This solution seeks to find the perimeter of an island in a grid. The problem considers a grid where each cell represents land (1) or water (0). In addition, the function calculates the total perimeter of the connected land (island) in the grid.

```
class Solution:
def islandPerimeter(self, grid: List[List[int]]) -> int:
"""
Calculates the perimeter of the island in a 2D grid.
Args:
grid (List[List[int]]): A 2D list representing the grid, where grid[i][j] is 1 for land and 0 for water.
Returns:
int: The total perimeter length of the island in the grid.
"""
visited = set() # Set to keep track of visited cells
def dfs(row: int, col: int) -> int:
"""
Performs a Depth-First Search (DFS) traversal to explore the island and calculate its perimeter.
Args:
row (int): The row index of the current cell.
col (int): The column index of the current cell.
Returns:
int: The perimeter contribution of the current cell (1 if on the edge, 0 otherwise).
"""
if (row >= len(grid) or col >= len(grid[0]) or # Check for out-of-bounds
row < 0 or col < 0 or
grid[row][col] == 0): # Check for water cell
return 1
if (row, col) in visited: # Skip already visited cell
return 0
visited.add((row, col)) # Mark cell as visited
perimeter = dfs(row, col + 1) # Explore right neighbor
perimeter += dfs(row, col - 1) # Explore left neighbor
perimeter += dfs(row + 1, col) # Explore down neighbor
perimeter += dfs(row - 1, col) # Explore up neighbor
return perimeter
# Start DFS from any land cell to find the island
for row in range(len(grid)):
for col in range(len(grid[0])):
if grid[row][col] == 1:
return dfs(row, col)
# If no land is found, the perimeter is 0
return 0
```

The thought process behind this solution is to calculate the perimeter of an island represented in a 2D grid using Depth-First Search (DFS). Here’s a detailed breakdown of the logic and reasoning:

**Island Definition**: The island is represented by cells with value`1`

, and water is represented by cells with value`0`

.- Each land cell contributes to the perimeter if it is adjacent to a water cell or the boundary of the grid. Specifically:
- A cell at the grid boundary contributes an edge to the perimeter.
- A land cell adjacent to a water cell contributes an edge to the perimeter for each water cell it touches.

**Traversal Strategy**: Use Depth-First Search (DFS) to traverse the island, ensuring that each cell is visited only once. DFS is chosen because it effectively explores all connected components of the island.**Edge Calculation**: For each land cell, check its four neighbors (up, down, left, right). If a neighbor is out of bounds or a water cell, it contributes to the perimeter.**Tracking Visited Cells**: Use a set to track visited cells to avoid redundant calculations and infinite loops.- Loop through each cell in the grid. Start DFS from the first land cell found, which initiates the traversal and perimeter calculation for the entire island.

**Time Complexity**

The time complexity of the solution is determined by how many times each cell in the grid is visited and processed.

**Initial Loop:** The initial double loop to find the first land cell runs in **O(m by n)**, where m is the number of rows and n is the number of columns in the grid.

**DFS Traversal:** Each land cell (with value 1) is visited exactly once, and for each land cell, the algorithm checks its four neighbors. Thus, the DFS traversal will also run in **O(m by n) **time in the worst case, where every cell in the grid island.

**Space Complexity**

The space complexity is influenced by the storage needed for the visited set and the call stack for the recursive DFS.

** Visited Set:** In the worst case, the visited set could store every cell in the grid if all cells are land. Thus, the space required for the visited set is **O(m by n)**.

** DFS Call Stack:** The depth of the recursion stack for DFS can go up to **O(m by n)** in the worst case, where the grid forms a single narrow strip of land.

**Unit Tests**

Here's the tests to examine the accuracy & robustness of this solution.

```
class TestislandPerimeter(unittest.TestCase):
def setUp(self):
self.sol = Solution()
def test_islandPerimeter(self):
grid = [[0,1,0,0],[1,1,1,0],[0,1,0,0],[1,1,0,0]]
self.assertEqual(self.sol.islandPerimeter(grid), 16)
def test_empty_grid(self):
"""Tests the function with an empty grid."""
solution = Solution()
grid = []
result = solution.islandPerimeter(grid)
self.assertEqual(result, 0)
def test_single_cell_island(self):
"""Tests the function with a grid containing a single land cell."""
solution = Solution()
grid = [[1]]
result = solution.islandPerimeter(grid)
self.assertEqual(result, 4) # Perimeter of a single cell
def test_horizontal_island(self):
"""Tests the function with a horizontal island of connected land cells."""
solution = Solution()
grid = [[1, 1, 1, 1]]
result = solution.islandPerimeter(grid)
self.assertEqual(result, 10) # 3 edges shared internally
def test_vertical_island(self):
"""Tests the function with a vertical island of connected land cells."""
solution = Solution()
grid = [[1],
[1],
[1]]
result = solution.islandPerimeter(grid)
self.assertEqual(result, 8) #
def test_complex_island(self):
"""Tests the function with a complex island with various shapes."""
solution = Solution()
grid = [[0, 1, 0, 0],
[1, 1, 1, 0],
[0, 1, 0, 1],
[0, 0, 0, 0]]
result = solution.islandPerimeter(grid)
# todo
# one island is 12 while another is 4
self.assertEqual(result, 16)
def test_no_island(self):
"""Tests the function with a grid containing only water cells."""
solution = Solution()
grid = [[0, 0, 0],
[0, 0, 0],
[0, 0, 0]]
result = solution.islandPerimeter(grid)
self.assertEqual(result, 0)
if __name__ == '__main__':
unittest.main()
```