<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:
        size = len(grid[0])

        dp =[[[-1]*size for i in range(size)] for i in range(size)]
        vis =[[[0]*size for i in range(size)] for i in range(size)]


        def isSafe(row, col, row2, col2):
            nonlocal size
            if row<0 or col<0 or row2<0 or col2<0 or row>=size or col>=size or row2>=size or col2>=size:
                return False
            if grid[row][col] == -1 or grid[row2][col2] == -1:
                return False
            return True
    
        def dfs(row, col, row2):
            nonlocal size
            n = size
            col2 = abs(row+col -row2)

            if not isSafe(row, col, row2, col2):
                return (False, 0)

            if row==n-1 and col==n-1 and row2==n-1:
                return (True, grid[row][col])

            if vis[row][col][row2]:
                if dp[row][col][row2] == -1:
                    return (False, 0)
                else:
                    return (True, dp[row][col][row2])

            vis[row][col][row2] = 1
            x = [(1,0,0), (0,1,1), (1,0,1), (0,1,0)]

            for inc in x:
                n_row = row + inc[0]
                n_col = col + inc[1]
                n_row2 = row2 + inc[2]
                val = dfs(n_row, n_col, n_row2)
                if val[0]:
                    dp[row][col][row2] = max(dp[row][col][row2], val[1])

            if dp[row][col][row2] == -1:
                return (False, 0)

            if row == row2:
                dp[row][col][row2] += grid[row][col]
            else:
                dp[row][col][row2] += grid[row][col] + grid[row2][col2]

            return (True, dp[row][col][row2])

        
        res = dfs(0,0,0)
        #print(res)
        #print(dp)
        return res[1]





        
        

            


        
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