<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)
        res = 0        

        @cache
        def dfs(x,y):
            if x == n - 1 and y == n - 1:
                return grid[x][y]
            else:
                maxc = -1
                for dx, dy in [[0,1],[1,0]]:
                    xx = dx + x
                    yy = dy + y
                    if xx < n and xx >= 0 and yy < n and yy >= 0 and grid[xx][yy] != -1:
                        maxc = max(dfs(xx,yy), maxc)
                if maxc == -1:
                    return maxc
                else:
                    return maxc + grid[x][y]
                
        res = dfs(0,0)
        if res == -1:
            return 0
        else:
            prefix_by_col = [[0 for _ in range(n)] for _ in range(n + 1)]
            for j in range(n):
                for i in range(n):
                    prefix_by_col[i+1][j] = prefix_by_col[i][j] + grid[i][j]
                    
            @cache  
            def dfser(x, f1, f2):
                if x == n - 1:
                    res = 0
                    for i in range(min(f1,f2),n):
                        if grid[i][x] == -1:
                            return -1
                        res += grid[i][x]
                    return res
                elif grid[f1][x] == -1 or grid[f2][x] == -1:
                    return -1
                else:
                    maxer = -1
                    for i in range(f1, n):
                        if grid[i][x] == -1:
                            break
                        for j in range(f2,n):
                            if grid[j][x] == -1 :
                                break
                            else:
                  
                                indices = set()
                                if i >= f2:
                                    loot = prefix_by_col[max(i,j) + 1][x] - prefix_by_col[f1][x]
                                else:
                                    loot = prefix_by_col[i + 1][x] - prefix_by_col[f1][x]
                                    loot += prefix_by_col[j + 1][x] - prefix_by_col[f2][x]
                                    
                            
                                    
                                cur = dfser(x + 1, min(i,j), max(i,j))
                                if cur != -1:
                                    maxer = max(maxer, cur + loot)
         
                    return maxer
            rez = dfser(0, 0, 0)

            if rez == -1:
                return 0
            else:
                return rez
                    
                            
                    
                
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