<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:
    void dfs(vector<vector<int>>& grid, vector<vector<bool>>& visited, int i, int j, int& maxGold, int curGold) {

        maxGold = max(maxGold, curGold);
        visited[i][j] = true;
        int goldVal = grid[i][j];
        grid[i][j] = -1;
        int n = grid.size();
        int m = grid[0].size();
        vector<pair<int,int>> kids({{1,0}, {0,-1}, {-1,0}, {0,1}});
        for (int k = 0; k < 4; k++) {
            int x = i + kids[k].first;
            int y = j + kids[k].second;

            if (x >= 0 && x < n && y >= 0 && y < m && grid[x][y] > 0) {
                
                dfs(grid, visited, x, y, maxGold, curGold + grid[x][y]);
                
            } 
        }
        grid[i][j] = goldVal;
    }
    int getMaximumGold(vector<vector<int>>& grid) {
        int n = grid.size();
        int m = grid[0].size();
        vector<vector<bool>> visited(n, vector<bool> (m, false));
        int result = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (grid[i][j] != -1 && !visited[i][j]) {
                    int maxGold = 0;
                    int curGold = grid[i][j];
                    dfs(grid, visited, i, j, maxGold, curGold);
                    result = max(result, maxGold);
                }
            }
        }
        return result;
    }
};
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