<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 result = 0;

        int rows = grid.size();
        int cols = grid[0].size();

        vector<vector<bool>> visited(rows, vector<bool>(cols));

        for(int i = 0; i < rows; i++) {
            for(int j = 0; j < cols; j++) {
                if(grid[i][j] == 1 && !visited[i][j]) {
                    result = max(bfs(i, j, grid, visited), result);
                }
            }
        }

        return result;
    }
private:
    int bfs(int i, int j, vector<vector<int>>& grid, vector<vector<bool>>& visited) {
        int rows = grid.size();
        int cols = grid[0].size();

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

        queue<pair<int, int>> q;

        q.push({i, j});
        visited[i][j] = true;

        int area = 1;

        while(!q.empty()) {
            pair<int, int> curr = q.front();

            q.pop();

            for(int k = 0; k < 4; k++) {
                int di = curr.first + dx[k];
                int dj = curr.second + dy[k];

                if(di >= 0 && di < rows && dj >= 0 && dj < cols && grid[di][dj] != 0 && !visited[di][dj]) {
                    visited[di][dj] = true;
                    q.push({di,  dj});
                    area++;
                }
            }
        }

        return area;
    }
};
32550