# Path with Maximum Gold

In a gold mine `grid` of size `m * n`, each cell in this mine has an integer representing the amount of gold in that cell, `0` if it is empty.

Return the maximum amount of gold you can collect under the conditions:

• Every time you are located in a cell you will collect all the gold in that cell.
• From your position you can walk one step to the left, right, up or down.
• You can’t visit the same cell more than once.
• Never visit a cell with `0` gold.
• You can start and stop collecting gold from any position in the grid that has some gold.

Example 1:

```Input: grid = [[0,6,0],[5,8,7],[0,9,0]]
Output: 24
Explanation:
[[0,6,0],
[5,8,7],
[0,9,0]]
Path to get the maximum gold, 9 -> 8 -> 7.
```

Example 2:

```Input: grid = [[1,0,7],[2,0,6],[3,4,5],[0,3,0],[9,0,20]]
Output: 28
Explanation:
[[1,0,7],
[2,0,6],
[3,4,5],
[0,3,0],
[9,0,20]]
Path to get the maximum gold, 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7.
```

Constraints:

• `1 <= grid.length, grid[i].length <= 15`
• `0 <= grid[i][j] <= 100`
• There are at most 25 cells containing gold.

Since the count of cells is at maximum 15×15 and we know at maximum there can be 25 cells with gold, we can perform a simple DFS starting with each cell with gold and find the maximum possible gold considering each step as a new start. We can then return the maximum gold that can we aquired of all starting points.

```class Solution {

public int getMaximumGold(int[][] grid) {

if(grid == null || grid.length == 0 || grid.length == 0){
return 0;
}
int m = grid.length;
int n = grid.length;
int max_gold = 0;
Set<String> visited = new HashSet<>();
for(int i = 0 ; i < m ; i++){
for(int j = 0 ; j < n ; j++){
max_gold = Math.max(max_gold, DFS(grid, i, j, m, n, visited));
}
}
return max_gold;
}

private int DFS(int[][] grid, int i, int j, int m, int n, Set<String> visited){
if(i < 0 || i == m || j < 0 || j == n || grid[i][j] == 0){
return 0;
}
String key = i + "_" + j;
if(visited.contains(key)){
return 0;
}
int cur_gold = grid[i][j];

int gold_in_top = DFS(grid, i-1, j, m, n, visited);
int gold_in_bottom = DFS(grid, i+1, j, m, n, visited);
int gold_in_left = DFS(grid, i, j-1, m, n, visited);
int gold_in_right = DFS(grid, i, j+1, m, n, visited);

int vertical_max = Math.max(gold_in_top, gold_in_bottom);
int side_max = Math.max(gold_in_left, gold_in_right);

cur_gold = cur_gold + Math.max(vertical_max,side_max);

visited.remove(key);
return cur_gold;
}
}
```

Complexity

• Space: `O(n)` for the algorithm. `O(1) `for this question. Since the depth of recursion can go upto max `25` cells.
• Time: `O(4n)`. Each cell is initiating 4 `DFS` in 4 directions.

Categories: DFS