<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:
        n=len(grid)
        #a[i] is the possible position after i moves
        a=[]
        for i in range(2*n-1):
            a.append([])
            for j in range(-1,i+2):
                a[i].append((j,i-j))
        #b keeps track of the grid:
        b=[]
        for i in range(2*n-1):
            b.append({})
            for item in a[i]:
                if 0<=item[0]<n and 0<=item[1]<n:
                    b[i][item]=grid[item[0]][item[1]] if grid[item[0]][item[1]]!=-1 else -float('inf')
                else:
                    b[i][item]=-float('inf')
        #c keeps track of the cartesian product of the grid:
        c=[]
        for i in range(2*n-1):
            c.append({})
            if i==0:
                for item1 in b[i]:
                    for item2 in b[i]:
                        c[i][(item1,item2)]=b[i][item1] if item1==item2==(0,0) else -float('inf')
            else:
                for item1 in b[i]:
                    for item2 in b[i]:
                        c[i][(item1,item2)]=b[i][item1] if item1==item2 else b[i][item1]+b[i][item2]
                        if c[i][(item1,item2)]!=-float('inf'):
                            c[i][(item1,item2)]+=max(c[i-1][((item1[0]-1,item1[1]),(item2[0]-1,item2[1]))],c[i-1][((item1[0]-1,item1[1]),(item2[0],item2[1]-1))],c[i-1][((item1[0],item1[1]-1),(item2[0]-1,item2[1]))],c[i-1][((item1[0],item1[1]-1),(item2[0],item2[1]-1))])
        
        return c[2*n-2][((n-1,n-1),(n-1,n-1))] if c[2*n-2][((n-1,n-1),(n-1,n-1))]>=0 else 0


        



        

            




        




            






        
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