Path with Maximum Gold

Problem Statement

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.

Leetcode link

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

Python Code

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def getMaximumGold(self, grid: List[List[int]]) -> int:
        def dfs(i, j, v):
            seen.add((i, j))
            dp[i][j] = max(dp[i][j], v)
            for x, y in (i - 1, j), (i + 1, j), (i, j - 1), (i, j + 1):
                if 0 <= x < m and 0 <= y < n and grid[x][y] and (x, y) not in seen:
                    dfs(x, y, v + grid[x][y])
            seen.discard((i, j))
        m, n = len(grid), len(grid[0])
        dp = [[0] * n for _ in range(m)]
        for i in range(m):
            for j in range(n):
                if grid[i][j]:
                    seen = set()
                    dfs(i, j, grid[i][j])
        return max(c for row in dp for c in row)