<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 {
public:
    int islandPerimeter(vector<vector<int>>& grid) {
        if (grid.size() == 0)
            return 0;

        int rows = grid.size();
        int cols = grid[0].size();
        int perimeter = 0;
        vector<vector<bool>> visit(rows, vector<bool>(cols, false));

        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < cols; j++) {
                // If current cell is land, hasn't visited
                if (grid[i][j] == 1 && visit[i][j] == false)
                    perimeter += dfs(i, j, grid, visit);
            }
        }
        return perimeter;
    }

    int dfs(int row, int col, vector<vector<int>>& grid, vector<vector<bool>>& visit) {
        int rows = grid.size();
        int cols = grid[0].size();

        // Mark visited
        visit[row][col] = true;

        int perimeter = 0;
        int directions[4][2] = {
            {-1,  0}, // Left
            { 1,  0}, // Right
            { 0, -1}, // Up
            { 0,  1}  // Down
        };

        for (auto dir : directions) {
            int newRow = row + dir[0];
            int newCol = col + dir[1];

            // If the neighbor is water or out of bounds, length of 1 is added
            if (newRow < 0 || newRow >= rows || newCol < 0 || newCol >= cols || grid[newRow][newCol] == 0)
                perimeter++;
            // If the neighbor is land, don't do anything
        }

        return perimeter;
    }
};
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