Number of Islands BFS and DFS in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to start BFS from the given cell.

DSA Go

func bfs(grid [][]byte, visited [][]bool, row int, col int) {
    queue := [][]int{{row, col}}
    visited[row][col] = true
    directions := [][]int{{-1, 0}, {1, 0}, {0, -1}, {0, 1}}

    for len(queue) > 0 {
        cell := queue[0]
        queue = queue[1:]
        r, c := cell[0], cell[1]

        for _, dir := range directions {
            nr, nc := r+dir[0], c+dir[1]
            if nr >= 0 && nr < len(grid) && nc >= 0 && nc < len(grid[0]) && grid[nr][nc] == '1' && !visited[nr][nc] {
                queue = append(queue, []int{nr, nc})
                visited[nr][nc] = [1]
            }
        }
    }
}

Drag options to blanks, or click blank then click option'

Afalse

Bvisited

Ctrue

Dgrid[nr][nc]

Attempts:

3 left

💡 Hint

Common Mistakes

Forgetting to mark the cell as visited.

Marking visited after popping from the queue instead of when adding.

We mark the neighbor cell as visited by setting visited[nr][nc] = true to avoid revisiting it.

2fill in blank

medium

Complete the DFS function to explore all connected land cells.

DSA Go

func dfs(grid [][]byte, visited [][]bool, row int, col int) {
    if row < 0 || row >= len(grid) || col < 0 || col >= len(grid[0]) || grid[row][col] == '0' || visited[row][col] {
        return
    }
    visited[row][col] = true
    directions := [][]int{{-1, 0}, {1, 0}, {0, -1}, {0, 1}}
    for _, dir := range directions {
        dfs(grid, visited, row+[1], col+dir[1])
    }
}

Drag options to blanks, or click blank then click option'

Adir[0]

Bdir[1]

Crow

Dcol

Attempts:

3 left

💡 Hint

Common Mistakes

Using the same direction index for both row and column.

Not checking boundaries before recursion.

We add dir[0] to row and dir[1] to col separately. Here, the blank is for row+dir[0].

3fill in blank

hard

Fix the error in the DFS call to correctly explore neighbors.

DSA Go

func dfs(grid [][]byte, visited [][]bool, row int, col int) {
    if row < 0 || row >= len(grid) || col < 0 || col >= len(grid[0]) || grid[row][col] == '0' || visited[row][col] {
        return
    }
    visited[row][col] = true
    directions := [][]int{{-1, 0}, {1, 0}, {0, -1}, {0, 1}}
    for _, dir := range directions {
        dfs(grid, visited, row+dir[0], col+[1])
    }
}

Drag options to blanks, or click blank then click option'

Adir[0]

Bcol

Crow

Ddir[1]

Attempts:

3 left

💡 Hint

Common Mistakes

Using dir[0] for both row and column updates.

Swapping row and column indices.

The column index should be updated by dir[1] to move correctly in the grid.

4fill in blank

hard

Fill both blanks to count islands using BFS.

DSA Go

func numIslands(grid [][]byte) int {
    if len(grid) == 0 {
        return 0
    }
    visited := make([][]bool, len(grid))
    for i := range visited {
        visited[i] = make([]bool, len(grid[0]))
    }
    count := 0
    for i := 0; i < len(grid); i++ {
        for j := 0; j < len(grid[0]); j++ {
            if grid[i][j] == '1' && !visited[i][j] {
                [1](grid, visited, i, j)
                count[2]
            }
        }
    }
    return count
}

Drag options to blanks, or click blank then click option'

Abfs

Bdfs

C++

D--

Attempts:

3 left

💡 Hint

Common Mistakes

Calling dfs instead of bfs.

Decrementing count instead of incrementing.

We call bfs to explore the island and increment count by 1 after each BFS.

5fill in blank

hard

Fill all three blanks to count islands using DFS.

DSA Go

func numIslandsDFS(grid [][]byte) int {
    if len(grid) == 0 {
        return 0
    }
    visited := make([][]bool, len(grid))
    for i := range visited {
        visited[i] = make([]bool, len(grid[0]))
    }
    count := 0
    for i := 0; i < len(grid); i++ {
        for j := 0; j < len(grid[0]); j++ {
            if grid[i][j] == '1' && !visited[i][j] {
                [1](grid, visited, i, j)
                count[2]
            }
        }
    }
    return count
}

Drag options to blanks, or click blank then click option'

Adfs

B++

C--

Dbfs

Attempts:

3 left

💡 Hint

Common Mistakes

Calling bfs instead of dfs.

Decrementing count instead of incrementing.

We call dfs to explore islands and increment count by 1 after each DFS.

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