<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 ans=0;
        int n = grid.size();
        int m = grid[0].size();
        int dx[] = {-1, 0, 1, 0};
        int dy[] = {0, -1, 0, 1};

        for(int i=0; i<n; i++) {
            for(int j=0; j<m; j++) {
                if(grid[i][j] == 1) {
                    int curr=0;
                    queue<pair<int, int>> q;
                    q.push({i, j});
                    grid[i][j] = 0;
                    curr++;
                    while(!q.empty()) {
                        int x = q.front().first;
                        int y = q.front().second;
                        q.pop();
                        for(int k=0; k<4; k++) {
                            int nx = x+dx[k];
                            int ny = y+dy[k];
                            if(nx>=0 && ny >=0 && nx<n && ny<m && grid[nx][ny] == 1) {
                                grid[nx][ny] = 0;
                                curr++;
                                q.push({nx, ny});
                            }
                        }
                    }
                    ans = max(ans, curr);
                }
            }
        }
        return ans;
    }
};
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