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

Time Complexity: DFS Depth First Search on Graph

O(n + e)

Understanding Time Complexity

We want to understand how the time taken by Depth First Search (DFS) changes as the graph size grows.

How does DFS scale when exploring nodes and edges in a graph?

Scenario Under Consideration

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

function dfs(node: number, graph: number[][], visited: boolean[]): void {
  visited[node] = true;
  for (const neighbor of graph[node]) {
    if (!visited[neighbor]) {
      dfs(neighbor, graph, visited);
    }
  }
}
    

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

Identify Repeating Operations

Look for loops and recursive calls that repeat work.

• Primary operation: Visiting each node and checking 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 all nodes and edges once.

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

Pattern observation: Operations grow roughly proportional to nodes plus edges.

Final Time Complexity

Time Complexity: O(n + e)

This means DFS takes time proportional to 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 does not repeat work unnecessarily. The time depends on nodes plus edges, not their square.

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 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 until all are visited

Solution

1. Step 1: Understand DFS traversal strategy

   DFS explores nodes 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 goes deep first, BFS goes wide first [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 TypeScript function signature for a DFS function on a graph represented as an adjacency list?

easy

A. function dfs(graph: Map, start: number): void

B. function dfs(graph: number[], start: number): void

C. function dfs(graph: number[][], start: number): void

D. function dfs(graph: string[][], start: string): void

Solution

1. Step 1: Identify graph representation type

   An adjacency list for a graph with numbered nodes is usually number[][], where each index holds neighbors.

2. Step 2: Check parameter types

   Start node is a number, so function signature with graph: number[][] and start: number is correct.

3. Final Answer:

   function dfs(graph: number[][], start: number): void -> Option C

4. Quick Check:

   Adjacency list = number[][], start node = number [OK]

Hint: Adjacency list uses array of arrays for neighbors [OK]

Common Mistakes:

• Using number[] instead of number[][] for adjacency list
• Using Map with incorrect value type
• Using string types for numeric graphs

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

const graph = [[1, 2], [2], [0, 3], [3]];
const visited: boolean[] = [];
const result: number[] = [];
function dfs(node: number) {
  if (visited[node]) return;
  visited[node] = true;
  result.push(node);
  for (const neighbor of graph[node]) {
    dfs(neighbor);
  }
}
dfs(0);
console.log(result);

medium

A. [0, 2, 3, 1]

B. [0, 2, 1, 3]

C. [0, 1, 2]

D. [0, 1, 2, 3]

Solution

1. Step 1: Start DFS at node 0

   Visit 0, mark visited, add to result: [0]. Neighbors: 1, 2.

2. Step 2: Visit neighbor 1

   Visit 1, mark visited, add: [0,1]. Neighbor: 2.

3. Step 3: Visit neighbor 2 from 1

   Visit 2, mark visited, add: [0,1,2]. Neighbors: 0 (visited), 3.

4. Step 4: Visit neighbor 3 from 2

   Visit 3, mark visited, add: [0,1,2,3]. Neighbor: 3 (already visited).

5. Final Answer:

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

6. Quick Check:

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

Hint: Follow neighbors recursively, mark visited to avoid repeats [OK]

Common Mistakes:

• Not marking nodes visited before recursion
• Visiting neighbors in wrong order
• Ignoring already visited nodes

4. Identify the bug in this DFS implementation and choose the fix:

function dfs(node: number, graph: number[][], visited: boolean[]) {
  visited[node] = true;
  for (const neighbor of graph[node]) {
    if (!visited[neighbor]) {
      dfs(neighbor, graph, visited);
    }
  }
}

const graph = [[1], [2], [0]];
const visited = new Array(graph.length).fill(false);
dfs(0, graph, visited);
console.log(visited);

medium

A. Mark visited[node] = true after recursive calls

B. Initialize visited array with true instead of false

C. Add a base case to return if node is already visited at start

D. No bug, code works correctly

Solution

1. Step 1: Check visited marking timing

   Visited[node] is marked true before recursion, preventing revisits.

2. Step 2: Verify recursion and base case

   Recursion only happens if neighbor not visited, so no infinite loops.

3. Final Answer:

   No bug, code works correctly -> Option D

4. Quick Check:

   Visited marked before recursion = correct DFS [OK]

Hint: Mark visited before recursion to avoid cycles [OK]

Common Mistakes:

• Marking visited after recursion causing infinite loops
• Initializing visited array incorrectly
• Missing base case check

5. Given a graph with possible cycles, how can you modify DFS to detect if a cycle exists starting from a given node? Choose the best approach.

hard

A. Use a recursion stack array to track nodes in the current path and detect back edges

B. Only mark nodes visited after all neighbors are visited

C. Use BFS instead of DFS to detect cycles

D. Ignore visited nodes and continue traversal

Solution

1. Step 1: Understand cycle detection in DFS

   Detecting cycles requires knowing if a node is in the current recursion path (stack).

2. Step 2: Use recursion stack tracking

   Maintain a separate array to track nodes currently in recursion; if a neighbor is in this stack, a cycle exists.

3. Step 3: Evaluate other options

   Marking visited after neighbors or using BFS won't reliably detect cycles; ignoring visited nodes causes infinite loops.

4. Final Answer:

   Use a recursion stack array to track nodes in the current path and detect back edges -> Option A

5. Quick Check:

   Cycle detection = recursion stack tracking [OK]

Hint: Track recursion path to find back edges indicating cycles [OK]

Common Mistakes:

• Not tracking recursion stack causing missed cycles
• Using BFS which doesn't track recursion path
• Marking visited too late or ignoring visited nodes