DFS Depth First Search on Graph in DSA Java - 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 using DFS changes as the graph grows.

Specifically, how does the number of steps grow when the graph has more nodes and edges?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

void dfs(int node, boolean[] visited, List<List<Integer>> graph) {
    visited[node] = true;
    for (int neighbor : graph.get(node)) {
        if (!visited[neighbor]) {
            dfs(neighbor, visited, graph);
        }
    }
}

This code visits all nodes reachable from the starting node using depth-first search on an adjacency list graph.

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.

How Execution Grows With Input

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

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

Pattern observation: The total steps grow roughly in proportion to the number of nodes plus edges.

Final Time Complexity

Time Complexity: O(n + m)

This means the time to run DFS grows linearly with the total number of nodes and edges in the graph.

Common Mistake

[X] Wrong: "DFS takes O(n²) time because it uses recursion inside a loop."

[OK] Correct: Each node and edge is visited only once, so the total work is linear, not quadratic.

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 was represented by 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. Randomly visit nodes until all are visited

B. Visit all neighbors of a node before moving deeper

C. Visit nodes in order of their distance from the start node

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

Solution

1. Step 1: Understand DFS traversal

   DFS explores a graph by going deep into one branch before backtracking to explore other branches.

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 D

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 way to start a DFS traversal from node 0 in Java using recursion?

easy

A. dfs(0, visited, graph);

B. DFS(visited, graph, 0);

C. dfs(graph, 0, visited);

D. dfs(visited, 0, graph);

Solution

1. Step 1: Identify typical DFS method signature

   Common DFS signature is dfs(currentNode, visitedArray, graph).

2. Step 2: Match options to this signature

   dfs(0, visited, graph); matches the order: node, visited, graph. Others have wrong parameter order.

3. Final Answer:

   dfs(0, visited, graph); -> Option A

4. Quick Check:

   Correct parameter order = dfs(node, visited, graph) [OK]

Hint: DFS method usually starts with node, visited, graph [OK]

Common Mistakes:

• Mixing parameter order
• Using uppercase method name 'DFS'
• Passing graph before node

3. Given the graph adjacency list: graph = {{1, 2}, {0, 3}, {0}, {1}} and starting DFS from node 0, what is the order of nodes visited?

medium

A. [0, 1, 3, 2]

B. [0, 2, 1, 3]

C. [0, 1, 2, 3]

D. [0, 3, 1, 2]

Solution

1. Step 1: Start DFS at node 0 and visit neighbors

   From 0, neighbors are 1 and 2. DFS visits 1 first (assuming adjacency order).

2. Step 2: From node 1, visit its neighbors

   Neighbors of 1 are 0 (visited) and 3. Visit 3 next.

3. Step 3: Backtrack and visit remaining neighbors

   After 3, backtrack to 0 and visit 2.

4. Final Answer:

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

5. Quick Check:

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

Hint: Follow neighbors deeply in adjacency order [OK]

Common Mistakes:

• Visiting neighbors in wrong order
• Not backtracking correctly
• Confusing BFS order with DFS

4. What is wrong with this DFS code snippet in Java?

void dfs(int node, boolean[] visited, List<Integer>[] graph) {
  visited[node] = true;
  for (int neighbor : graph[node]) {
    if (visited[neighbor]) {
      dfs(neighbor, visited, graph);
    }
  }
}

medium

A. It does not mark nodes as visited

B. It calls dfs on already visited neighbors causing infinite recursion

C. It uses wrong data types for graph

D. It misses the base case for recursion

Solution

1. Step 1: Check visited condition before recursive call

   The code calls dfs(neighbor) only if visited[neighbor] is true, which means already visited.

2. Step 2: Identify the problem caused

   This causes infinite recursion because it revisits nodes already visited instead of skipping them.

3. Final Answer:

   It calls dfs on already visited neighbors causing infinite recursion -> Option B

4. Quick Check:

   Call dfs only if neighbor not visited [OK]

Hint: Call dfs only if neighbor is NOT visited [OK]

Common Mistakes:

• Calling dfs on visited nodes
• Forgetting to mark visited nodes
• Incorrect loop over neighbors

5. You want to find if there is a path between node 0 and node 4 in this graph:

graph = {{1, 2}, {0, 3}, {0, 4}, {1}, {2}}

Which DFS approach correctly determines if node 4 is reachable from node 0?

hard

A. Run DFS from 4, stop and return true if node 0 is found during traversal

B. Run DFS from 0 and count all visited nodes without checking for 4

C. Run DFS from 0, stop and return true if node 4 is found during traversal

D. Run DFS from 0 but skip node 2 to avoid cycles

Solution

1. Step 1: Understand the goal

   We want to check if node 4 is reachable from node 0, so start DFS at 0.

2. Step 2: Use DFS to find target node

   During DFS, if node 4 is found, stop and return true immediately.

3. Step 3: Evaluate options

   Run DFS from 0, stop and return true if node 4 is found during traversal matches this approach. Run DFS from 4, stop and return true if node 0 is found during traversal starts from wrong node. Run DFS from 0 and count all visited nodes without checking for 4 ignores target. Run DFS from 0 but skip node 2 to avoid cycles skips nodes incorrectly.

4. Final Answer:

   Run DFS from 0, stop and return true if node 4 is found during traversal -> Option C

5. Quick Check:

   DFS from start, stop on target found [OK]

Hint: Stop DFS early when target node is found [OK]

Common Mistakes:

• Starting DFS from wrong node
• Not checking for target during traversal
• Skipping nodes that are part of path