### Approach To effectively answer the question **“How do you calculate the number of islands in a 2D grid?”**, follow this structured framework: 1. **Understand the Problem**: Define what constitutes an island. 2. **Choose an Algorithm**: Identify suitable algorithms for grid traversal. 3. **Implement the Solution**: Write and explain the code. 4. **Analyze Complexity**: Discuss time and space complexities. 5. **Provide Edge Cases**: Mention any special scenarios to consider. ### Key Points - **Definition of an Island**: An island is formed by connecting adjacent lands (1s) horizontally or vertically surrounded by water (0s). - **Traversal Methods**: Depth First Search (DFS) or Breadth First Search (BFS) are commonly used to explore the grid. - **Iterative vs. Recursive**: Understand the trade-offs between using an iterative approach and recursion. - **Complexity Analysis**: Clearly articulate the efficiency of your solution. - **Edge Cases**: Consider grids with no land, all land, or varying dimensions. ### Standard Response To calculate the number of islands in a 2D grid, you can use the Depth First Search (DFS) algorithm. Here’s how you can approach the problem step-by-step: 1. **Initialize the Grid**: Represent the grid as a 2D array where '1' indicates land and '0' indicates water. 2. **Count Islands**: - Create a variable to keep track of the number of islands. - Loop through each cell in the grid. - If a cell contains '1', increment your island count and perform DFS from that cell to mark all connected lands. 3. **DFS Implementation**: - In your DFS function, change the current cell from '1' to '0' to mark it as visited. - Recursively call DFS for all four directions (up, down, left, right). Here's a sample code implementation in Python: ```python def numIslands(grid): if not grid: return 0 def dfs(i, j): # Check for boundaries and if the cell is water 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' # Recursively visit all adjacent cells 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) return count ``` ### Complexity Analysis - **Time Complexity**: O(M * N), where M is the number of rows and N is the number of columns in the grid. Each cell is visited once. - **Space Complexity**: O(M * N) in the worst case due to the recursion stack in DFS. ### Tips & Variations #### Common Mistakes to Avoid - **Ignoring Diagonal Connections**: Only consider horizontal and vertical connections when defining islands. - **Not Marking Visited Cells**: Failing to mark cells as visited can lead to counting the same island multiple times. - **Misunderstanding Input**: Ensure you understand whether the grid can be empty or consists solely of water or land. #### Alternative Ways to Answer - **Using BFS**: Instead of DFS, you can use a queue to implement a BFS approach, which iteratively explores the grid. ```python from collections import deque def numIslandsBFS(grid): if not grid: return 0 def bfs(i, j): queue = deque([(i, j)]) grid[i][j] = '0' # Mark as visited while queue: x, y = queue.popleft() for dx, dy in [(1, 0), (-1, 0), (0, 1), (0, -1)]: nx, ny = x + dx, y + dy if 0 <= nx < len(grid) and 0 <= ny < len(grid[0]) and grid[nx][ny] == '1': grid[nx][ny] = '0' # Mark as visited queue.append((nx, ny)) count = 0 for i in range(len(grid)): for j in range(len(grid[0])): if grid[i][j] == '1': count += 1 bfs(i, j) return count