<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:
    // Main function to find the maximum amount of gold
    int getMaximumGold(vector<vector<int>>& grid) {
        int maxNum = 0;
        int rows = grid.size();
        int cols = grid[0].size();
        
        // Loop through every cell in the grid
        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < cols; j++) {
                // If the cell contains gold, start DFS
                if (grid[i][j] != 0) {
                    vector<vector<bool>> visited(rows, vector<bool>(cols, false));
                    maxNum = max(maxNum, calcsum(grid, i, j, visited));
                }
            }
        }
        
        return maxNum;
    }

    // DFS function to explore all possible paths and collect gold
    int calcsum(vector<vector<int>>& grid, int i, int j, vector<vector<bool>>& visited) {
        // Boundary check and whether the cell is visited or contains no gold
        if (i < 0 || i >= grid.size() || j < 0 || j >= grid[0].size() || grid[i][j] == 0 || visited[i][j]) {
            return 0;
        }
        
        // Mark the current cell as visited
        visited[i][j] = true;
        
        // Store the amount of gold in the current cell
        int currentGold = grid[i][j];
        
        // Explore all 4 possible directions (left, right, up, down)
        int left = calcsum(grid, i, j - 1, visited);
        int right = calcsum(grid, i, j + 1, visited);
        int up = calcsum(grid, i - 1, j, visited);
        int down = calcsum(grid, i + 1, j, visited);
        
        // Mark the cell as unvisited (for backtracking)
        visited[i][j] = false;
        
        // Return the maximum gold collected from all 4 directions plus the current cell's gold
        return currentGold + max({left, right, up, down});
    }
};

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