<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();
        int n = grid[0].size();
        int ans = 0;

        vector<vector<int>> vis(m , vector<int>(n , 0));

        vector<vector<int>> moves = {{-1,0} , {1 , 0} , {0,1} , {0,-1}};

        for (int i=0 ; i<m ; i++){
            for (int j=0 ; j<n ; j++){
                if(!vis[i][j] && grid[i][j] == 1){
                    vis[i][j] = 1;
                    queue<pair<int , int>> q;
                    q.push({i , j});
                    int temp = 0;
                    while(!q.empty()){
                        auto it = q.front();
                        q.pop();
                        temp++;
                        for(int k=0 ; k<moves.size() ; k++){
                            int newx = it.first + moves[k][0];
                            int newy = it.second + moves[k][1];

                            if(newx >= 0 && newy >= 0 && newx < m && newy < n && vis[newx][newy] == 0 && grid[newx][newy] == 1){
                                vis[newx][newy] = 1;
                                q.push({newx , newy});
                            }
                        }
                    }
                    ans = max(ans , temp);
                }
            }
        }
        return ans;
    }
};
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