# Island Perimeter

## 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. 

![](/files/-MQZBPj_TtZJIdpp_trZ)

{% hint style="info" %}
Another example:

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

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

{% endhint %}

### 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


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