<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 {
private:
    void dfs(int r, int c, vector<vector<int>>& grid, vector<vector<int>> &vis,
            map<pair<int, int>, int> &sides, int rows, int cols) {
        vis[r][c] = 1;

        int delRow[] = {-1,0,1,0};
        int delCol[] = {0,-1,0,1};
        for(int i=0; i<4; i++) {
            int newR = r + delRow[i];
            int newC = c + delCol[i];

            if(newR>=0 && newR<rows && newC>=0 && newC<cols && grid[newR][newC]==1) {
                if(!vis[newR][newC]) {
                    sides[{newR,newC}] = sides[{newR, newC}]-1;
                    dfs(newR, newC, grid, vis, sides, rows, cols);
                }
                else {
                    sides[{newR,newC}]--;
                }
            }
        }
    }
public:
    int islandPerimeter(vector<vector<int>>& grid) {
        map<pair<int, int>, int> sides;

        int rows = grid.size();
        int cols = grid[0].size();
        
        vector<vector<int>> vis(rows, vector<int>(cols, 0));
        int r,c;
        for(int i=0; i<rows; i++) {
            for(int j=0; j<cols; j++) {
                if(grid[i][j] == 1) {
                    r = i;
                    c = j;
                    sides[{i,j}]  =4;
                }
            }
        }
        dfs(r, c, grid, vis, sides, rows, cols);
        int peri = 0;

        for(auto itr: sides) {
            peri += itr.second;
        }

        return peri;
    }
};
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