<p>You are given an <code>n x n</code> <code>grid</code> representing a field of cherries, each cell is one of three possible integers.</p>

<ul>
	<li><code>0</code> means the cell is empty, so you can pass through,</li>
	<li><code>1</code> means the cell contains a cherry that you can pick up and pass through, or</li>
	<li><code>-1</code> means the cell contains a thorn that blocks your way.</li>
</ul>

<p>Return <em>the maximum number of cherries you can collect by following the rules below</em>:</p>

<ul>
	<li>Starting at the position <code>(0, 0)</code> and reaching <code>(n - 1, n - 1)</code> by moving right or down through valid path cells (cells with value <code>0</code> or <code>1</code>).</li>
	<li>After reaching <code>(n - 1, n - 1)</code>, returning to <code>(0, 0)</code> by moving left or up through valid path cells.</li>
	<li>When passing through a path cell containing a cherry, you pick it up, and the cell becomes an empty cell <code>0</code>.</li>
	<li>If there is no valid path between <code>(0, 0)</code> and <code>(n - 1, n - 1)</code>, then no cherries can be collected.</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/12/14/grid.jpg" style="width: 242px; height: 242px;" />
<pre>
<strong>Input:</strong> grid = [[0,1,-1],[1,0,-1],[1,1,1]]
<strong>Output:</strong> 5
<strong>Explanation:</strong> The player started at (0, 0) and went down, down, right right to reach (2, 2).
4 cherries were picked up during this single trip, and the matrix becomes [[0,1,-1],[0,0,-1],[0,0,0]].
Then, the player went left, up, up, left to return home, picking up one more cherry.
The total number of cherries picked up is 5, and this is the maximum possible.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> grid = [[1,1,-1],[1,-1,1],[-1,1,1]]
<strong>Output:</strong> 0
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= n &lt;= 50</code></li>
	<li><code>grid[i][j]</code> is <code>-1</code>, <code>0</code>, or <code>1</code>.</li>
	<li><code>grid[0][0] != -1</code></li>
	<li><code>grid[n - 1][n - 1] != -1</code></li>
</ul>

from typing import List

class Solution:
    def cherryPickup(self, grid: List[List[int]]) -> int:
        rows = len(grid)
        cols = len(grid[0])

        # Initialize visited array to store the maximum cherries collected
        # at each state (r0, c0, r1, c1) where -1 means not yet visited
        visited = [[[-1 for _ in range(rows)] for _ in range(cols)] for _ in range(rows)]

        def searchPaths(r0, c0, r1, c1):
            # Time Complexity: O((rows * cols)^2) due to state space exploration
            # Space Complexity: O(rows * cols * rows) due to memoization

            # If a path goes off the grid
            if r0 >= rows or r1 >= rows or c0 >= cols or c1 >= cols:
                return -1
            # If a path moves into a thorn
            elif grid[r0][c0] == -1 or grid[r1][c1] == -1:
                return -1
            # If the state (r0, c0, r1, c1) has been computed before
            elif visited[r0][c0][r1] != -1:
                return visited[r0][c0][r1]
            else:
                pickups = 0

                # If first path crosses a cherry
                if grid[r0][c0] == 1:
                    pickups += 1
                # If second path crosses a cherry that is distinct from the first
                if grid[r1][c1] == 1 and (r1 != r0 or c1 != c0):
                    pickups += 1

                # If both paths reach the bottom-right corner of the grid
                if r0 == rows - 1 and c0 == cols - 1:
                    return pickups

                # Explore all 4 possible moves for both paths
                maxFuturePickups = max(
                    searchPaths(r0 + 1, c0, r1 + 1, c1),   # Both paths move down
                    searchPaths(r0 + 1, c0, r1, c1 + 1),   # First path moves down, second path moves right
                    searchPaths(r0, c0 + 1, r1 + 1, c1),   # First path moves right, second path moves down
                    searchPaths(r0, c0 + 1, r1, c1 + 1)    # Both paths move right
                )

                # If no valid path exists
                if maxFuturePickups == -1:
                    visited[r0][c0][r1] = -100  # Mark as not reachable
                    return -1
                else:
                    pickups += maxFuturePickups
                    visited[r0][c0][r1] = pickups
                    return pickups

        # Compute the maximum cherries collected starting from (0, 0) for both paths
        maxCherries = searchPaths(0, 0, 0, 0)
        
        # If no valid path found, return 0
        return max(maxCherries, 0)

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