<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 {
public:

    int f(int r1, int c1, int r2, int c2, vector<vector<int>> &grid, vector<vector<vector<vector<int>>>> &dp, int n, vector<int> &dr1, vector<int> &dc1, vector<int> &dr2, vector<int> &dc2){

        if(r1 >= n || r2 >= n || c1 >= n || c2 >= n) return -1e9;

        if(grid[r1][c1] == -1 || grid[r2][c2] == -1) return -1e9;

        if(r1 == n-1 && c1 == n-1 && r2 == n-1 && c2 == n-1) return grid[r1][c1];

        if(dp[r1][c1][r2][c2] != -1) return dp[r1][c1][r2][c2];

        int cherries = -1e9;

        if(r1 == r2 && c1 == c2){  // same cell, only pick cherries once

            for(int i = 0;  i < 4; i++){

                int nr1 = r1 + dr1[i];
                int nc1 = c1 + dc1[i];
                int nr2 = r2 + dr2[i];
                int nc2 = c2 + dc2[i];

                cherries = max(cherries, grid[r1][c1] + f(nr1, nc1, nr2, nc2, grid, dp, n, dr1, dc1, dr2, dc2));
            }
        }

        else{  // diff cell, pick cherries from both cells

            for(int i = 0;  i < 4; i++){

                int nr1 = r1 + dr1[i];
                int nc1 = c1 + dc1[i];
                int nr2 = r2 + dr2[i];
                int nc2 = c2 + dc2[i];

                cherries = max(cherries, grid[r1][c1] + grid[r2][c2] + f(nr1, nc1, nr2, nc2, grid, dp, n, dr1, dc1, dr2, dc2));
            }
        }

        if(cherries == -1e9 || cherries == -1e9 + 1 || cherries == -1e9 + 2) return dp[r1][c1][r2][c2] = -1e9;

        return dp[r1][c1][r2][c2] = cherries;
    }

    int cherryPickup(vector<vector<int>>& grid) {

        ios_base::sync_with_stdio(false);
        cin.tie(nullptr);

        int n = grid.size();

        vector<vector<vector<vector<int>>>> dp(n, vector<vector<vector<int>>> (n, vector<vector<int>> (n, vector<int> (n, -1))));

        vector<int> dr1 = {0, 0, 1, 1};
        vector<int> dc1 = {1, 1, 0, 0};
        vector<int> dr2 = {0, 1, 0, 1};
        vector<int> dc2 = {1, 0, 1, 0};

        int val = f(0, 0, 0, 0, grid, dp, n, dr1, dc1, dr2, dc2);

        return val == -1e9 ? 0 : val; 
    }
};
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