Approach

When answering the question, "How would you count the number of islands in a 2D grid map consisting of '1's (land) and '0's (water), where an island is defined as a group of adjacent lands connected horizontally or vertically?", follow this structured framework:

1. Understand the Problem: Identify what constitutes an island and the grid representation.

2. Choose a Method: Decide on a traversal technique (Depth-First Search (DFS) or Breadth-First Search (BFS)).

3. Implement the Algorithm: Write pseudocode or actual code to demonstrate your solution.

4. Explain Your Solution: Walk through the logic of your implementation.

5. Discuss Complexity: Address time and space complexity of your approach.

Key Points

• Definition Clarity: Clearly define what an island is.

• Algorithm Selection: Be prepared to justify your choice of DFS or BFS.

• Edge Cases: Consider scenarios like a grid full of water or land.

• Efficiency: Discuss the use of visited data structures to avoid counting the same island multiple times.

Standard Response

To tackle the problem of counting the number of islands in a 2D grid, we can use the Depth-First Search (DFS) approach. Below is a structured explanation and implementation.

Step 1: Understand the Problem

An island is defined as a group of connected '1's (land) in the grid, where connectivity is only vertical or horizontal. Water is represented by '0's. Our goal is to count the distinct islands.

Step 2: Choose a Method

• Start from any unvisited land cell (1).

• Mark all connected land cells as visited.

• Increment the island count every time we initiate a DFS from an unvisited land cell.

• We can use Depth-First Search (DFS) to explore each island:

Step 3: Implement the Algorithm

Here is a sample implementation in Python:

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

 def dfs(i, j):
 if i < 0 or i >= len(grid) or j < 0 or j >= len(grid[0]) or grid[i][j] == '0':
 return
 # Mark the current cell as visited by converting '1' to '0'
 grid[i][j] = '0'
 # Explore all adjacent cells (up, down, left, right)
 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': # Found an unvisited land
 dfs(i, j) # Start DFS to mark all connected lands
 count += 1 # Increment island count

 return count

Step 4: Explain Your Solution

• Grid Traversal: We loop through each cell in the 2D grid.

• DFS Execution: Upon finding an unvisited land cell (1), we initiate a DFS which marks all connected parts of the island as visited (changing '1' to '0').

• Count Increment: Each time we start a DFS from an unvisited land, we increment our island count.

Step 5: Discuss Complexity

• Time Complexity: O(M * N), where M is the number of rows and N is the number of columns. Each cell is visited once.

• Space Complexity: O(M * N) in the worst case due to the recursion stack.

Tips & Variations

Common Mistakes to Avoid

• Forgetting to mark cells as visited, leading to infinite loops.

• Not considering edge cases, such as empty grids or grids with no islands.

Alternative Ways to Answer

• BFS Approach: Instead of DFS, you could use BFS to explore the grid:

• Implement BFS using a queue to track cells to visit.

Example BFS implementation: