1219. Path with Maximum Gold

Problem Link

Description

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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

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Python3

class Solution:
    def getMaximumGold(self, grid: List[List[int]]) -> int:
        rows, cols = len(grid), len(grid[0])
        res = 0
        visited = [[False] * cols for _ in range(rows)]
 
        def backtrack(x, y, cost):
            nonlocal res
 
            if cost > res:
                res = cost
 
            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 not visited[dx][dy] and grid[dx][dy] != 0:
                    visited[dx][dy] = True
                    backtrack(dx, dy, cost + grid[dx][dy])
                    visited[dx][dy] = False
 
        for x in range(rows):
            for y in range(cols):
                if grid[x][y] != 0:
                    visited[x][y] = True
                    backtrack(x, y, grid[x][y])
                    visited[x][y] = False
 
        return res