<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>

class Solution:
    def cherryPickup(self, grid: List[List[int]]) -> int:
        """
        dp (r1, c1, r2, c2) is the max number of cherries if p1 goes from origin to (r1, c1)
        and p2 goes from origin to (r2, c2)
        r1 + c1 = r2 + c2 = t --> c2 = r1 - r2 + c1
        ans = dp(n - 1, n - 1, n - 1, n - 1)

        To calculate: (note r1 + c1 = r2 + c2 = t)
        
        if (r1, c1) == -1 or (r2, c2) == -1:
            return 0
        dp (r1, c1, r2, c2) = max(
            dp (r1 - 1, c1, r2 - 1, c2), dp (r1 - 1, c1, r2, c2 - 1), 
            dp (r1, c1 - 1, r2 - 1, c2), dp (r1, c1 - 1, r2, c2 - 1)
        ) + (1 if (r1, c1) == 1 else 0) + (1 if (r2, c2) == 1 and (r1, c1) != (r2, c2) else 0)
        """
        cache: dict[tuple[int, int, int]: int] = {}
        n: int = len(grid)
        def dp(r1: int, c1: int, r2: int) -> int:
            c2: int = r1 - r2 + c1
            grid_sum: int = grid[r1][c1] + grid[r2][c2] if (r1, c1) != (r2, c2) else grid[r1][c1]
            
            # Base case (i forgor :skull:)
            if r1 == c1 == r2 == 0:
                return grid_sum
            if not (0 <= r1 < n and 0 <= r2 < n and 0 <= c1 < n and 0 <= c2 < n):
                return float('-inf')
            key: tuple[int, int, int] = (r1, c1, r2)
            if key in cache:
                return cache[key]
            cache[key] = float('-inf')
            if grid[r1][c1] == -1 or grid[r2][c2] == -1:
                return cache[key]
            # If any one if neg inf then it's neg inf
            vals: list[int] = [dp (r1 - 1, c1, r2 - 1), dp (r1 - 1, c1, r2), 
                dp (r1, c1 - 1, r2 - 1), dp (r1, c1 - 1, r2)]

            cache[key] = max(
                vals
            ) + grid_sum
            return cache[key]
        
        val: int = dp(n - 1, n - 1, n - 1)

        return val if val >= 0 else 0

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