<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();
        int n = grid[0].size();
        vector<vector<bool>> visited(m, vector<bool>(n, false));
        vector<vector<int>> dir = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
        int res = 0;
        queue<pair<int, int>> q;
        for(int i = 0; i < m; ++i){
            for(int j = 0; j < n; ++j){
                if(grid[i][j] == 1 && visited[i][j] == false){
                    q.push({i, j});
                    visited[i][j] = true;
                    int count = 0;
                    while(q.size() > 0){
                        auto temp = q.front();
                        q.pop();
                        count++;
                        for(auto d : dir){
                            int new_i = temp.first + d[0];
                            int new_j = temp.second + d[1];
                            if(new_i < 0 || new_i >= m || new_j < 0 || new_j >= n || visited[new_i][new_j] == true || grid[new_i][new_j] == 0){
                                continue;
                            }
                            q.push({new_i, new_j});
                            visited[new_i][new_j] = true;
                        }
                    }
                    res = max(res, count);
                }
            }
        }
        return res;
    }
};
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