<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 {
    int dr[4] = {1,0,-1,0};
    int dc[4] = {0,1,0,-1};
    vector<pair<int,int>>path;
    void dfs(int r,int c,vector<vector<int>>&vis,int n,int m)
    {
        vis[r][c] = 1;
        path.push_back({r,c});
        for(int i=0;i<4;i++)
        {
            int nr = r+dr[i];
            int nc = c+dc[i];
            if(nr>=0 && nr<n && nc>=0 && nc<m && !vis[nr][nc])
            {
                dfs(nr,nc,vis,n,m);
            }
        }
    }
public:
    int maxAreaOfIsland(vector<vector<int>>& grid) 
    {
         int n = grid.size();
         int m = grid[0].size();
         vector<pair<int,int>>nodes;
         vector<vector<int>>vis(n,vector<int>(m,0));
         for(int i=0;i<n;i++)
         {
            for(int j=0;j<m;j++)
            {
                 if(grid[i][j]==1)
                 {
                     nodes.push_back({i,j});
                 }
                 else
                 vis[i][j] = 1;
            }
         }
         vector<vector<pair<int,int>>>com;
         for(int i=0;i<nodes.size();i++)
         {
            auto x = nodes[i];
            if(!vis[x.first][x.second])
            {
                dfs(x.first,x.second,vis,n,m);
                com.push_back(path);
                path.clear();
            }
         }
        int cnt = 0;
        for(auto x:com)
        {
            int a = x.size();
            cnt = max(cnt,a);
        }
        return cnt;   
    }
};
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