DFS code snippet and select the fix:

graph = {0: [1], 1: [2], 2: []}
visited = set()
def dfs(node):
    if node in visited:
        return
    print(node)
    for neighbor in graph[node]:
        dfs(neighbor)
dfs(0)

medium

A. Add 'visited.add(node)' after printing node to mark it visited.

B. Change 'if node in visited' to 'if node not in visited'.

C. Remove the 'return' statement inside the if condition.

D. Add 'visited.remove(node)' after recursive calls.

Solution

1. Step 1: Identify missing visited marking

   The code never adds nodes to visited, so it revisits nodes infinitely.

2. Step 2: Fix by adding visited marking

   Adding 'visited.add(node)' after printing marks node as visited to avoid revisits.

3. Final Answer:

   Add 'visited.add(node)' after printing node to mark it visited. -> Option A

4. Quick Check:

   Mark visited after visiting node [OK]

Hint: Always add node to visited set after visiting [OK]

Common Mistakes:

• Forgetting to mark nodes visited
• Incorrect if condition logic
• Removing nodes from visited incorrectly

5. Given a directed graph represented as adjacency list:

graph = {0: [1, 2], 1: [2], 2: [3], 3: [1]}

Write the DFS traversal order starting from node 0 that detects cycles by tracking recursion stack. What is the order of nodes visited before detecting a cycle?

hard

A. 0 1 2

B. 0 2 3 1

C. 0 1 2 3

D. 0 2 1 3

Solution

1. Step 1: Start DFS from node 0 and track recursion stack

   Visit 0, then 1, then 2, then 3. At 3, neighbor 1 is in recursion stack, cycle detected.

2. Step 2: List nodes visited before cycle detection

   Nodes visited in order: 0, 1, 2, 3 before cycle found.

3. Final Answer:

   0 1 2 3 -> Option C

4. Quick Check:

   DFS with cycle detection visits 0 1 2 3 [OK]

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

Common Mistakes:

• Ignoring recursion stack for cycle detection
• Stopping traversal too early
• Visiting nodes in wrong order