<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:

    bool isValid(int i, int j, vector<vector<int>> &g, vector<vector<bool>> &v)
    {
        int m = g.size();
        int n = g[0].size();

        return (i >= 0 && i<m && j>=0 && j<n && g[i][j] >0 && !v[i][j]);
    }

    void DFS(int i, int j, vector<vector<int>> &g, vector<vector<bool>> &v, int currSum, int &maxSum)
    {
        static int dx[] = { 0, 0, -1, 1};
        static int dy[] = {-1, 1, 0 , 0};
        v[i][j] = true;
        currSum+=g[i][j];
        maxSum = max(maxSum, currSum);
        //cout<<"maxSum "<<maxSum<<endl;

        for (int k = 0; k < 4; k++)
        {
            int x = i+dx[k];
            int y = j+dy[k];

            if (isValid(x, y, g, v))
                DFS(x, y, g, v, currSum, maxSum);
        }
        
        currSum-=g[i][j];
        v[i][j] = false;
    }

    int getMaximumGold(vector<vector<int>>& grid) {
        
        int currSum = 0;
        int maxSum = 0;
        int ans = 0;
        int m = grid.size();
        int n = grid[0].size();


        for (int i = 0; i < m; i++)
        {
            for (int j = 0; j < n; j++)
            {
                if (grid[i][j] != 0)
                {
                    maxSum = 0;
                    vector<vector<bool>>visited(m, vector<bool>(n, false));

                    DFS(i, j, grid, visited, 0, maxSum);
                  //  cout<<endl<<endl;
                }

                ans = max(ans, maxSum);
            }
        }

        return ans;
    }
};
12000