<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:
    vector<pair<int,int>> dir = { {1,0},{0,1},{-1,0},{0,-1} };
    void dfs(vector<vector<int>>& grid, int r, int c, int &maxi )
    {
        grid[r][c] = 0;
        maxi += 1;

        for( auto it : dir )
        {
            int x = r + it.first;
            int y = c + it.second;
            if( x >= 0 && x<grid.size() && y >= 0 && y<grid[0].size() && grid[x][y] == 1 )
            {
                dfs(grid,x,y,maxi);
            }
        }
    }
    int maxAreaOfIsland(vector<vector<int>>& grid) {
        int maxAns = 0;
        int n = grid.size();
        int m = grid[0].size();

        for( int i = 0; i<n; i++ )
        {
            for( int j = 0; j<m; j++ )
            {
                if( grid[i][j] == 1 )
                {
                    int maxi = 0;
                    dfs(grid,i,j,maxi);
                    maxAns = max(maxAns,maxi);
                }
            }
        }
    return maxAns;
    }
};
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