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.

class Solution {
    public int getMaximumGold(int[][] grid) {
        int m = grid.length, n = grid[0].length;
        int maxGold = 0;
        for (int r = 0; r < m; r++)
            for (int c = 0; c < n; c++)
                maxGold = Math.max(maxGold, findMaxGold(grid, m, n, r, c));
        return maxGold;
    }

    int[][] DIR = { { 0, 1 }, { 1, 0 }, { 0, -1 }, { -1, 0 } };

    int findMaxGold(int[][] grid, int m, int n, int r, int c) {
        if (r < 0 || r == m || c < 0 || c == n || grid[r][c] == 0)
            return 0;
        int origin = grid[r][c];
        grid[r][c] = 0; // mark as visited
        int maxGold = 0;
        for (int i = 0; i < 4; i++)
            maxGold = Math.max(maxGold, findMaxGold(grid, m, n, DIR[i][0] + r, DIR[i][1] + c));
        grid[r][c] = origin; // backtrack
        return maxGold + origin;
    }
}