<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 maxAreaOfIsland(vector<vector<int>>& grid) {

        int m = grid.size(), n = grid[0].size();
        vector<vector<bool>> isVisited(m, vector<bool>(n, false));

        int maxArea = 0;

        function<int(int, int)> dfs = [&](int i, int j) {

            if(i < 0 || j < 0 || i >= m || j >= n) return 0;
            if(grid[i][j] == 0 || isVisited[i][j]) return 0;

            int area = 1;
            isVisited[i][j] = true;
            area += (   dfs(i-1, j) + 
                        dfs(i+1, j) +
                        dfs(i, j-1) +
                        dfs(i, j+1)
                    );
            
            return area;
        };

        for(int i = 0; i < m; i++) {
            for(int j = 0; j < n; j++) {
                if(grid[i][j] == 1 && !isVisited[i][j]) {
                    int area = dfs(i, j);
                    if(area > maxArea) maxArea = area;
                }
            }
        }

        return maxArea;
    }
};
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