Why Number of Islands BFS and DFS in DSA Go?

The Big Idea

What if you could find all islands on a map without getting lost or confused?

The Scenario

Imagine you have a big map made of water and land squares. You want to count how many separate islands are on the map. Doing this by looking at each square and trying to remember which ones you already counted is very confusing.

The Problem

Manually checking each square and trying to track connected land pieces is slow and easy to mess up. You might count the same island twice or miss some parts because it's hard to remember all connections without a clear method.

The Solution

Using BFS (Breadth-First Search) or DFS (Depth-First Search) helps you explore each island fully before moving on. These methods let you start from one land square and find all connected land squares automatically, so you count each island exactly once.

Before vs After

✗ Before

count = 0
for each cell in grid:
  if cell is land and not visited:
    count += 1
    mark connected land manually

✓ After

count = 0
for each cell in grid:
  if cell is land and not visited:
    count += 1
    DFS or BFS to mark all connected land

What It Enables

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

Real Life Example

Counting groups of connected trees in a forest map to plan fire control, or finding clusters of buildings in satellite images for city planning.

Key Takeaways

Manual counting is confusing and error-prone.

BFS and DFS explore connected land automatically.

They help count islands accurately and efficiently.

Practice

1. What does the "Number of Islands" problem ask you to find in a grid of '1's and '0's?

easy

A. Count how many groups of connected '1's (islands) are in the grid

B. Find the largest island size

C. Count all '0's in the grid

D. Calculate the perimeter of islands

Solution

1. Step 1: Understand the grid content

   The grid contains '1's representing land and '0's representing water.

2. Step 2: Define the problem goal

   We need to count how many separate groups of connected '1's (islands) exist.

3. Final Answer:

   Count how many groups of connected '1's (islands) are in the grid -> Option A

4. Quick Check:

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

Hint: Islands = connected '1's groups in grid [OK]

Common Mistakes:

• Counting individual '1's instead of groups
• Confusing '0's as islands
• Trying to find island size instead of count

2. Which of the following Go code snippets correctly starts a DFS from cell (i, j) in a grid to mark connected land?

easy

A. func dfs(grid [][]byte, i, j int) { if grid[i][j] == '1' { grid[i][j] = '1' } }

B. func dfs(grid [][]byte, i, j int) { for i < len(grid) { for j < len(grid[0]) { grid[i][j] = '0' } } }

C. func dfs(grid [][]byte, i, j int) { if i < 0 || j < 0 || i >= len(grid) || j >= len(grid[0]) || grid[i][j] == '0' { return } grid[i][j] = '0'; dfs(grid, i+1, j); dfs(grid, i-1, j); dfs(grid, i, j+1); dfs(grid, i, j-1) }

D. func dfs(grid [][]byte, i, j int) { if i > len(grid) || j > len(grid[0]) { return } grid[i][j] = '1' }

Solution

1. Step 1: Check boundary and water conditions

   func dfs(grid [][]byte, i, j int) { if i < 0 || j < 0 || i >= len(grid) || j >= len(grid[0]) || grid[i][j] == '0' { return } grid[i][j] = '0'; dfs(grid, i+1, j); dfs(grid, i-1, j); dfs(grid, i, j+1); dfs(grid, i, j-1) } correctly checks if indices are out of bounds or cell is water ('0'), then returns.

2. Step 2: Mark current land as visited and recurse

   func dfs(grid [][]byte, i, j int) { if i < 0 || j < 0 || i >= len(grid) || j >= len(grid[0]) || grid[i][j] == '0' { return } grid[i][j] = '0'; dfs(grid, i+1, j); dfs(grid, i-1, j); dfs(grid, i, j+1); dfs(grid, i, j-1) } marks current cell as '0' to avoid revisiting and calls dfs on neighbors up, down, left, right.

3. Final Answer:

   func dfs(grid [][]byte, i, j int) { if i < 0 || j < 0 || i >= len(grid) || j >= len(grid[0]) || grid[i][j] == '0' { return } grid[i][j] = '0'; dfs(grid, i+1, j); dfs(grid, i-1, j); dfs(grid, i, j+1); dfs(grid, i, j-1) } -> Option C

4. Quick Check:

   DFS marks visited land and explores neighbors [OK]

Hint: Check boundaries and mark visited before recursive calls [OK]

Common Mistakes:

• Not checking boundaries before access
• Not marking visited cells
• Using wrong loop instead of recursion

3. What is the output of the following Go code that counts islands using BFS on this grid?

grid := [][]byte{
  {'1','1','0','0','0'},
  {'1','1','0','0','0'},
  {'0','0','1','0','0'},
  {'0','0','0','1','1'},
}
count := numIslands(grid)
fmt.Println(count)

func numIslands(grid [][]byte) int {
  rows, cols := len(grid), len(grid[0])
  count := 0
  directions := [][]int{{1,0},{-1,0},{0,1},{0,-1}}
  for i := 0; i < rows; i++ {
    for j := 0; j < cols; j++ {
      if grid[i][j] == '1' {
        count++
        queue := [][]int{{i,j}}
        grid[i][j] = '0'
        for len(queue) > 0 {
          r, c := queue[0][0], queue[0][1]
          queue = queue[1:]
          for _, d := range directions {
            nr, nc := r+d[0], c+d[1]
            if nr >= 0 && nr < rows && nc >= 0 && nc < cols && grid[nr][nc] == '1' {
              queue = append(queue, []int{nr,nc})
              grid[nr][nc] = '0'
            }
          }
        }
      }
    }
  }
  return count
}

medium

A. 1

B. 3

C. 4

D. 2

Solution

1. Step 1: Identify islands in the grid

   There are three groups of connected '1's: top-left block, middle single '1', bottom-right block.

2. Step 2: BFS marks visited and counts islands

   Each BFS call marks one island, so count increments 3 times.

3. Final Answer:

   3 -> Option B

4. Quick Check:

   Count islands = 3 [OK]

Hint: Count BFS runs triggered by unvisited '1's [OK]

Common Mistakes:

• Counting connected '1's individually
• Not marking visited cells
• Miscounting overlapping islands

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

func dfs(grid [][]byte, i, j int) {
  if i < 0 || j < 0 || i > len(grid) || j > len(grid[0]) || grid[i][j] == '0' {
    return
  }
  grid[i][j] = '0'
  dfs(grid, i+1, j)
  dfs(grid, i-1, j)
  dfs(grid, i, j+1)
  dfs(grid, i, j-1)
}

medium

A. Boundary checks use '>' instead of '>=' causing out-of-range errors

B. It does not mark visited cells

C. It uses BFS instead of DFS

D. It does not check if grid[i][j] is '1'

Solution

1. Step 1: Analyze boundary conditions

   The code uses 'i > len(grid)' and 'j > len(grid[0])' which allows i or j to equal length, causing index out of range.

2. Step 2: Correct boundary check

   It should use 'i >= len(grid)' and 'j >= len(grid[0])' to prevent invalid access.

3. Final Answer:

   Boundary checks use '>' instead of '>=' causing out-of-range errors -> Option A

4. Quick Check:

   Use '>=' for boundary checks to avoid errors [OK]

Hint: Use '>=' for boundary checks in arrays [OK]

Common Mistakes:

• Using '>' instead of '>=' in boundary checks
• Forgetting to mark visited cells
• Mixing BFS and DFS logic

5. Given a grid where diagonal connections also count as connected land, how would you modify the BFS or DFS approach to count islands correctly?

hard

A. Count islands without marking visited cells

B. Only check horizontal neighbors, ignore vertical and diagonal

C. Use BFS only without any direction changes

D. Add diagonal directions (top-left, top-right, bottom-left, bottom-right) to the neighbors list

Solution

1. Step 1: Understand connection rules

   Diagonal connections mean neighbors include 8 directions, not just 4 (up, down, left, right).

2. Step 2: Modify neighbor directions

   Add diagonal directions: (-1,-1), (-1,1), (1,-1), (1,1) to the directions list used in BFS/DFS.

3. Final Answer:

   Add diagonal directions (top-left, top-right, bottom-left, bottom-right) to the neighbors list -> Option D

4. Quick Check:

   Include diagonals in neighbors for diagonal connectivity [OK]

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

Common Mistakes:

• Ignoring diagonal neighbors
• Not updating directions list
• Counting without marking visited