<p>You are given <code>row x col</code> <code>grid</code> representing a map where <code>grid[i][j] = 1</code> represents&nbsp;land and <code>grid[i][j] = 0</code> represents water.</p>

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

<p>The island doesn&#39;t have &quot;lakes&quot;, meaning the water inside isn&#39;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&#39;t exceed 100. Determine the perimeter of the island.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img src="https://assets.leetcode.com/uploads/2018/10/12/island.png" style="width: 221px; height: 213px;" />
<pre>
<strong>Input:</strong> grid = [[0,1,0,0],[1,1,1,0],[0,1,0,0],[1,1,0,0]]
<strong>Output:</strong> 16
<strong>Explanation:</strong> The perimeter is the 16 yellow stripes in the image above.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> grid = [[1]]
<strong>Output:</strong> 4
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> grid = [[1,0]]
<strong>Output:</strong> 4
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>row == grid.length</code></li>
	<li><code>col == grid[i].length</code></li>
	<li><code>1 &lt;= row, col &lt;= 100</code></li>
	<li><code>grid[i][j]</code> is <code>0</code> or <code>1</code>.</li>
	<li>There is exactly one island in <code>grid</code>.</li>
</ul>

class Solution:
    def islandPerimeter(self, grid: List[List[int]]) -> int:
        DIRS = [(-1,0),(1,0),(0,-1),(0,1)]
        def dfs(i,j,visited):

            if not grid[i][j]:
                return 0

            def _count_neighbors(i,j):
                cnt = 0
                for d in DIRS:
                    if (i+d[0]<0) or (i+d[0]==len(grid)) or (j+d[1]<0) or (j+d[1]==len(grid[0])):
                        cnt+=1
                    else:
                        cnt+=0 if grid[i+d[0]][j+d[1]] else 1
                return cnt
            visited.add((i,j))
            total = _count_neighbors(i,j)
            for d in DIRS:
                if (i+d[0]>=0) and (i+d[0]<len(grid)) and (j+d[1]>=0) and (j+d[1]<len(grid[0])) and (i+d[0],j+d[1]) not in visited:
                    total+=dfs(i+d[0],j+d[1],visited)
            return total
        
        visited = set()
        for i in range(len(grid)):
            for j in range(len(grid[0])):
                if grid[i][j]==1:
                    return dfs(i,j,visited)
        return 0
19464