Island Perimeter Python Solution with Tests

Tested Python solution for LeetCode 463 with 15 pytest cases. Generate a practice environment with lcpy.

LeetCode 463, Easy. Topics: Array, Depth-First Search, Breadth-First Search, Matrix. View on LeetCode.

Generate this problem as a practice environment: tested reference solution, 15 parametrized pytest cases, and a playground notebook:

lcpy gen -n 463   # by problem number
lcpy gen -s island_perimeter   # by problem name

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Problem

You are given row x col grid representing a map where grid[i][j] = 1 represents land and grid[i][j] = 0 represents water.

Grid cells are connected horizontally/vertically (not diagonally). The grid is completely surrounded by water, and there is exactly one island (i.e., one or more connected land cells).

The island doesn’t have “lakes”, meaning the water inside isn’t connected to the water around the island. One cell is a square with side length 1. The grid is rectangular, width and height don’t exceed 100.

Determine the perimeter of the island.

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Examples

Input: grid = [[0,1,0,0],[1,1,1,0],[0,1,0,0],[1,1,0,0]]
Output: 16
Explanation: The perimeter is the 16 yellow stripes in the image above.

Input: grid = [[1]]
Output: 4

Input: grid = [[1,0]]
Output: 4

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Constraints

• row == grid.length
• col == grid[i].length
• 1 <= row, col <= 100
• grid[i][j] is 0 or 1.
• There is exactly one island in grid.

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Solution

Reference implementation from solution.py on GitHub, full suite in test_solution.py:

class Solution:
    # Time: O(row * col) every cell inspected once
    # Space: O(1)
    def island_perimeter(self, grid: list[list[int]]) -> int:
        rows = len(grid)
        cols = len(grid[0])
        perimeter = 0
        for r in range(rows):
            for c in range(cols):
                if grid[r][c] == 0:
                    continue
                # Each land cell starts with 4 exposed sides; subtract shared edges.
                exposed = 4
                for dr, dc in ((-1, 0), (1, 0), (0, -1), (0, 1)):
                    nr, nc = r + dr, c + dc
                    if 0 <= nr < rows and 0 <= nc < cols and grid[nr][nc] == 1:
                        exposed -= 1
                perimeter += exposed
        return perimeter

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Complexity

Time	Space
O(row * col) every cell inspected once	O(1)

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Tags

NeetCode 250, NeetCode All.