<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>

#include <vector>
#include <queue>
#include <utility>

using namespace std;

class Solution {
public:
    // Function to check if the cell is valid (within grid bounds)
    bool isvalid(int new_i, int new_j, int m ,int n){
        return new_i >= 0 && new_i < m && new_j >= 0 && new_j < n;
    }
    
    int islandPerimeter(vector<vector<int>>& grid) {
        int m = grid.size();       // Number of rows
        int n = grid[0].size();    // Number of columns
        int perimeter = 0;         // Initialize perimeter
        
        // Direction vectors for up, right, down, left
        int r[] = {-1, 0, 1, 0};   // Row changes
        int c[] = {0, 1, 0, -1};   // Column changes
        
        // BFS initialization
        queue<pair<int, int>> q;
        vector<vector<bool>> visited(m, vector<bool>(n, false));
        
        // Find the first land cell (1) to start BFS
        for (int i = 0; i < m; i++) {
            bool found = false;
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == 1) {
                    q.push({i, j});
                    visited[i][j] = true;
                    found = true;
                    break;
                }
            }
            if (found) break;
        }
        
        // Perform BFS
        while (!q.empty()) {
            int x = q.front().first;
            int y = q.front().second;
            q.pop();
            
            // Check the four possible directions (up, right, down, left)
            for (int k = 0; k < 4; k++) {
                int new_x = x + r[k];
                int new_y = y + c[k];
                
                // If the neighboring cell is out of bounds or water, it contributes to the perimeter
                if (!isvalid(new_x, new_y, m, n) || grid[new_x][new_y] == 0) {
                    perimeter++;
                } 
                // If the neighboring cell is land and not visited, add it to the BFS queue
                else if (grid[new_x][new_y] == 1 && !visited[new_x][new_y]) {
                    visited[new_x][new_y] = true;
                    q.push({new_x, new_y});
                }
            }
        }
        
        return perimeter;
    }
};

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