<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:
        ROW, COL = len(grid), len(grid[0])
        memo = {}

        def dfs(r1, c1, c2):
            # optimization that r2 can be deduced from other state
            r2 = r1 + c1 - c2

            if not (
                r1 < ROW and c1 < COL and
                r2 < ROW and c2 < COL
            ) or grid[r1][c1] == -1 or grid[r2][c2] == -1:
                return float('-inf')

            # SPEED UP! basically the 2 robots can swap place
            # and have the same amount of cherry collected.
            # whenever that happens, we can refers to the calculated result
            if r2 > r1:
                return dfs(r2, c2, c1)
                
            if (r1, c1) == (ROW-1, COL-1) or (r2, c2) == (ROW-1, COL-1):
                return grid[ROW-1][COL-1]


            if (r1, c1, c2) in memo:
                return memo[(r1, c1, c2)]
            
            # c1 -> right, c2 -> right
            # c1 -> right, c2 -> down
            # c1 -> down, c2 -> right
            # c1 -> down, c2-> down

            cur = grid[r1][c1] + (grid[r2][c2] if (r1, c1) != (r2, c2) else 0)
            cur += max(
                dfs(r1, c1 + 1,c2 + 1),
                dfs(r1, c1 + 1, c2),
                dfs(r1 + 1, c1, c2 + 1),
                dfs(r1 + 1, c1, c2)
            )

            memo[(r1, c1, c2)] = cur

            return cur 

        return max(0, dfs(0,0,0))
26903