1219. Path with Maximum Gold

# Time:  O(m^2 * n^2)
# Space: O(m * n)

class Solution(object):
    def getMaximumGold(self, grid):
        """
        :type grid: List[List[int]]
        :rtype: int
        """
        directions = [(0, 1), (1, 0), (0, -1), (-1, 0)]
        def backtracking(grid, i, j):
            result = 0
            grid[i][j] *= -1
            for dx, dy in directions:
                ni, nj = i+dx, j+dy
                if not (0 <= ni < len(grid) and
                        0 <= nj < len(grid[0]) and
                        grid[ni][nj] > 0):
                    continue
                result = max(result, backtracking(grid, ni, nj))
            grid[i][j] *= -1
            return grid[i][j] + result

        result = 0
        for i in xrange(len(grid)):
            for j in xrange(len(grid[0])):
                if grid[i][j]:
                    result = max(result, backtracking(grid, i, j))
        return result

Solution from kamyu104/LeetCode-Solutions · MIT