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

Practice - 5 Tasks

1fill in blank

easy

Complete the code to start BFS from the given cell.

DSA Python

def bfs(grid, start_row, start_col):
    queue = [(start_row, start_col)]
    grid[start_row][start_col] = '0'
    while queue:
        row, col = queue.pop(0)
        for dr, dc in [(1,0), (-1,0), (0,1), (0,-1)]:
            r, c = row + dr, col + dc
            if 0 <= r < len(grid) and 0 <= c < len(grid[0]) and grid[r][c] == [1]:
                queue.append((r, c))
                grid[r][c] = '0'

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

A'1'

B'0'

C1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Checking for '0' instead of '1' causes no land to be found.

Using integer 1 instead of string '1' causes type mismatch.

We look for land cells marked as '1' to add to the BFS queue and mark them visited by changing to '0'.

2fill in blank

medium

Complete the code to count islands using DFS.

DSA Python

def dfs(grid, row, col):
    if row < 0 or row >= len(grid) or col < 0 or col >= len(grid[0]) or grid[row][col] == [1]:
        return
    grid[row][col] = '0'
    dfs(grid, row+1, col)
    dfs(grid, row-1, col)
    dfs(grid, row, col+1)
    dfs(grid, row, col-1)

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

A0

B'1'

C'0'

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Stopping DFS on land cells instead of water cells.

Not checking boundaries before accessing grid.

We stop DFS when we hit water cells marked as '0' or out of bounds.

3fill in blank

hard

Fix the error in the island counting loop.

DSA Python

def num_islands(grid):
    count = 0
    for r in range(len(grid)):
        for c in range(len(grid[0])):
            if grid[r][c] == [1]:
                dfs(grid, r, c)
                count += 1
    return count

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

A'1'

B'0'

C1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Starting DFS on water cells causes wrong island count.

Using integer 1 instead of string '1' causes errors.

We start DFS only on land cells marked as '1' to count islands.

4fill in blank

hard

Fill both blanks to implement BFS island counting correctly.

DSA Python

def num_islands_bfs(grid):
    count = 0
    for r in range(len(grid)):
        for c in range(len(grid[0])):
            if grid[r][c] == [1]:
                bfs(grid, r, c)
                count [2] 1
    return count

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

A'1'

B+=

C-=

D'0'

Attempts:

3 left

💡 Hint

Common Mistakes

Using '-=' decreases count incorrectly.

Checking for '0' instead of '1' misses islands.

We check for land '1' and increment count by 1 after BFS.

5fill in blank

hard

Fill all three blanks to create a dictionary comprehension that maps each island's row to its count if count is positive.

DSA Python

island_map = { [1]: [2] for [1] in range(len(grid)) if [2] > 0 }

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

Ar

Bcount

Attempts:

3 left

💡 Hint

Common Mistakes

Using the same variable for key and value incorrectly.

Not filtering rows with count > 0.

We map each row 'r' to its island count 'count' only if count > 0.

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 all '0's in the grid

B. Find the largest island size

C. Count how many groups of connected '1's (islands) exist

D. Calculate the perimeter of all islands

Solution

1. Step 1: Understand the grid representation

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

2. Step 2: Identify what the problem asks

   It asks to count how many separate groups of connected '1's (islands) are present.

3. Final Answer:

   Count how many groups of connected '1's (islands) exist -> Option C

4. Quick Check:

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

Hint: Islands = groups of connected '1's, count them [OK]

Common Mistakes:

• Confusing island count with island size
• Counting '0's instead of '1's
• Calculating perimeter instead of count

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

easy

A. def dfs(i, j): if i<0 or j<0 or i>=len(grid) or j>=len(grid[0]) or grid[i][j] == '0': return; grid[i][j] = '0'; dfs(i+1,j); dfs(i-1,j); dfs(i,j+1); dfs(i,j-1)

B. def dfs(i, j): while grid[i][j] == '1': grid[i][j] = '0'; dfs(i+1,j); dfs(i-1,j)

C. def dfs(i, j): if grid[i][j] == '1': return; dfs(i+1,j); dfs(i-1,j)

D. def dfs(i, j): for x in range(len(grid)): for y in range(len(grid[0])): dfs(x,y)

Solution

1. Step 1: Check boundary and water conditions

   def dfs(i, j): if i<0 or j<0 or i>=len(grid) or j>=len(grid[0]) or grid[i][j] == '0': return; grid[i][j] = '0'; dfs(i+1,j); dfs(i-1,j); dfs(i,j+1); dfs(i,j-1) correctly checks if indices are out of bounds or cell is water ('0') and returns.

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

   def dfs(i, j): if i<0 or j<0 or i>=len(grid) or j>=len(grid[0]) or grid[i][j] == '0': return; grid[i][j] = '0'; dfs(i+1,j); dfs(i-1,j); dfs(i,j+1); dfs(i,j-1) marks grid[i][j] = '0' to avoid revisiting and calls dfs on all four neighbors.

3. Final Answer:

   Option A code correctly implements DFS with boundary checks and marking visited -> Option A

4. Quick Check:

   DFS with boundary and visited marking = def dfs(i, j): if i<0 or j<0 or i>=len(grid) or j>=len(grid[0]) or grid[i][j] == '0': return; grid[i][j] = '0'; dfs(i+1,j); dfs(i-1,j); dfs(i,j+1); dfs(i,j-1) [OK]

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

Common Mistakes:

• Not checking boundaries before recursion
• Not marking visited cells
• Using loops incorrectly inside DFS

3. What is the output of the following code snippet for the given grid?

grid = [
  ['1','1','0','0'],
  ['1','0','0','1'],
  ['0','0','1','1'],
  ['0','0','0','0']
]

count = 0
for i in range(len(grid)):
    for j in range(len(grid[0])):
        if grid[i][j] == '1':
            count += 1
            stack = [(i,j)]
            while stack:
                x,y = stack.pop()
                if 0 <= x < len(grid) and 0 <= y < len(grid[0]) and grid[x][y] == '1':
                    grid[x][y] = '0'
                    stack.extend([(x+1,y),(x-1,y),(x,y+1),(x,y-1)])
print(count)

medium

A. 2

B. 3

C. 4

D. 5

Solution

1. Step 1: Identify islands in the grid

   The grid has three islands: top-left (cells (0,0),(0,1),(1,0)), single cell island at (1,3), and bottom-right island (cells (2,2),(2,3)).

2. Step 2: Trace the DFS using stack

   Each '1' found triggers DFS marking connected '1's as '0'. The count increments for each new island found. Total islands counted = 3.

3. Final Answer:

   3 -> Option B

4. Quick Check:

   Number of islands found by DFS = 3 [OK]

Hint: Count connected '1' groups using DFS stack traversal [OK]

Common Mistakes:

• Counting each '1' without marking visited
• Misidentifying connected cells
• Off-by-one errors in loops

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

def dfs(i, j):
    if i < 0 or j < 0 or i >= len(grid) or j >= len(grid[0]):
        return
    if grid[i][j] == '0':
        return
    dfs(i+1, j)
    dfs(i-1, j)
    dfs(i, j+1)
    dfs(i, j-1)

count = 0
for i in range(len(grid)):
    for j in range(len(grid[0])):
        if grid[i][j] == '1':
            count += 1
            dfs(i, j)
print(count)

medium

A. The DFS only checks horizontal neighbors

B. The boundary checks are incorrect

C. The count variable is not incremented properly

D. The code does not mark visited cells, causing infinite recursion

Solution

1. Step 1: Analyze DFS function behavior

   The DFS checks boundaries and water but never marks visited land cells as '0' or visited.

2. Step 2: Identify consequence of missing visited marking

   Without marking visited, DFS revisits same cells repeatedly, causing infinite recursion or stack overflow.

3. Final Answer:

   The code does not mark visited cells, causing infinite recursion -> Option D

4. Quick Check:

   Missing visited marking causes infinite recursion [OK]

Hint: Always mark visited cells to avoid infinite loops in DFS [OK]

Common Mistakes:

• Forgetting to mark visited cells
• Incorrect boundary conditions
• Incrementing count inside DFS

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. Add diagonal neighbors (i+1,j+1), (i+1,j-1), (i-1,j+1), (i-1,j-1) to the search directions

B. Only count horizontal neighbors and ignore vertical and diagonal

C. Use BFS but skip marking visited cells

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

Solution

1. Step 1: Understand diagonal connectivity

   Diagonal connections mean neighbors include cells diagonally adjacent, not just up/down/left/right.

2. Step 2: Modify search directions

   Add diagonal neighbors (i+1,j+1), (i+1,j-1), (i-1,j+1), (i-1,j-1) to BFS or DFS neighbor checks to include diagonals.

3. Final Answer:

   Add diagonal neighbors (i+1,j+1), (i+1,j-1), (i-1,j+1), (i-1,j-1) to the search directions -> Option A

4. Quick Check:

   Include diagonals in neighbors for diagonal connectivity [OK]

Hint: Include diagonal neighbors in BFS/DFS directions [OK]

Common Mistakes:

• Ignoring diagonal neighbors
• Not marking visited cells
• Counting without traversal