<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:
    def getMaximumGold(self, grid: List[List[int]]) -> int:
        n_rows, n_cols = len(grid), len(grid[0])

        def backtrack(x: int, y: int, total: int, seen: set[tuple[int, int]]) -> None:
            max_gold = total

            for (nx, ny) in (
                (x - 1, y),
                (x, y - 1),
                (x, y + 1),
                (x + 1, y)
            ):
                if 0 <= nx < n_rows and 0 <= ny < n_cols \
                    and (nx, ny) not in seen \
                    and grid[nx][ny] > 0:
                    seen.add((nx, ny))
                    max_gold = max(max_gold, backtrack(nx, ny, total + grid[nx][ny], seen))
                    seen.remove((nx, ny))

            return max_gold

        res = 0

        for x in range(n_rows):
            for y in range(n_cols):
                res = max(res, backtrack(x, y, grid[x][y], {(x, y)}))

        return res

        
        
16700