<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 r = grid.size();
        int c = grid[0].size();

        vector<pair<int, int>> directions = {{1, 0}, {-1, 0}, {0, -1}, {0, 1}};
        int maxii = 0;
        for (int i = 0; i < r; i++) {
            for (int j = 0; j < c; j++) {
                if (grid[i][j] == 1) {
                    // perform bfs from here
                    queue<pair<int, int>> q;
                    int area = 0;

                    q.push({i, j});
                    while (!q.empty()) {
                        auto [ii, jj] = q.front();
                        q.pop();

                        if (ii < 0 || ii >= r || jj < 0 || jj >= c)
                            continue;

                        if (grid[ii][jj] != 1)
                            continue;

                        grid[ii][jj] = '0';
                        area++;

                        for (auto d : directions) {
                            int iid = ii + d.first;
                            int jjd = jj + d.second;

                            q.push({iid, jjd});
                        }
                    }
                    maxii = max(maxii, area);
                }
            }
        }
        return maxii;
    }
};

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