<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 f(int i,int j,vector<vector<int>>&arr){
        int n=arr.size();int m=arr[0].size();
        if(i>=n || i<0 || j>=m || j<0 || arr[i][j]==-1 || arr[i][j]==0)return 0;
        int s=0;int maxi=0;
        int k=arr[i][j];
        arr[i][j]=-1;
        int l=k+f(i,j+1,arr);
        int r=k+f(i,j-1,arr);
        int u=k+f(i-1,j,arr);
        int d=k+f(i+1,j,arr);
        maxi=max({maxi,l,r,u,d});
        arr[i][j]=k;
        return maxi;
    }
    int getMaximumGold(vector<vector<int>>& grid) {
       int maxi=0;
       int n=grid.size();int m=grid[0].size();
       for(int i=0;i<n;i++){
        for(int j=0;j<m;j++){
            maxi=max(maxi,f(i,j,grid));
        }
       }
       return maxi;
    }
};
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