<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) {
        // Visited areas
        set<pair<int, int>> visited;
        
        // Answer
        int out = 0;

        // Move through each cell
        for(int y = 0; y < grid.size(); y++) {
            for(int x = 0; x < grid[0].size(); x++) {
                // dfs if it is unvisited land
                if(grid[y][x] == 1 && !visited.contains({y, x})) {
                    // int currArea = ;
                    out = max(dfs(y, x, visited, grid), out);
                }
            }
        }

        return out;
    }

    int dfs(int y, int x, set<pair<int, int>>& visited, vector<vector<int>>& grid) {
        // Check out of bounds
        if(y < 0 || x < 0 || y >= grid.size() || x >= grid[0].size()) {
            return 0;
        }

        // Check water
        if(grid[y][x] == 0) {
            return 0;
        }

        // Check visited already
        if(visited.contains({y, x})) {
            return 0;
        }

        // Visit the place
        visited.insert({y, x});

        // Launch in all directions
        int up = dfs(y - 1, x, visited, grid);
        int down = dfs(y + 1, x, visited, grid);
        int left = dfs(y, x - 1, visited, grid);
        int right = dfs(y, x + 1, visited, grid);

        return up + down + left + right + 1;
    }
};
35070