Why Number of Islands BFS and DFS in DSA C++?

The Big Idea

What if you could find every island on a map without missing a single piece or counting twice, all with a simple trick?

The Scenario

Imagine you have a big map made of water and land squares. You want to count how many separate islands are there. Doing this by looking at each square and trying to remember which ones you already counted is like trying to find all your friends in a huge crowd without any help.

The Problem

Manually checking each land square and trying to remember if it belongs to an island you already counted is slow and confusing. You might count the same island twice or miss some parts. It's easy to get lost and make mistakes.

The Solution

Using BFS (Breadth-First Search) or DFS (Depth-First Search) helps you explore each island completely before moving to the next. These methods systematically visit all connected land squares, so you never count the same island twice and never miss any part.

Before vs After

✗ Before

int countIslands(vector<vector<char>>& grid) {
  int count = 0;
  for (int i = 0; i < grid.size(); i++) {
    for (int j = 0; j < grid[0].size(); j++) {
      if (grid[i][j] == '1') {
        // Manually try to find connected land
        count++;
      }
    }
  }
  return count;
}

✓ After

void dfs(int i, int j, vector<vector<char>>& grid) {
  if (i < 0 || j < 0 || i >= grid.size() || j >= grid[0].size() || grid[i][j] == '0') return;
  grid[i][j] = '0';
  dfs(i+1, j, grid);
  dfs(i-1, j, grid);
  dfs(i, j+1, grid);
  dfs(i, j-1, grid);
}

int numIslands(vector<vector<char>>& grid) {
  int count = 0;
  for (int i = 0; i < grid.size(); i++) {
    for (int j = 0; j < grid[0].size(); j++) {
      if (grid[i][j] == '1') {
        dfs(i, j, grid);
        count++;
      }
    }
  }
  return count;
}

What It Enables

This lets you quickly and correctly count all islands in a map, no matter how big or complex, without missing or double counting any.

Real Life Example

Think about a drone scanning a flooded area to find all dry land patches. Using BFS or DFS helps the drone's software identify each separate dry patch accurately and fast.

Key Takeaways

Manually counting islands is confusing and error-prone.

BFS and DFS explore connected land squares fully to avoid mistakes.

This method makes counting islands fast, reliable, and easy to implement.

Practice

1. What does the "Number of Islands" problem ask you to find in a 2D grid of '1's (land) and '0's (water)?

easy

A. Count total number of '1's in the grid

B. Count how many separate groups of connected '1's exist

C. Find the largest island size

D. Find the shortest path between two islands

Solution

1. Step 1: Understand the problem definition

   The problem asks to count distinct islands, where an island is a group of connected '1's horizontally or vertically.

2. Step 2: Clarify what is being counted

   We count how many separate clusters of '1's exist, not the total number of '1's or sizes.

3. Final Answer:

   Count how many separate groups of connected '1's exist -> Option B

4. Quick Check:

   Number of islands = count of connected '1' groups [OK]

Hint: Islands = groups of connected '1's, count groups not cells [OK]

Common Mistakes:

• Counting total '1's instead of groups
• Including diagonal connections as connected
• Confusing island size with island count

2. Which of the following C++ code snippets correctly initializes a 2D vector grid of chars with given rows and columns filled with '0'?

easy

A. std::vector grid(rows, std::vector<char>(cols, '0'));

B. std::vector<char> grid(rows, std::vector<char>(cols, '0'));

C. std::vector grid(cols, std::vector<char>(rows, '0'));

D. std::vector grid(rows, cols, '0');

Solution

1. Step 1: Understand 2D vector initialization syntax

   To create a 2D vector of chars, use vector of vectors: outer vector size = rows, inner vector size = cols.

2. Step 2: Check each option

   std::vector grid(rows, std::vector<char>(cols, '0')); correctly creates 'rows' vectors each with 'cols' chars initialized to '0'. Others have wrong types or parameters.

3. Final Answer:

   std::vector grid(rows, std::vector<char>(cols, '0')); -> Option A

4. Quick Check:

   2D vector init = vector of vectors with sizes [OK]

Hint: Use vector(rows, vector<char>(cols, '0')) [OK]

Common Mistakes:

• Swapping rows and cols sizes
• Using wrong vector types
• Trying to initialize 2D vector with three parameters

3. Given the following grid and BFS-based code snippet to count islands, what is the output?

std::vector grid = {
  {'1','1','0','0'},
  {'1','0','0','1'},
  {'0','0','1','1'},
  {'0','0','0','0'}
};
int countIslands(const std::vector& grid) {
  int count = 0;
  int rows = grid.size();
  int cols = grid[0].size();
  std::vector visited(rows, std::vector<bool>(cols, false));
  std::queue> q;
  int directions[4][2] = {{1,0},{-1,0},{0,1},{0,-1}};
  for(int i=0; i=0 && nr=0 && nc

medium

A. 3

B. 4

C. 2

D. 1

Solution

1. Step 1: Identify islands in the grid

   Islands are groups of connected '1's. Here, islands are: top-left cluster (3 cells), single '1' at (1,3), and bottom cluster (2 cells at (2,2),(2,3)). Total 3 islands.

2. Step 2: Understand BFS traversal

   The BFS marks all connected '1's visited per island, so count increments once per island found.

3. Final Answer:

   3 -> Option A

4. Quick Check:

   Count islands = 3 [OK]

Hint: Count connected groups by BFS marking visited [OK]

Common Mistakes:

• Counting diagonal connections as islands
• Missing isolated '1's as separate islands
• Not marking visited correctly causing overcount

4. The following DFS code for counting islands has a bug. What is the bug?

void dfs(int r, int c, std::vector& grid) {
  int rows = grid.size();
  int cols = grid[0].size();
  if(r < 0 || c < 0 || r >= rows || c >= cols || grid[r][c] == '0')
    return;
  grid[r][c] = '0';
  dfs(r+1, c, grid);
  dfs(r-1, c, grid);
  dfs(r, c+1, grid);
  dfs(r, c-1, grid);
}

int numIslands(std::vector& grid) {
  int count = 0;
  for(int i=0; i

medium

A. The base case does not check for visited cells

B. The DFS does not check boundaries correctly

C. The DFS does not mark visited cells, causing infinite recursion

D. The grid is modified during DFS, which is incorrect

Solution

1. Step 1: Analyze DFS grid modification

   The DFS marks visited cells by changing '1' to '0' in the original grid, modifying input.

2. Step 2: Identify why this is a bug

   Modifying input grid may be unintended side effect; better to use a separate visited array to avoid altering input data.

3. Final Answer:

   The grid is modified during DFS, which is incorrect -> Option D

4. Quick Check:

   Modifying input grid = bug [OK]

Hint: Avoid changing input grid; use visited array instead [OK]

Common Mistakes:

• Ignoring that grid is changed in-place
• Assuming no visited tracking needed
• Misreading boundary checks as incorrect

5. You want to count islands in a grid where diagonal connections also count as connected (8 directions). How should you modify the BFS or DFS approach?

hard

A. Count islands by summing all '1's without traversal

B. Only check horizontal neighbors, ignore vertical and diagonal

C. Add diagonal directions (like (1,1), (1,-1), (-1,1), (-1,-1)) to the directions array

D. Use BFS for horizontal and DFS for vertical connections separately

Solution

1. Step 1: Understand connectivity change

   Diagonal connections mean neighbors include 8 directions: up, down, left, right, and 4 diagonals.

2. Step 2: Modify directions array

   Add diagonal moves (1,1), (1,-1), (-1,1), (-1,-1) to directions used in BFS or DFS to explore all connected neighbors.

3. Final Answer:

   Add diagonal directions (like (1,1), (1,-1), (-1,1), (-1,-1)) to the directions array -> Option C

4. Quick Check:

   Include all 8 neighbors for diagonal connectivity [OK]

Hint: Add diagonal moves to directions for 8-way connectivity [OK]

Common Mistakes:

• Ignoring diagonal neighbors
• Counting only horizontal or vertical neighbors
• Trying to count without traversal