<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>

#if 1
class Solution {

    int get_area_dfs(vector<vector<int>>& grid, int r, int c, vector<vector<bool>>& visited) {

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

        if (r < 0 || c < 0 || r >= m || c >= n || grid[r][c] == 0 || visited[r][c] == true) {
            return 0;
        }

        visited[r][c] = true;

        int count = 0;

        for (auto vc : dir) {
            count += get_area_dfs(grid, r+vc[0], c+vc[1], visited);
        }

        return count + 1;
    }

    int get_area_dfs_no_visited(vector<vector<int>>& grid, int r, int c) {

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

        if (r < 0 || c < 0 || r >= m || c >= n || grid[r][c] == 0) {
            return 0;
        }

        grid[r][c] = 0;

        int count = 0;

        for (auto vc : dir) {
            count += get_area_dfs_no_visited(grid, r+vc[0], c+vc[1]);
        }

        return count + 1;
    }

    vector<vector<int>> dir = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};

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

        int m = grid.size();
        
        if (m == 0)
            return 0;

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

        //vector<vector<bool>> visited(m, vector<bool>(n, false));

        for (int r = 0; r < m; r++) {
            for (int c= 0; c< n; c++) {

                //max_area = max(max_area, get_area_dfs(grid, r, c, visited));

                max_area = max(max_area, get_area_dfs_no_visited(grid, r, c));
            }
        }

        return max_area;
    }
};


#else

class Solution {
public:
    int maxAreaOfIsland(vector<vector<int>>& grid) {
        int m = grid.size(), n = m ? grid[0].size() : 0, area = 0, offsets[] = {0, 1, 0, -1, 0};
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j]) {
                    int a = 1;
                    grid[i][j] = 0;
                    queue<pair<int, int>> todo;
                    todo.push({i, j});
                    while (!todo.empty()) {
                        pair<int, int> p = todo.front();
                        todo.pop();
                        for (int k = 0; k < 4; k++) {
                            int r = p.first + offsets[k], c = p.second + offsets[k + 1];
                            if (r >= 0 && r < m && c >= 0 && c < n && grid[r][c]) {
                                grid[r][c] = 0;
                                a++;
                                todo.push({r, c});
                            }
                        }
                    }
                    area = max(area, a);
                }
            }
        }
        return area;
    }
};

#endif
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