<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:
        res = 0
        x = len(grid[0])
        y = len(grid)
        visited = []
        def dfs(i,j):
            nonlocal res
            if i<0 or j<0 or i>=y or j>=x or grid[i][j] == 0 or (i,j) in visited:
                return
            visited.append((i,j))
            # if grid[i][j] == 0:
            #     return

            if i-1 < 0 or grid[i-1][j] == 0:
                res += 1
            if i+1 >= y or grid[i+1][j] == 0:
                res += 1
            if j-1 < 0 or grid[i][j-1] == 0:
                res += 1
            if j+1 >= x or grid[i][j+1] == 0:
                res += 1

            dfs(i+1,j)
            dfs(i-1,j)
            dfs(i,j+1)
            dfs(i,j-1)
            return


        for i in range(x):
            for j in range(y):
                if grid[j][i] == 1:
                    dfs(j,i)
                    return res
        # return res
22784