Cherry Pickup
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0700-0799/0741.Cherry%20Pickup/images/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>
Array; Dynamic Programming; Matrix
function cherryPickup(grid: number[][]): number {
    const n: number = grid.length;
    const f: number[][][] = Array.from({ length: n * 2 - 1 }, () =>
        Array.from({ length: n }, () => Array.from({ length: n }, () => -Infinity)),
    );
    f[0][0][0] = grid[0][0];
    for (let k = 1; k < n * 2 - 1; ++k) {
        for (let i1 = 0; i1 < n; ++i1) {
            for (let i2 = 0; i2 < n; ++i2) {
                const [j1, j2]: [number, number] = [k - i1, k - i2];
                if (
                    j1 < 0 ||
                    j1 >= n ||
                    j2 < 0 ||
                    j2 >= n ||
                    grid[i1][j1] == -1 ||
                    grid[i2][j2] == -1
                ) {
                    continue;
                }
                const t: number = grid[i1][j1] + (i1 != i2 ? grid[i2][j2] : 0);
                for (let x1 = i1 - 1; x1 <= i1; ++x1) {
                    for (let x2 = i2 - 1; x2 <= i2; ++x2) {
                        if (x1 >= 0 && x2 >= 0) {
                            f[k][i1][i2] = Math.max(f[k][i1][i2], f[k - 1][x1][x2] + t);
                        }
                    }
                }
            }
        }
    }
    return Math.max(0, f[n * 2 - 2][n - 1][n - 1]);
}
