<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 btHelper(const vector<vector<int>>& grid, vector<vector<bool>>& visited, int m, int n, int i, int j, int val) {
        int maxGold(val);
        if (i > 0 && grid[i-1][j] > 0 && !visited[i-1][j]) {
            visited[i-1][j] = true;
            maxGold = max(maxGold, btHelper(grid, visited, m, n, i-1, j, val + grid[i-1][j]));
            visited[i-1][j] = false;
        }
        if (i < m-1 && grid[i+1][j] > 0 && !visited[i+1][j]) {
            visited[i+1][j] = true;
            maxGold = max(maxGold, btHelper(grid, visited, m, n, i+1, j, val + grid[i+1][j]));
            visited[i+1][j] = false;
        }
        if (j > 0 && grid[i][j-1] > 0 && !visited[i][j-1]) {
            visited[i][j-1] = true;
            maxGold = max(maxGold, btHelper(grid, visited, m, n, i, j-1, val + grid[i][j-1]));
            visited[i][j-1] = false;
        }
        if (j < n-1 && grid[i][j+1] > 0 && !visited[i][j+1]) {
            visited[i][j+1] = true;
            maxGold = max(maxGold, btHelper(grid, visited, m, n, i, j+1, val + grid[i][j+1]));
            visited[i][j+1] = false;
        }

        return maxGold;
    }
public:
    int getMaximumGold(vector<vector<int>>& grid) {
        int m(grid.size()), n(grid[0].size());
        vector<vector<bool>> visited(m, vector<bool>(n, false));
        int maxGold(0);
        for (int i = 0; i < m; i ++) {
            for (int j = 0; j < n; j ++) {
                if (grid[i][j] != 0) {
                    visited[i][j] = true;
                    maxGold = max(maxGold, btHelper(grid, visited, m, n, i, j, grid[i][j]));
                    visited[i][j] = false;
                }
            }
        }
        return maxGold;
    }
};
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