<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 row, int col, int &ans, vector<vector<int>>& grid, vector<vector<int>>& vis) {
        int n = grid.size();
        int m = grid[0].size();
        vis[row][col] = 1;  // Mark as visited

        int delrow[] = {0, 1, 0, -1};
        int delcol[] = {-1, 0, 1, 0};

        // Explore neighbors in 4 directions
        for (int i = 0; i < 4; i++) {
            int nrow = row + delrow[i];
            int ncol = col + delcol[i];

            // Check if the neighbor is within bounds
            if (nrow >= 0 && ncol >= 0 && nrow < n && ncol < m && grid[nrow][ncol] == 1) {  
                if (!vis[nrow][ncol]) {
                    dfs(nrow, ncol, ans, grid, vis);
                }
            }
            else{
                ans++;
            }
        }
    }

public:
    int islandPerimeter(vector<vector<int>>& grid) {
        int n = grid.size();
        int m = grid[0].size();
        vector<vector<int>> vis(n, vector<int>(m, 0));  // Visited array
        int ans = 0;

        // Loop through the grid to find the first land cell
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (grid[i][j] == 1 && !vis[i][j]) {
                    // Start DFS from the first unvisited land cell
                    dfs(i, j, ans, grid, vis);
                }
            }
        }
        return ans;
    }
};

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