<p>You are given an <code>m x n</code> binary matrix <code>grid</code>. An island is a group of <code>1</code>&#39;s (representing land) connected <strong>4-directionally</strong> (horizontal or vertical.) You may assume all four edges of the grid are surrounded by water.</p>

<p>The <strong>area</strong> of an island is the number of cells with a value <code>1</code> in the island.</p>

<p>Return <em>the maximum <strong>area</strong> of an island in </em><code>grid</code>. If there is no island, return <code>0</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/05/01/maxarea1-grid.jpg" style="width: 500px; height: 310px;" />
<pre>
<strong>Input:</strong> grid = [[0,0,1,0,0,0,0,1,0,0,0,0,0],[0,0,0,0,0,0,0,1,1,1,0,0,0],[0,1,1,0,1,0,0,0,0,0,0,0,0],[0,1,0,0,1,1,0,0,1,0,1,0,0],[0,1,0,0,1,1,0,0,1,1,1,0,0],[0,0,0,0,0,0,0,0,0,0,1,0,0],[0,0,0,0,0,0,0,1,1,1,0,0,0],[0,0,0,0,0,0,0,1,1,0,0,0,0]]
<strong>Output:</strong> 6
<strong>Explanation:</strong> The answer is not 11, because the island must be connected 4-directionally.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> grid = [[0,0,0,0,0,0,0,0]]
<strong>Output:</strong> 0
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 50</code></li>
	<li><code>grid[i][j]</code> is either <code>0</code> or <code>1</code>.</li>
</ul>

class Solution {
public:
    int bfs(int r, int c, vector<vector<int>>& grid, vector<vector<int>>& vis,
            int n, int m) {

        vis[r][c] = 1;
        int ar=1;

        queue<pair<int, int>> q;
        q.push({r, c});

        int dx[4] = {0, 1, 0, -1};
        int dy[4] = {1, 0, -1, 0};

        while (!q.empty()) {
            int x = q.front().first;
            int y = q.front().second;
            q.pop();

            for (int i = 0; i < 4; i++) {
                int dr = x + dx[i];
                int dc = y + dy[i];

                if(dr>=0 && dr<n && dc>=0 && dc<m && grid[dr][dc]==1 && vis[dr][dc]==0){
                    q.push({dr, dc});
                    vis[dr][dc]=1;
                    ar++;
                }
            }
        }
        return ar;
    }

    int maxAreaOfIsland(vector<vector<int>>& grid) {
        int n = grid.size();
        int m = grid[0].size();

        vector<vector<int>> vis(n, vector<int>(m, 0));

        if (m == 0 || n == 0) {
            return 0;
        }
        int maxarea=0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {

                if (grid[i][j] == 1 && vis[i][j] == 0) {
                    maxarea = max(maxarea, bfs(i, j, grid, vis, n, m));
                }
            }
        }
        return maxarea;
    }
};
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