<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)
        if N == 1:
            return grid[0][0]
        memo = {}
        self.pickCherry2(grid,N,0,0,0,0,memo)
        print(memo)
        if len(memo) == 0 or memo[(0,0,0,0)] < 0:
            return 0
        return  memo[(0,0,0,0)]
    
    def pickCherry2(self, grid, N, x0, y0, x1, y1, memo):
        
        if x0 >= N or y0 >= N or x1 >= N or y1 >= N :
            return float('-inf') 
        
        if (x0, y0, x1, y1) in memo:
            return memo[(x0, y0, x1, y1)]
        
        if grid[x0][y0] == -1 or grid[x1][y1] == -1:
            return float('-inf')
        
        elif x0 == x1 and y0 == y1:
            cherryPicked = grid[x1][y1]
        else :
            cherryPicked = grid[x0][y0] + grid[x1][y1]
        
        #print(x0, y0, x1, y1 , cherryPicked)
        if x0 == N-1 and y0 == N-1 :
            return grid[x0][y0]
        if x1 == N-1 and y1 == N-1 :
            return grid[x1][y1]
            
        cherryPicked += max([self.pickCherry2(grid,N,x0+1,y0,x1+1,y1,memo),
                             self.pickCherry2(grid,N,x0+1,y0,x1,y1+1,memo),
                             self.pickCherry2(grid,N,x0,y0+1,x1+1,y1,memo),
                             self.pickCherry2(grid,N,x0,y0+1,x1,y1+1,memo)])
        
        memo[(x0, y0, x1, y1)] = cherryPicked
        
        return cherryPicked
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