1219. Path with Maximum Gold

In a gold mine grid of size m x 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:

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

Code

class Solution:
    def getMaximumGold(self, grid: List[List[int]]) -> int:
        m, n = len(grid), len(grid[0])
        DIR = [0, 1, 0, -1, 0]

        def dfs(r, c):
            if r < 0 or r == m or c < 0 or c == n or grid[r][c] == 0: return 0
            ans = 0
            orgGold = grid[r][c]
            grid[r][c] = 0
            for i in range(4):
                ans = max(ans, dfs(r + DIR[i], c + DIR[i+1]))
            grid[r][c] = orgGold
            return ans + grid[r][c]

        ans = 0
        for r in range(m):
            for c in range(n):
                ans = max(ans, dfs(r, c))
        return ans