<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) {

        std::queue <pair<int,int>> bfs_queue;
        vector<vector<bool>> visited(grid.size(),vector<bool>(grid[0].size(), 0));
        
        //Push the first occuring land into the bfs 

        for (int i = 0; i< grid.size(); i++)
        {
            for (int j = 0; j< grid[0].size(); j++)
            {
                if (grid[i][j] == 1)
                {
                    bfs_queue.push({i,j});
                    visited[i][j] = true;
                    break;
                }
            }
        }

        vector<pair<int,int>> direction{{0,1}, {0,-1},{1,0},{-1,0}};
        int perimeter = 0;
        //find the sides of square adjacent to water or its out of bound.
        while (!bfs_queue.empty())
        {
            int qsize = bfs_queue.size();
            for (int i = 0; i< qsize; i++)
            {
                auto [cur_row, cur_col] = bfs_queue.front();
                bfs_queue.pop();
                for (auto delta: direction)
                {
                    int next_row = cur_row + delta.first;
                    int next_col = cur_col + delta.second;

                    if (next_row< 0 || next_row >= grid.size() || next_col < 0 || next_col >= grid[0].size() || 
                        grid[next_row][next_col] == 0)
                    {
                        perimeter++;
                    }
                    else if (next_row >=0 && next_row < grid.size() &&
                        next_col >= 0 && next_col < grid[0].size() && !visited[next_row][next_col])
                    {
                            bfs_queue.push({next_row,next_col});
                            visited[next_row][next_col] = true;
                    }
                } 
            }
        }
        return perimeter;
    }
};
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