Island Perimeter

DFS

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.

Another example:

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

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

Thought Process

• We can count the number of water cells that surround the island cell

• Since we are (in a way) dealing with connected components, we can apply DFS

Soultion

class Solution:
    def islandPerimeter(self, grid: List[List[int]]) -> int:
        
        
        for i in range(len(grid)):
            for j in range(len(grid[0])):
                if grid[i][j] == 1:
                    return self.dfsearch(i,j,grid)
        return 0
                    
    def dfsearch(self, i, j, grid):
        if i < 0 or i >= len(grid) or j < 0 or j >= len(grid[0]):
            return 1
        if grid[i][j] == 0:
            return 1
        if grid[i][j] == -1:
            return 0
        
        grid[i][j] = -1
        
        
        count = 0
        count += self.dfsearch(i+1, j, grid)
        count += self.dfsearch(i-1, j, grid)
        count += self.dfsearch(i, j-1, grid)
        count += self.dfsearch(i, j+1, grid)
        
        return count
                
        

Time Complexity

• Time: O(m*n)

• Space: O(m*n) for the recursion call stack