<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 n;
    int solve(int x1, int y1 , int x2 , int y2 , vector<vector<int>>& grid , vector<vector<vector<vector<int>>>> &dp){
        if(x1 < 0 || y1 < 0 || x2 < 0 || y2 < 0)  return -1e6;
        if(grid[x1][y1] == -1 || grid[x2][y2] == -1) return -1e6;

        if(x1 == 0 && y1 == 0) return grid[x1][y1] == 1 ? 1 : 0;
        if(dp[x1][y1][x2][y2] != -1)    return dp[x1][y1][x2][y2];

        int ans = 0;

        if(x1 == x2 && y1 == y2){
            if(grid[x1][y1] == 1)   
                ans += grid[x1][y1];
        }else{
            if(grid[x1][y1] == 1)  ans++;
            if(grid[x2][y2] == 1)  ans++;
        }

        int u1u2 = solve(x1-1, y1, x2-1, y2, grid, dp);
        int u1l2 = solve(x1-1, y1, x2, y2-1, grid, dp);
        int l1u2 = solve(x1, y1-1, x2-1, y2, grid, dp);
        int l1l2 = solve(x1, y1-1, x2, y2-1, grid, dp);

        ans += max({u1u2,u1l2,l1u2,l1l2});

        return dp[x1][y1][x2][y2] = ans;
    }

    int cherryPickup(vector<vector<int>>& grid) {
        n = grid.size();
        vector<vector<vector<vector<int>>>> dp(52 , vector<vector<vector<int>>>(52 , vector<vector<int>>(52 , vector<int>(52 , -1))));

        int ans = solve(n-1,n-1,n-1,n-1,grid,dp); 
        return (ans < 0) ? 0 : ans;
    }
};
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