<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 m;
    int n;
    vector<vector<int>> direction{{1,0},{-1,0}, {0,1},{0,-1}};
    int bfs(vector<vector<int>>& grid, int i, int j) {
        int perim =0;
        queue<pair<int,int>>q;
        q.push({i,j});
        grid[i][j] =-1;
        while(!q.empty()){
            auto it = q.front();
            q.pop();
            for(auto &dir:direction){
                int i_ = it.first + dir[0]; 
                int j_ = it.second + dir[1];

                if (i_ < 0 || i_ >= m || j_ < 0 || j_ >= n || grid[i_][j_] == 0) {
                    perim++;
                }
                else if(grid[i_][j_]==-1){
                    continue;
                }else{
                    q.push({i_,j_});
                    grid[i_][j_]=-1;
                }
            }
        }
        return perim;
    }

    int islandPerimeter(vector<vector<int>>& grid) {
        m = grid.size();       
        n = grid[0].size();

        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == 1) {
                    return bfs(grid, i, j);       
                }
            }
        }
        return -1;  
    }
};

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