📘 Blog 3: DFS Pattern (Path Exploration, Connectivity, Components) 🌐

🔹 Why DFS Matters?

Depth-First Search (DFS) explores a graph/tree as far as possible along one branch before backtracking.
Think of a maze explorer: keep going forward until you hit a wall, then backtrack and try a different path.

DFS is us…


This content originally appeared on DEV Community and was authored by Dev Cookies

🔹 Why DFS Matters?

Depth-First Search (DFS) explores a graph/tree as far as possible along one branch before backtracking.
Think of a maze explorer: keep going forward until you hit a wall, then backtrack and try a different path.

DFS is used when:

  • You want to explore all possible paths (e.g., backtracking problems).
  • You need connectivity detection (components, cycles).
  • Graph is huge, and recursion helps break it naturally.

🔹 DFS Core Idea

  1. Start from a source node.
  2. Recursively (or using a stack) visit an unvisited neighbor.
  3. Backtrack when no unvisited neighbors remain.

🔹 DFS Template in Java

Recursive DFS 

import java.util.*;

public class DFSExample {
    public static void dfs(int node, boolean[] visited, List<List<Integer>> graph) {
        visited[node] = true;
        System.out.print(node + " ");

        for (int neighbor : graph.get(node)) {
            if (!visited[neighbor]) {
                dfs(neighbor, visited, graph);
            }
        }
    }

    public static void main(String[] args) {
        int n = 5; 
        List<List<Integer>> graph = new ArrayList<>();
        for (int i = 0; i < n; i++) graph.add(new ArrayList<>());

        // Undirected edges
        graph.get(0).add(1); graph.get(1).add(0);
        graph.get(0).add(2); graph.get(2).add(0);
        graph.get(1).add(3); graph.get(3).add(1);
        graph.get(2).add(4); graph.get(4).add(2);

        boolean[] visited = new boolean[n];
        dfs(0, visited, graph); // Output: 0 1 3 2 4
    }
}

Iterative DFS (using stack) 

public void dfsIterative(int start, List<List<Integer>> graph) {
    boolean[] visited = new boolean[graph.size()];
    Stack<Integer> stack = new Stack<>();
    stack.push(start);

    while (!stack.isEmpty()) {
        int node = stack.pop();
        if (!visited[node]) {
            visited[node] = true;
            System.out.print(node + " ");

            for (int neighbor : graph.get(node)) {
                if (!visited[neighbor]) {
                    stack.push(neighbor);
                }
            }
        }
    }
}

🔹 DFS Use Cases (Patterns)

1. Path Exploration

DFS is the go-to for exploring all possible paths.

👉 Example: Path from source to destination in DAG

public boolean hasPath(List<List<Integer>> graph, int src, int dest, boolean[] visited) {
    if (src == dest) return true;
    visited[src] = true;
    for (int neighbor : graph.get(src)) {
        if (!visited[neighbor] && hasPath(graph, neighbor, dest, visited)) {
            return true;
        }
    }
    return false;
}

👉 Related Problems:

  • Word Search (LeetCode 79)
  • N-Queens
  • All Paths from Source to Target (LeetCode 797)

2. Connected Components

DFS helps find how many disconnected parts exist in a graph.

👉 Example: Count connected components in undirected graph

public int countComponents(int n, int[][] edges) {
    List<List<Integer>> graph = new ArrayList<>();
    for (int i = 0; i < n; i++) graph.add(new ArrayList<>());
    for (int[] e : edges) {
        graph.get(e[0]).add(e[1]);
        graph.get(e[1]).add(e[0]);
    }

    boolean[] visited = new boolean[n];
    int count = 0;

    for (int i = 0; i < n; i++) {
        if (!visited[i]) {
            dfs(i, visited, graph);
            count++;
        }
    }
    return count;
}

👉 Related Problems:

  • Number of Islands (LeetCode 200)
  • Connected Components in Graph (LeetCode 323)

3. Cycle Detection

DFS helps detect cycles in both directed and undirected graphs.

👉 Cycle detection in undirected graph

public boolean hasCycleUndirected(int node, int parent, boolean[] visited, List<List<Integer>> graph) {
    visited[node] = true;
    for (int neighbor : graph.get(node)) {
        if (!visited[neighbor]) {
            if (hasCycleUndirected(neighbor, node, visited, graph)) return true;
        } else if (neighbor != parent) {
            return true;
        }
    }
    return false;
}

👉 Cycle detection in directed graph (using recursion stack)

public boolean hasCycleDirected(int node, boolean[] visited, boolean[] recStack, List<List<Integer>> graph) {
    if (recStack[node]) return true;
    if (visited[node]) return false;

    visited[node] = true;
    recStack[node] = true;

    for (int neighbor : graph.get(node)) {
        if (hasCycleDirected(neighbor, visited, recStack, graph)) return true;
    }

    recStack[node] = false;
    return false;
}

👉 Related Problems:

  • Course Schedule (LeetCode 207)
  • Detect Cycle in Graph

4. Topological Sorting (DFS-based)

DFS can be used to generate topological order in DAGs.

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

👉 Related Problems:

  • Course Schedule II (LeetCode 210)

5. Backtracking (DFS with undo step)

DFS naturally extends to backtracking problems.

👉 Example: Word Search (LeetCode 79)

public boolean exist(char[][] board, String word) {
    int m = board.length, n = board[0].length;
    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
            if (dfs(board, word, 0, i, j)) return true;
        }
    }
    return false;
}

private boolean dfs(char[][] board, String word, int idx, int i, int j) {
    if (idx == word.length()) return true;
    if (i<0 || j<0 || i>=board.length || j>=board[0].length || board[i][j] != word.charAt(idx)) 
        return false;

    char temp = board[i][j];
    board[i][j] = '#'; // mark visited

    boolean found = dfs(board, word, idx+1, i+1, j) || dfs(board, word, idx+1, i-1, j) ||
                    dfs(board, word, idx+1, i, j+1) || dfs(board, word, idx+1, i, j-1);

    board[i][j] = temp; // undo (backtrack)
    return found;
}

🔹 DFS Interview Problem Map

  • Path Exploration: Word Search, N-Queens, All Paths.
  • Connectivity: Number of Islands, Connected Components.
  • Cycle Detection: Course Schedule, Detect Cycle in Graph.
  • Topological Sort: Course Schedule II.
  • Backtracking: Sudoku Solver, Word Search, Combinations/Permutations.

🔹 DFS Complexity

  • Time: O(V + E) (Vertices + Edges).

  • Space:

    • Recursive stack: O(V) in worst case.
    • Iterative DFS: O(V) for stack + visited.

✅ Key Takeaway:
DFS is the Swiss Army knife for exploration problems:

  • Useful when you need to explore all paths, detect cycles, or analyze connectivity.
  • Natural fit for backtracking problems (try → explore → undo).


This content originally appeared on DEV Community and was authored by Dev Cookies


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