<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 DFS(vector<vector<int>> grid){
        int res=0;
        int x[]={0,-1,0,1};
        int y[]={-1,0,1,0};
        for(int i=0;i<grid.size();i++){
            for(int j=0;j<grid[0].size();j++){
                if(grid[i][j]){
                    stack<pair<int,int>> st;
                    st.push({i,j});
                    grid[i][j]=0;
                    int cnt=0;
                    while(!st.empty()){
                        pair<int,int> p=st.top();
                        st.pop();
                        cnt++;
                        res=max(res,cnt);
                        for(int k=0;k<4;k++){
                            int x1=p.first+x[k];
                            int y1=p.second+y[k];
                            if(x1>=0 && y1>=0 && x1<=grid.size()-1 && y1<=grid[0].size()-1 && grid[x1][y1] ){
                                st.push({x1,y1});
                                grid[x1][y1]=0;
                            }
                        }
                    }
                }
            }
        }
        return res;
    }
    int maxAreaOfIsland(vector<vector<int>>& grid) {
        return DFS(grid);
    }
};
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