<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)
        
        # Initialize the dp array with -inf
        dp = [[[-10**9 for _ in range(n + 1)] for _ in range(n + 1)] for _ in range(2 * n + 1)]
        
        # Initialize the starting position
        dp[0][0][0] = grid[0][0]
        
        # Iterate over all possible steps (k = i + j)
        for k in range(1, 2 * n - 1):
            for i1 in range(max(0, k - n + 1), min(n - 1, k) + 1):
                for i2 in range(max(0, k - n + 1), min(n - 1, k) + 1):
                    j1, j2 = k - i1, k - i2
                    
                    # If the current cells are thorns, skip
                    if grid[i1][j1] == -1 or grid[i2][j2] == -1:
                        continue
                    
                    # Initialize the current dp value
                    current = grid[i1][j1]
                    if i1 != i2:
                        current += grid[i2][j2]
                    
                    # Update the dp value with the maximum of possible previous states
                    dp[k][i1][i2] = max(dp[k-1][i1-1][i2-1], dp[k-1][i1-1][i2], dp[k-1][i1][i2-1], dp[k-1][i1][i2]) + current
        
        # The answer will be at dp[2*n-2][n-1][n-1]
        return max(0, dp[2 * n - 2][n - 1][n - 1])

# Example usage:
solution = Solution()
grid = [[0, 1, -1], [1, 0, -1], [1, 1, 1]]
print(solution.cherryPickup(grid))  # Output: 5
20446