<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 bfs(int i, int j, vector<vector<int>>& grid, vector<vector<int>>& vis){
        queue<pair<int, int>> q;
        q.push({i, j});
        vis[i][j] = 1;

        int n = grid.size(), m = grid[0].size(), ans = 0;

        int rc[] = {-1, 0, 1, 0};
        int cc[] = {0, 1, 0, -1};

        while(!q.empty()){
            int row = q.front().first;
            int col = q.front().second;
            q.pop();

            for(int k = 0; k < 4; k++){
                int newRow = row + rc[k];
                int newCol = col + cc[k];

                if(min(newRow, newCol) >= 0 && newRow < n && newCol < m && grid[newRow][newCol] == 1){
                    if(!vis[newRow][newCol]){
                        q.push({newRow, newCol});
                        vis[newRow][newCol] = 1;
                    }
                }
                else ans++;
            }
        }

        return ans;
    }
public:
    int islandPerimeter(vector<vector<int>>& grid) {
        int n = grid.size(), m = grid[0].size(), ans = 0;
        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){
                    return bfs(i, j, grid, vis);
                }
            }
        }

        return 0;
    }
};
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