<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>

inline const auto optimize = []() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    std::cout.tie(nullptr);
    return 0;
}();

class Solution {

    static constexpr int solve(const std::vector<std::vector<int>>& grid,
                               const int N, std::vector<int>& dp, const int r1,
                               const int c1, const int r2) {
        const int key = r1 * N * N + c1 * N + r2;
        if (dp[key] != -2) {
            return dp[key];
        }
        // We know that r1 + c1 = r2 + c2 <=> c2 = r1 + c1 - r2
        const int c2 = r1 + c1 - r2;

        if (grid[r1][c1] == -1 || grid[r2][c2] == -1) {
            dp[key] = -1;
            return -1;
        }

        const bool sameSpot = (r1 == r2) && (c1 == c2);

        int sum = grid[r1][c1];
        if (!sameSpot) {
            sum += grid[r2][c2];
        } else if (r1 == N - 1 && c1 == N - 1) {
            dp[key] = sum;
            return sum;
        }

        int ans = -1;
        if (r1 < N - 1) {
            if (r2 < N - 1) {
                ans = std::max(ans, solve(grid, N, dp, r1 + 1, c1, r2 + 1));
            }

            if (c2 < N - 1) {
                ans = std::max(ans, solve(grid, N, dp, r1 + 1, c1, r2));
            }
        }

        if (c1 < N - 1) {
            if (c1 < c2 && r2 < N - 1) {
                ans = std::max(ans, solve(grid, N, dp, r1, c1 + 1, r2 + 1));
            }

            if (c2 < N - 1) {
                ans = std::max(ans, solve(grid, N, dp, r1, c1 + 1, r2));
            }
        }

        if (ans != -1) {
            dp[key] = ans + sum;
        } else {
            dp[key] = -1;
        }
        return dp[key];
    }

public:
    static int cherryPickup(const std::vector<std::vector<int>>& grid) {
        const int N = grid.size();
        std::vector<int> dp(N * N * N, -2);

        return std::max(0, solve(grid, N, dp, 0, 0, 0));
    }
};
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