<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:
void dfs(vector<vector<int>>&grid,vector<vector<int>>&vis,int row,int col,int &cnt,int delrow[],int delcol[],int n,int m){
   vis[row][col]=1;
   cnt++;
   for(int i=0;i<4;i++){
    int nrow=row+delrow[i];
    int ncol=col+delcol[i];
    if(nrow>=0&&nrow<n&&ncol>=0&&ncol<m&&!vis[nrow][ncol]&&grid[nrow][ncol]==1){
        dfs(grid,vis,nrow,ncol,cnt,delrow,delcol,n,m);
    }
   }
}
    int maxAreaOfIsland(vector<vector<int>>& grid) {
        int n=grid.size();
        int m=grid[0].size();
        vector<vector<int>>vis(n,vector<int>(m,0));
        int max_count=0;
        int delrow[4]={-1,0,1,0};
        int delcol[4]={0,1,0,-1};
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                int cnt=0;
                if(!vis[i][j]&&grid[i][j]==1){
                    dfs(grid,vis,i,j,cnt,delrow,delcol,n,m);
                    max_count=max(max_count,cnt);
                }
            }
        }
        return max_count;
    }
};
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