Why DFS Depth First Search on Graph in DSA C?

The Big Idea

What if you could explore every path in a maze without ever getting lost or repeating steps?

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 certain room starting from the entrance. Trying to check every possible path by hand would be confusing and take forever.

The Problem

Manually exploring each path can be slow and easy to get lost. You might repeat the same paths again and again or miss some rooms. It is hard to keep track of where you have been and where to go next without a clear plan.

The Solution

Depth First Search (DFS) helps by exploring one path deeply before backtracking. It uses a simple rule: go as far as possible along one path, then back up and try another. This way, you systematically visit all connected rooms without missing or repeating.

Before vs After

✗ Before

void explore(int current) {
  // try all neighbors manually
  if (neighbor1 not visited) explore(neighbor1);
  if (neighbor2 not visited) explore(neighbor2);
  // ...
}

✓ After

void dfs(int current) {
  visited[current] = 1;
  for (int i = 0; i < graph[current].size(); i++) {
    int neighbor = graph[current][i];
    if (!visited[neighbor]) dfs(neighbor);
  }
}

What It Enables

DFS enables efficient and complete exploration of all connected parts of a graph or maze, making complex problems manageable.

Real Life Example

Finding all friends connected to you in a social network by exploring friend connections deeply before moving to other groups.

Key Takeaways

Manual path checking is slow and error-prone.

DFS explores deeply along one path before backtracking.

DFS systematically visits all connected nodes without repetition.

Practice

1. What is the main idea behind Depth First Search (DFS) on a graph?

easy

A. Visit all nodes at the current depth before moving deeper

B. Visit nodes randomly without any order

C. Explore as far as possible along each branch before backtracking

D. Only visit nodes with no outgoing edges

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 visits nodes level by level, but DFS goes deep first.

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 nodes level-wise
• Assuming DFS visits nodes randomly

2. Which of the following is the correct way to start a DFS function in C for a graph represented by adjacency list adj and visited array visited?

easy

A. void dfs(int node) { visited[node] = 1; for(int i=0; i

B. void dfs(int node) { visited[node] = true; for(int i=0; i

C. void dfs(int node) { visited[node] = 0; for(int i=0; i

D. void dfs(int node) { visited[node] = 1; for(int i=0; i

Solution

1. Step 1: Mark current node as visited

   Set visited[node] = 1 (or true) to avoid revisiting.

2. Step 2: Traverse neighbors only if not visited

   Check if neighbor is unvisited before recursive DFS call to prevent infinite loops.

3. Final Answer:

   void dfs(int node) { visited[node] = 1; for(int i=0; i -> Option D

4. Quick Check:

   Mark visited and check before recursion [OK]

Hint: Always check visited before recursive DFS call [OK]

Common Mistakes:

• Not checking if neighbor is visited
• Using wrong visited value (0 instead of 1)
• Using .size() instead of .length in C arrays

3. Consider the following graph adjacency list:
0: [1, 2]
1: [2]
2: [0, 3]
3: [3]
Starting DFS from node 2, what is the order of nodes visited?

medium

A. [2, 3, 0, 1]

B. [2, 0, 1, 3]

C. [2, 0, 3, 1]

D. [2, 1, 0, 3]

Solution

1. Step 1: Start DFS at node 2

   Visit 2, mark visited. Neighbors: 0 and 3.

2. Step 2: Visit neighbor 0 first

   From 2, go to 0. Visit 0, neighbors: 1 and 2 (2 already visited).

3. Step 3: Visit neighbor 1 from 0

   Visit 1, neighbor: 2 (visited). Backtrack.

4. Step 4: Back to 2, visit neighbor 3

   Visit 3, neighbor: 3 (already visited). End.

5. Final Answer:

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

6. Quick Check:

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

Hint: Follow neighbors depth-first, mark visited to avoid repeats [OK]

Common Mistakes:

• Visiting nodes multiple times
• Visiting neighbors in wrong order
• Ignoring visited checks causing loops

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

void dfs(int node) {
  visited[node] = 1;
  for(int i=0; i<adj[node].length; i++) {
    dfs(adj[node][i]);
  }
}

medium

A. Missing check if neighbor is already visited before recursive call

B. Visited array is not initialized

C. Incorrect loop condition syntax

D. Visited[node] should be set after recursion

Solution

1. Step 1: Analyze DFS recursion calls

   The code calls dfs on all neighbors without checking if they are visited.

2. Step 2: Identify risk of infinite recursion

   Without visited check, DFS may revisit nodes endlessly causing stack overflow.

3. Final Answer:

   Missing check if neighbor is already visited before recursive call -> Option A

4. Quick Check:

   Always check visited before recursion [OK]

Hint: Check visited before recursive DFS call to avoid loops [OK]

Common Mistakes:

• Assuming visited check is optional
• Setting visited after recursion
• Ignoring infinite recursion risk

5. Given a directed graph with possible cycles, how can DFS be modified to detect if a cycle exists?

hard

A. Use a recursion stack array to track nodes in the current path and check if a node is revisited in this stack

B. Only mark nodes visited after all neighbors are processed

C. Use BFS instead of DFS to detect cycles

D. Ignore visited array and rely on node values

Solution

1. Step 1: Understand cycle detection in DFS

   Detect cycles by tracking nodes currently in recursion (call) stack.

2. Step 2: Use recursion stack array

   Maintain an array marking nodes in current DFS path; if a neighbor is in this stack, a cycle exists.

3. Step 3: Differentiate from visited array

   Visited marks all visited nodes; recursion stack tracks current path only.

4. Final Answer:

   Use a recursion stack array to track nodes in the current path and check if a node is revisited in this stack -> Option A

5. Quick Check:

   Recursion stack detects cycles in DFS [OK]

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

Common Mistakes:

• Confusing visited with recursion stack
• Using BFS for cycle detection in directed graphs
• Not tracking current path nodes