<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:
        '''
        Idea: Iterate through each square of the grid. If a square has a land cell, then just check if there is any land cells above, below, to the left and to the right of it. If so, subtract one each time from the max of 4 possible perimeter points you could add to the perimeter.
        Estimated time complexity O(nm)
        '''
        perimeter = 0
        for i in range(len(grid)):
            for j in range(len(grid[i])):
                if grid[i][j] == 0:
                    continue
                if j != 0 and grid[i][j-1] == 0:
                    perimeter += 1
                if i != 0 and grid[i-1][j] == 0:
                    perimeter += 1
                if j != len(grid[i]) - 1 and grid[i][j+1] == 0:
                    perimeter += 1
                if i != len(grid) - 1 and grid[i+1][j] == 0:
                    perimeter += 1
                if j == 0:
                    perimeter += 1
                if i == 0:
                    perimeter += 1
                if j == len(grid[i]) - 1:
                    perimeter += 1
                if i == len(grid) - 1:
                    perimeter += 1
        return perimeter


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