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

    void dfs(int i, int j, int n, int m, vector<vector<int>>& grid, int & area){
        if(i<0 || i>=n || j<0 || j>=m || grid[i][j]==0) return;

        grid[i][j]=0;
        area +=1;

        for(int k=0; k<4; k++){
            int nx=i+dx[k];
            int ny=j+dy[k];

            dfs(nx,ny,n,m,grid,area);
        }
        
    }
    int maxAreaOfIsland(vector<vector<int>>& grid) {
        int n=grid.size();
        int m=grid[0].size();
        int ans=0;

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

        return ans;
    }
};
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