<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 {
public:
    int dx[4] = {-1, 0, 1, 0};
    int dy[4] = {0, -1, 0, 1};
     int islandPerimeter(vector<vector<int>> & grid) {
        ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        int n = grid.size(), m = grid[0].size();
        vector<vector<int>> vis(n, vector<int>(m));
        int per = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (!vis[i][j] and grid[i][j]) {
                    queue<pair<int, int>> q;
                    q.push({i, j});
                    while (!q.empty()) {
                        auto [x, y] = q.front();
                        vis[x][y]=1;
                        q.pop();
                        int ans = 0;
                        for (int k = 0; k < 4; k++) {
                            int nx = x + dx[k];
                            int ny = y + dy[k];
                            if (nx >= 0 and nx < n and ny >= 0 and ny < m and
                                !vis[nx][ny] and grid[nx][ny]) {
                                vis[nx][ny] = 1;
                                q.push({nx, ny});
                            } 
                            else if((nx < 0 or nx >= n or ny < 0 or ny >= m) or !vis[nx][ny])ans++;
                        }
                        per += ans;
                    }
                }
            }
        }
        return per;
    }
};
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