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:

1. Every time you are located in a cell you will collect all the gold in that cell.
2. From your position, you can walk one step to the left, right, up, or down.
3. You can't visit the same cell more than once.
4. Never visit a cell with 0 gold.
5. You can start and stop collecting gold from any position in the grid that has some gold.

For example:

grid = [[0,6,0],[5,8,7],[0,9,0]]

Output: 24 Explanation: Path to get the maximum gold, 9 -> 8 -> 7.

grid = [[1,0,7],[2,0,6],[3,4,5],[0,3,0],[9,0,20]]

Output: 28 Explanation: 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.

def getMaximumGold(grid):
    if not grid or not grid[0]:
        return 0

    m, n = len(grid), len(grid[0])
    max_gold = 0

    def dfs(row, col, current_gold, visited):
        nonlocal max_gold
        max_gold = max(max_gold, current_gold)

        # Possible directions: up, down, left, right
        directions = [(0, 1), (0, -1), (1, 0), (-1, 0)]

        for dr, dc in directions:
            new_row, new_col = row + dr, col + dc

            if 0 <= new_row < m and 0 <= new_col < n and \
               grid[new_row][new_col] != 0 and (new_row, new_col) not in visited:

                dfs(new_row, new_col, current_gold + grid[new_row][new_col], visited | {(new_row, new_col)})

    for row in range(m):
        for col in range(n):
            if grid[row][col] != 0:
                dfs(row, col, grid[row][col], {(row, col)})

    return max_gold

Explanation:

1. Problem Understanding:

   • The problem requires finding the maximum gold that can be collected from a grid by moving up, down, left, or right.
   • The path can start from any cell with gold and end at any cell.
   • A cell cannot be visited more than once, and cells with 0 gold cannot be visited.

2. Naive Approach (Brute Force):

   • Try every possible path starting from each cell containing gold.
   • Use Depth-First Search (DFS) to explore all possible paths.
   • Keep track of visited cells to avoid cycles.
   • Calculate the total gold collected for each path and update the maximum gold found so far.

3. Optimal Solution:

   • The DFS approach described above is already quite optimal for the given constraints (grid size <= 15x15).
   • There isn't a significantly faster algorithm due to the need to explore all possible paths.
   • Optimization can involve careful coding and avoiding unnecessary computations, but the core idea remains the same.

def getMaximumGold(grid):
    if not grid or not grid[0]:
        return 0

    m, n = len(grid), len(grid[0])
    max_gold = 0

    def dfs(row, col, current_gold, visited):
        nonlocal max_gold
        max_gold = max(max_gold, current_gold)

        # Possible directions: up, down, left, right
        directions = [(0, 1), (0, -1), (1, 0), (-1, 0)]

        for dr, dc in directions:
            new_row, new_col = row + dr, col + dc

            if 0 <= new_row < m and 0 <= new_col < n and \
               grid[new_row][new_col] != 0 and (new_row, new_col) not in visited:

                dfs(new_row, new_col, current_gold + grid[new_row][new_col], visited | {(new_row, new_col)})

    for row in range(m):
        for col in range(n):
            if grid[row][col] != 0:
                dfs(row, col, grid[row][col], {(row, col)})

    return max_gold

1. Big(O) Time Complexity Analysis:

   • O(4^(mn)): In the worst-case scenario, each cell contains gold, and from each cell, we can explore up to 4 directions. Since we can have at most mn cells, the time complexity is exponential.
   • Explanation: The dfs function explores all possible paths. In the worst case, the algorithm explores all possible paths in the grid. This can be seen as a tree where each node has up to 4 children, and the depth is at most the number of cells in the grid (m * n). Hence the exponential nature.

2. Big(O) Space Complexity Analysis:

   • O(mn): In the worst case, the recursion depth can go up to mn (the number of cells in the grid). The visited set can also contain up to m*n elements.
   • Explanation: The space complexity is dominated by the visited set, which stores the coordinates of visited cells, and the recursion stack depth. In the worst case, we might visit all cells, hence O(m*n).

3. Edge Cases and Handling:

   • Empty Grid: If the input grid is empty or has no rows, return 0.
   • Grid with No Gold: If the grid contains only 0s, return 0.
   • Single Cell Grid: If the grid has only one cell, return the gold in that cell (if it's not 0).
   • Large Grids: While the constraints limit the grid size to 15x15, larger grids would make the algorithm significantly slower due to exponential time complexity. Optimization techniques (like memoization, although not directly applicable here) would be needed for larger grids.

4. Alternative Approaches:

   • Dynamic Programming: Dynamic programming is not directly applicable here because the path can start and end at any cell. Standard DP approaches require defining clear start and end points.
   • Iterative DFS (Stack-based): An iterative DFS approach using a stack can be used to avoid recursion depth issues for very large grids, but it doesn't change the fundamental time complexity.

Summary:

The optimal solution for this problem is a DFS-based approach that explores all possible paths from each cell containing gold, while keeping track of visited cells to avoid cycles. The time complexity is O(4^(mn)), and the space complexity is O(mn). The algorithm handles edge cases such as empty grids, grids with no gold, and single-cell grids.