<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:
        m = len(grid)
        n = len(grid[0])
        dp = {}

        def recurMax(r1, c1, r2):
            c2 = r1 + c1 - r2
            curr = 0
            if (r1,c1,r2) in dp:
                return dp[(r1,c1,r2)]

            elif r1 >= m or c1 >= n or r2 >= m or c2 >= n or grid[r1][c1] == -1 or grid[r2][c2] == -1:
                dp[(r1,c1,r2)] = -float('inf')
                return -float('inf')

            elif r1 == m-1 and c1 == n-1:
                dp[(r1,c1,r2)] = grid[r1][c1]
                return grid[r1][c1]

            elif r1 == r2 and c1 == c2:
                curr = grid[r1][c1]

            else:
                curr += grid[r1][c1]
                curr += grid[r2][c2]

            option1 = recurMax(r1+1, c1, r2+1)
            option2 = recurMax(r1+1, c1, r2)
            option3 = recurMax(r1, c1+1, r2+1)
            option4 = recurMax(r1, c1+1, r2)
            maxVal = curr + max(option1, option2, option3, option4)
            dp[(r1,c1,r2)] = maxVal
            return maxVal

        result = recurMax(0,0,0)
        return result if result > 0 else 0


        


        
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