<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:
/* Its almost the same question as of number of islands.
 but in this we have only have to count the ones of each islands and save
  it to the maxarea variable using INT_MIN. */
  int m,n;
  int DFS(vector<vector<int>>&grid,int i,int j){
    // out of bound condition
    if(i<0||j<0||i>=m||j>=n||grid[i][j]==0){
        return 0;
    }
    // visited mark karna h
    grid[i][j]=0;
    // addding the recursion call as we getting the 1s
    return DFS(grid,i+1,j)+DFS(grid,i-1,j)+DFS(grid,i,j+1)+DFS(grid,i,j-1)+1;
  }
    int maxAreaOfIsland(vector<vector<int>>& grid) {
        int maxi=0;
        m=grid.size();
        n=grid[0].size();
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                if(grid[i][j]==1){
                    // calculating the number of ones on the basis of the count of 1s present in the shape.
                    maxi=max(maxi,DFS(grid,i,j));
                }
            }
        }
        return maxi;
    }
};
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