<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 {

    int dfs(int i,int j,vector<vector<int>>& grid,vector<vector<bool>>& vis){
        // For boundry of grid or water neighbour we return 1
        if(i<0 or i>=grid.size() or j<0 or j>=grid[0].size() or grid[i][j] == 0) return 1 ;
        if(vis[i][j]) return 0;
        vis[i][j] = true;
        int perimeter  = 0;

        perimeter += dfs(i+1,j,grid,vis);
        perimeter += dfs(i-1,j,grid,vis);
        perimeter += dfs(i,j+1,grid,vis);
        perimeter += dfs(i,j-1,grid,vis);
        

        return perimeter;
    }
public:
    int islandPerimeter(vector<vector<int>>& grid) {
        int n=grid.size(),c=grid[0].size();
        vector<vector<bool>>vis(n,vector<bool>(c,false));
        for(int i=0;i<n;i++)
        for(int j=0;j<c;j++){
            if(grid[i][j] == 1) return dfs(i,j,grid,vis);
        }
        return -1;
    }
};
47057