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1219. Path with Maximum Gold

Difficulty Topics

Description

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.

Solution

path-with-maximum-gold.py
class Solution:
    def getMaximumGold(self, grid: List[List[int]]) -> int:
        rows, cols = len(grid), len(grid[0])

        res = 0

        def dfs(x, y):
            tmp = grid[x][y]
            grid[x][y] = 0

            total = 0
            for dx,dy in ((x+1, y), (x-1, y), (x, y-1), (x, y+1)):
                if 0 <= dx < rows and 0 <= dy < cols and grid[dx][dy] != 0:
                    total = max(total, grid[dx][dy] + dfs(dx,dy))

            grid[x][y] = tmp

            return total


        for i in range(rows):
            for j in range(cols):
                if grid[i][j] > 0:
                    res = max(res, grid[i][j] + dfs(i,j))

        return res