DFS Depth First Search on Graph in DSA Go - 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 Depth First Search (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.

func DFS(graph map[int][]int, start int, visited map[int]bool) {
    visited[start] = true
    for _, neighbor := range graph[start] {
        if !visited[neighbor] {
            DFS(graph, neighbor, visited)
        }
    }
}
    

This code visits all nodes reachable from the start node by exploring as far as possible along each branch before backtracking.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Visiting each node and exploring its neighbors recursively.
• How many times: Each node is visited once, and each edge is checked once during the traversal.

How Execution Grows With Input

As the graph grows, the DFS visits every node and edge once, so the work grows with the total nodes and edges.

Input Size (n nodes, m edges)	Approx. Operations
10 nodes, 15 edges	About 25 visits (10 nodes + 15 edges)
100 nodes, 200 edges	About 300 visits (100 nodes + 200 edges)
1000 nodes, 3000 edges	About 4000 visits (1000 nodes + 3000 edges)

Pattern observation: The total steps increase roughly by the sum of nodes and edges as the graph grows.

Final Time Complexity

Time Complexity: O(n + m)

This means the time to finish DFS 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 and loops."

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

Interview Connect

Understanding DFS time complexity helps you explain how graph algorithms scale, 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. 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 without any order

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 options with DFS behavior

   Explore as far as possible along each branch before backtracking matches this behavior exactly, while others describe BFS or incorrect methods.

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 function in Go for a graph represented as adjacency list graph map[int][]int?

easy

A. func dfs(node int, visited map[int]bool) {}

B. func dfs(node int, visited []int) {}

C. func dfs(node int, visited map[int]int) {}

D. func dfs(node int, visited []bool) {}

Solution

1. Step 1: Identify correct visited type for DFS

   Visited should be a map[int]bool to mark nodes visited or not.

2. Step 2: Check options for visited parameter type

   Only func dfs(node int, visited map[int]bool) {} uses map[int]bool, others use incorrect types.

3. Final Answer:

   func dfs(node int, visited map[int]bool) {} -> Option A

4. Quick Check:

   Visited map with bool = correct DFS parameter [OK]

Hint: Use map[int]bool to track visited nodes in Go DFS [OK]

Common Mistakes:

• Using slice instead of map for visited
• Using wrong map value type
• Ignoring visited parameter

3. What is the output of the following Go code snippet using DFS on the graph?

graph := map[int][]int{
  1: {2, 3},
  2: {4},
  3: {},
  4: {},
}
visited := map[int]bool{}
var result []int
var dfs func(int)
dfs = func(node int) {
  if visited[node] {
    return
  }
  visited[node] = true
  result = append(result, node)
  for _, n := range graph[node] {
    dfs(n)
  }
}
dfs(1)
fmt.Println(result)

medium

A. [1 2 4 3]

B. [1 3 2 4]

C. [1 2 3 4]

D. [1 4 2 3]

Solution

1. Step 1: Trace DFS starting at node 1

   Visit 1, then go to 2, then 4 (no children), backtrack, then visit 3.

2. Step 2: Record order of visited nodes

   Order is 1, 2, 4, 3 as DFS explores deep before backtracking.

3. Final Answer:

   [1 2 4 3] -> Option A

4. Quick Check:

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

Hint: Follow each branch fully before moving to next neighbor [OK]

Common Mistakes:

• Visiting neighbors in wrong order
• Confusing BFS order with DFS
• Not marking visited nodes

4. The following DFS code in Go has a bug. What is the error?

func dfs(node int, graph map[int][]int, visited map[int]bool) {
  if visited[node] {
    return
  }
  visited[node] = true
  for _, n := range graph[node] {
    dfs(n, graph, visited)
  }
}

// Called as dfs(1, graph, visited)

medium

A. Missing base case to stop recursion

B. Visited map is not updated correctly

C. Incorrect parameter types in function

D. No code to process or print the node

Solution

1. Step 1: Analyze DFS function behavior

   Function marks visited and recurses but does not process or print nodes.

2. Step 2: Identify missing action

   DFS usually processes nodes (e.g., print or collect), missing here causes no visible output.

3. Final Answer:

   No code to process or print the node -> Option D

4. Quick Check:

   DFS missing node processing step [OK]

Hint: DFS must process nodes inside function [OK]

Common Mistakes:

• Assuming marking visited is enough
• Forgetting to print or collect nodes
• Thinking recursion stops without base case

5. Given a graph with possible cycles, how can you modify DFS in Go to detect if a cycle exists?

hard

A. Only mark nodes visited after all neighbors are processed

B. Use a recursion stack map to track nodes currently in the call stack

C. Use BFS instead of DFS to detect cycles

D. Ignore visited nodes and traverse all edges

Solution

1. Step 1: Understand cycle detection in DFS

   Detect cycles by tracking nodes currently in recursion stack to find back edges.

2. Step 2: Choose correct method

   Using a recursion stack map helps detect if a node is revisited while still in the current path.

3. Final Answer:

   Use a recursion stack map to track nodes currently in the call stack -> Option B

4. Quick Check:

   Cycle detection = recursion stack tracking [OK]

Hint: Track recursion stack to find cycles in DFS [OK]

Common Mistakes:

• Marking visited too late
• Using BFS which doesn't detect cycles easily
• Ignoring visited nodes causing infinite loops