<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 {
private:
    int dx[4] = {0,0,1,-1};
    int dy[4] = {1,-1,0,0};
public:
    bool isPossible(int x, int y, vector<vector<int>> &grid)
    {
        int n = grid.size();
        int m = grid[0].size();
        if( x >= 0 && x <n && y >=0 && y < m)
            return 1;
        return 0;
    }
    int Dfs(int x, int y,vector<vector<int>> &grid)
    {
        if(grid[x][y] == 0)
            return 0;
        int temp = grid[x][y];
        int ans = 0;
        grid[x][y] = 0;

        for(int i = 0; i < 4; ++i)
        {
            if(isPossible(x+dx[i], y+dy[i], grid))
            {
                
                    ans = max(ans, Dfs(x+dx[i], y+dy[i], grid));
            }
        }
        grid[x][y] = temp;
        return temp + ans;
    }
    int getMaximumGold(vector<vector<int>>& grid) {
        int n = grid.size();
        int m = grid[0].size();
        int ans = 0;
        for(int i = 0; i < n ; ++i)
        {
            for(int j = 0; j < m; ++j)
            {
                    ans = max(ans, Dfs(i,j,grid));
            }
        }
        return ans;
    }
};
9100