<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) {
         int cnt=0;
        int x[]={-1,0,1,0};
        int y[]={0,1,0,-1};
        queue<pair<int,int>>q;
        int n=grid.size();
        int m=grid[0].size();
        vector<vector<int>>vis(n,vector<int>(m,0));
         for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(grid[i][j]==1 && !vis[i][j]){
                    vis[i][j]=1;
                    q.push({i,j});

                
            
         

        

         while(!q.empty()){
                        int r =q.front().first;
                        int c=q.front().second;
                        q.pop();
         
                for(int i=0;i<4;i++){
            int nrow=r+x[i];
            int ncol=c+y[i];

 if (nrow < 0 || nrow >= n || ncol < 0 || ncol >= m || grid[nrow][ncol] ==0) {                          
    cnt++;   
    } 
  else if( vis[nrow][ncol]==0 && grid[nrow][ncol]==1 ){
          
           vis[nrow][ncol]=1;
           q.push({nrow,ncol});
     }

                }
                     }
                } 
            } 
         }
    return cnt;
    }

};
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