<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 {
    int n;
vector<vector<int>>arr;
int dp[50][50][50][50];
long long find_path(int i1,int j1,int i2,int j2)
{
    if(i1<0 || i1>n-1 || j1<0 || j1>n-1 || i2<0 || i2>n-1 || j2<0 || j2>n-1 ||arr[i1][j1]==-1 || arr[i2][j2]==-1)
    {
        return -1e9;
    }
    if(i1==n-1 && j1==n-1)
    {
        if(i2==n-1 && j2==n-1)
        {
            return arr[i1][j1];
        }
        else
        {
            return -1e9;
        }
    }
    if(dp[i1][j1][i2][j2]!=-1) return dp[i1][j1][i2][j2];
    long long maxi=INT_MIN;
    if(i1==i2 && j1==j2)
    {
        maxi=max(maxi,arr[i1][j1]+find_path(i1+1,j1,i2+1,j2));
        maxi=max(maxi,arr[i1][j1]+find_path(i1+1,j1,i2,j2+1));
        maxi=max(maxi,arr[i1][j1]+find_path(i1,j1+1,i2,j2+1));
        maxi=max(maxi,arr[i1][j1]+find_path(i1,j1+1,i2+1,j2));
    }
    else
    {
        maxi=max(maxi,arr[i1][j1]+arr[i2][j2]+find_path(i1+1,j1,i2+1,j2));
        maxi=max(maxi,arr[i1][j1]+arr[i2][j2]+find_path(i1+1,j1,i2,j2+1));
        maxi=max(maxi,arr[i1][j1]+arr[i2][j2]+find_path(i1,j1+1,i2,j2+1));
        maxi=max(maxi,arr[i1][j1]+arr[i2][j2]+find_path(i1,j1+1,i2+1,j2));
    }
    return dp[i1][j1][i2][j2]=maxi;
}
public:
    int cherryPickup(vector<vector<int>>& grid) {
        memset(dp,-1,sizeof(dp));
        n= grid.size();
        arr=grid;
        int ans = find_path(0,0,0,0);
        return ans<0?0:ans;
    }
};
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