Why DFS Depth First Search on Graph in DSA Typescript?

The Big Idea

What if you could explore every path without ever getting lost or repeating yourself?

The Scenario

Imagine you have a huge maze with many paths and rooms. You want to find if there is a way to reach a special room starting from the entrance. If you try to check every path by guessing randomly or writing down every step manually, it will take forever and you might get lost or repeat the same paths.

The Problem

Manually exploring each path in a maze is slow and confusing. You can easily miss some paths or go in circles. Writing down every step without a clear plan leads to mistakes and wastes time. It is hard to remember which rooms you already checked.

The Solution

Depth First Search (DFS) is like having a smart helper who explores one path deeply before backtracking and trying another. It remembers where it has been, so it never repeats rooms. This way, it quickly finds if the special room is reachable or not, without getting lost or confused.

Before vs After

✗ Before

function findRoom(maze, start, target) {
  // Try all paths randomly without tracking
  // May repeat rooms or miss paths
}

✓ After

function dfs(graph, node, visited) {
  if (visited.has(node)) return;
  visited.add(node);
  for (const neighbor of graph[node]) {
    dfs(graph, neighbor, visited);
  }
}

What It Enables

DFS enables systematic and efficient exploration of all connected parts of a graph or maze, ensuring no path is missed or repeated.

Real Life Example

Finding all friends in a social network starting from one person by exploring their friends, then friends of friends, and so on.

Key Takeaways

Manual path checking is slow and error-prone.

DFS explores deeply and remembers visited nodes.

DFS helps find connections or paths efficiently in graphs.

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