<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:
    int maxAreaOfIsland(vector<vector<int>>& grid) {
        int m=grid.size(),n=grid[0].size();
        vector<pair<int,int>>d={{0,-1},{0,1},{-1,0},{1,0}};
        bool vis[m][n];
        memset(vis,false,sizeof(vis));
        int area,mx=0;
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                area=0;
                if(!vis[i][j]&&grid[i][j]){
                    queue<pair<int,int>>q;
                    q.push({i,j});
                    vis[i][j]=true;
                    while(!q.empty()){
                        area++;
                        pair<int,int>par=q.front();
                        q.pop();
                        int ii=par.first,jj=par.second;
                        for(int it=0;it<4;it++){
                            int iii=ii+d[it].first,jjj=jj+d[it].second;
                            bool vld=0<=iii&&iii<m&&0<=jjj&&jjj<n;
                            if(vld&&!vis[iii][jjj]&&grid[iii][jjj]){
                                q.push({iii,jjj});
                                vis[iii][jjj]=true;
                            }
                        }
                    }
                }
                mx=max(mx,area);
            }
        }
        return mx;
    }
};
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