<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 bfs(int r,int c,vector<vector<int>>&visit,vector<vector<int>>& grid){
    int sum=1;
    visit[r][c]=1;
    queue<pair<int,int>>q;
    q.push({r,c});
    int dr[]={-1,0,1,0};
    int dc[]={0,1,0,-1};
    while(!q.empty()){
        int row=q.front().first;
        int col=q.front().second;
        q.pop();
        for(int i=0;i<4;i++){
            int nr=row+dr[i];
            int nc=col+dc[i];
            if(nr>=0 && nc>=0 && nr<grid.size() && nc<grid[0].size() && grid[nr][nc]==1 && visit[nr][nc]==0){
                visit[nr][nc]=1;
                q.push({nr,nc});
                sum++;
            }
        }
    }
    return sum;
}
public:
    
int maxAreaOfIsland(vector<vector<int>>& grid) {
    int n=grid.size();
    int m=grid[0].size();
    int ans=0;
    vector<vector<int>>visit(n,vector<int>(m,0));
    for(int i=0;i<n;i++){
        for(int j=0;j<m;j++){
            if(grid[i][j]==1 && visit[i][j]==0){
                ans=max(ans,bfs(i,j,visit,grid));
            }
        }
    }
    return ans;
}
};
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