<p>In a gold mine <code>grid</code> of size <code>m x n</code>, each cell in this mine has an integer representing the amount of gold in that cell, <code>0</code> if it is empty.</p>

<p>Return the maximum amount of gold you can collect under the conditions:</p>

<ul>
	<li>Every time you are located in a cell you will collect all the gold in that cell.</li>
	<li>From your position, you can walk one step to the left, right, up, or down.</li>
	<li>You can&#39;t visit the same cell more than once.</li>
	<li>Never visit a cell with <code>0</code> gold.</li>
	<li>You can start and stop collecting gold from <strong>any </strong>position in the grid that has some gold.</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> grid = [[0,6,0],[5,8,7],[0,9,0]]
<strong>Output:</strong> 24
<strong>Explanation:</strong>
[[0,6,0],
 [5,8,7],
 [0,9,0]]
Path to get the maximum gold, 9 -&gt; 8 -&gt; 7.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> grid = [[1,0,7],[2,0,6],[3,4,5],[0,3,0],[9,0,20]]
<strong>Output:</strong> 28
<strong>Explanation:</strong>
[[1,0,7],
 [2,0,6],
 [3,4,5],
 [0,3,0],
 [9,0,20]]
Path to get the maximum gold, 1 -&gt; 2 -&gt; 3 -&gt; 4 -&gt; 5 -&gt; 6 -&gt; 7.
</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;= 15</code></li>
	<li><code>0 &lt;= grid[i][j] &lt;= 100</code></li>
	<li>There are at most <strong>25 </strong>cells containing gold.</li>
</ul>

class Solution {
public:
    
    int dfs(int r,int c,vector<vector<int>>&grid,vector<vector<bool>>&vis)
	{
		if(r>=grid.size() || c>=grid[0].size() || r<0 || c < 0 || vis[r][c]== true || grid[r][c] ==0)
		return 0;
		
		int sum=grid[r][c];
		vis[r][c]=true;

		int dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}};
        int mx=0;

		for(int i=0;i<4;i++)
		{
			int nr=r+dir[i][0];
			int nc=c + dir[i][1];
		mx=max(mx,dfs(nr,nc,grid,vis));
        }
        vis[r][c]=false;
		return sum+mx;
		
}
	int getMaximumGold(vector<vector<int>>&grid)
	{
		//run dfs + not vis+ !0
		int mx=0;
		int tr=grid.size();
		int tc=grid[0].size();
		
		vector<vector<bool>>vis(tr,vector<bool>(tc,false));
		for(int r=0;r<tr;r++)
		{
			for(int c=0;c<tc;c++)
			{
				if(vis[r][c]==false && grid[r][c]!=0)
				{
					int temp=dfs(r,c,grid,vis);
					mx=max(mx,temp);
}
}
}
		return mx;
}

};
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