Approach

To effectively answer the question "How would you implement an algorithm to count the number of distinct islands in a grid?", follow this structured framework:

1. Understand the Problem: Clearly define what constitutes an island in the context of the grid. Islands are typically formed by connecting '1's (land) surrounded by '0's (water).

2. Choose an Algorithm: Decide on an appropriate algorithm, such as Depth-First Search (DFS) or Breadth-First Search (BFS), to traverse the grid.

3. Implement the Algorithm: Write the code to count the islands using the chosen algorithm, keeping track of visited cells.

4. Test the Implementation: Validate the algorithm with different grid configurations to ensure its accuracy and efficiency.

Key Points

• Definition: An island is formed by adjacent '1's (horizontally or vertically).

• Algorithm Choice: Both DFS and BFS are suitable for this problem, with the choice depending on personal comfort.

• Data Structures: Use a stack or queue for DFS or BFS, respectively, and a 2D array to track visited cells.

• Complexity: Aim for an O(m * n) time complexity, where m is the number of rows and n is the number of columns in the grid.

Standard Response

Here’s a sample answer that encompasses the necessary components for a compelling response:

To implement an algorithm to count the number of distinct islands in a grid, I would follow these steps:

• Define the Grid and Island Structure:

An island is defined as a group of adjacent '1's that form a continuous landmass, where adjacency is considered in four directions (up, down, left, right).

• Choose an Algorithm:

I would use the Depth-First Search (DFS) algorithm for this implementation. DFS is intuitive for this kind of traversal and allows for easy marking of visited cells.

• Initialize Data Structures:

• A 2D array to represent the grid.

• A visited set or array to keep track of cells we have already counted as part of an island.

• A counter variable to store the number of distinct islands.

• Implement the DFS Function:

The DFS function will recursively explore all connected land cells from a starting cell and mark them as visited.

• Iterate Through the Grid:

For each cell in the grid, if the cell is '1' and not visited, I would invoke the DFS function, increment the island counter, and explore all its connected land.

Here’s a sample implementation in Python:

def numIslands(grid):
 if not grid:
 return 0

 def dfs(i, j):
 # Check boundaries
 if i < 0 or i >= len(grid) or j < 0 or j >= len(grid[0]) or grid[i][j] == '0':
 return
 # Mark the cell as visited
 grid[i][j] = '0'
 # Explore all four directions
 dfs(i + 1, j)
 dfs(i - 1, j)
 dfs(i, j + 1)
 dfs(i, j - 1)

 island_count = 0
 for i in range(len(grid)):
 for j in range(len(grid[0])):
 if grid[i][j] == '1':
 dfs(i, j) # Start DFS from this cell
 island_count += 1 # Increment the count for each distinct island

 return island_count

• Testing the Implementation:

• A grid with no islands.

• A grid with multiple separate islands.

• A grid that forms a single large island.

• I would test the implementation with various grid configurations to ensure it accurately counts distinct islands. For example:

Tips & Variations

Common Mistakes to Avoid:

• Ignoring Edge Cases: Ensure to test grids with no land or all land.

• Not Marking Visited Cells: Failing to mark cells can lead to incorrect counts.

Alternative Ways to Answer:

• For a BFS approach, explain how to use a queue instead of recursion, which can help avoid stack overflow for large grids.

Role-Specific Variations:

• Technical Positions: Focus on algorithmic complexity and optimization.

• Managerial Roles: Emphasize team collaboration and code review processes.

• Creative Roles: Highlight the importance of innovative solutions and problem-solving approaches.

Follow-Up Questions:

• Can you explain the time and space complexity of your algorithm?

• How would you modify your algorithm if diagonally connected '1's also counted as part of the same island?

• What changes would you make for a larger grid or if the