Graphs and Network Algorithms · Graph Traversal

DFS Algorithm

Depth-first search explores as far as possible along a path before backtracking.

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Student Focus

We connect DFS to recursion so students can transfer earlier skills into graph problems.

Guided Lesson Notes

Understanding DFS Algorithm

DFS Algorithm focuses on relationships, reachability, paths, cycles, connectivity, and optimization over edges. Depth-first search explores as far as possible along a path before backtracking.

Model: draw vertices as dots and edges as connections; then decide whether direction, weight, or capacity matters.

Invariant: visited, distance, parent, component, or flow arrays must match what has actually been discovered so far.

Implementation: adjacency lists or matrices, queues, stacks, priority queues, union-find, and repeated edge relaxation.

Use DFS Algorithm when the problem mentions roads, networks, prerequisites, components, shortest paths, spanning costs, or dependency order.

Visual Model

A small picture for DFS Algorithm

Key Ideas

Visited tracking
Recursive or stack-based traversal
Backtracking behavior

Practice Prompts

Trace DFS order on a graph with a cycle.
Use DFS to count connected components.

Vocabulary

Terms students should be able to say clearly

Vertex

An object or state in the graph.

Edge

A relationship or move between vertices.

Path

A sequence of edges from one vertex to another.

Cycle

A path that returns to a previous vertex.

Component

A group of vertices connected by reachability.

Relaxation

Trying to improve a known distance or cost through an edge.

Worked Example

Worked example: tracing DFS Algorithm

Trace a small input, focusing on visited tracking.

1Write a small input where DFS Algorithm is clearly useful.
2Label the part of the input related to Visited tracking.
3Perform one operation and explain which invariant is still true afterward.
4Run a second operation that touches an edge case, such as an empty side, duplicate value, boundary index, disconnected vertex, or tie.
5Finish by saying which operation dominates the runtime and why.

Complexity Check

Costs students should be able to explain

Operation	Typical Cost	Reason
Adjacency-list traversal	O(V + E)	DFS and BFS visit vertices and inspect edges.
Adjacency matrix storage	O(V^2)	A matrix reserves space for every possible pair.
Weighted optimization	Varies	MST and shortest-path algorithms depend on sorting, heaps, and graph density.

Common Mistakes

What to watch while practicing

Coding before drawing the structure or state changes.
Forgetting the invariant that makes the algorithm correct.
Testing only the sample input and skipping boundary cases.
Giving Big-O without explaining which operation dominates the work.

Interview-Style Coding Problem

DFS Algorithm interview problem: Number of Islands

Interview medium

Problem

Given a grid of 0s and 1s, count how many connected islands of 1s exist using four-direction adjacency.

Input

The first line contains n and m. The next n lines contain m values, each 0 or 1.

Output

Print the island count.

Sample Input

Sample Output

Why the sample works

There are four separate groups of land cells.

Approach

1Scan every cell in the grid.
2When an unseen land cell is found, start a DFS or BFS and mark its entire island.
3Increase the island count once per new traversal.
4Water cells and already visited land cells are skipped.

O(nm) time and O(nm) extra space in the worst case.

Java Solution

import java.util.*;

public class Main {
  static int n;
  static int m;
  static int[][] grid;
  static boolean[][] seen;
  static int[] dr = {1, -1, 0, 0};
  static int[] dc = {0, 0, 1, -1};

  static void dfs(int r, int c) {
    seen[r][c] = true;
    for (int d = 0; d < 4; d++) {
      int nr = r + dr[d];
      int nc = c + dc[d];
      if (nr < 0 || nr >= n || nc < 0 || nc >= m) continue;
      if (grid[nr][nc] == 1 && !seen[nr][nc]) {
        dfs(nr, nc);
      }
    }
  }

  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    n = sc.nextInt();
    m = sc.nextInt();
    grid = new int[n][m];
    seen = new boolean[n][m];

    for (int r = 0; r < n; r++) {
      for (int c = 0; c < m; c++) {
        grid[r][c] = sc.nextInt();
      }
    }

    int islands = 0;
    for (int r = 0; r < n; r++) {
      for (int c = 0; c < m; c++) {
        if (grid[r][c] == 1 && !seen[r][c]) {
          islands++;
          dfs(r, c);
        }
      }
    }

    System.out.println(islands);
  }
}

Python Solution

n, m = map(int, input().split())
grid = [list(map(int, input().split())) for _ in range(n)]
seen = [[False] * m for _ in range(n)]

def dfs(r, c):
    seen[r][c] = True
    for dr, dc in ((1, 0), (-1, 0), (0, 1), (0, -1)):
        nr, nc = r + dr, c + dc
        if 0 <= nr < n and 0 <= nc < m:
            if grid[nr][nc] == 1 and not seen[nr][nc]:
                dfs(nr, nc)

islands = 0
for r in range(n):
    for c in range(m):
        if grid[r][c] == 1 and not seen[r][c]:
            islands += 1
            dfs(r, c)

print(islands)

Practice Challenge

Make the idea your own

Create a two-minute explanation of DFS Algorithm: define it, trace one example, name one edge case, and give the runtime of the main operation.

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