<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 dp[50][50][50];
    int cherries (int i1,int j1, int i2, vector<vector<int>>&grid)
    {  
        int j2 = i1+j1-i2;
         int n = grid.size();
       int m = grid[0].size();

       if(i1>=n or j1>=m or i2>=n or j2>=m or grid[i1][j1]==-1 or grid[i2][j2]==-1) return INT_MIN;

        if(i1==n-1 and j1==m-1 and i2==n-1 and j2==m-1)
        {
            return grid[i1][j1];
        }

        if(dp[i1][j1][i2]!=-1) return dp[i1][j1][i2];
        vector<pair<int,int>>v = {{1,0},{0,1}};
        int result = INT_MIN;

         for(int i=0;i<2;i++)
         {
            for(int j=0;j<2;j++)
            {   
                result= max(result,cherries(i1+v[i].first,j1+v[i].second,i2+v[j].first,grid));
            }
         }
         
         if(result==INT_MIN) return dp[i1][j1][i2]= INT_MIN;
         if(i1!=i2 or j1!=j2) result+=grid[i1][j1]+grid[i2][j2];
         else result+= grid[i1][j1];   
         return dp[i1][j1][i2] =result;
}

    int cherryPickup(vector<vector<int>>& grid) {
       int n = grid.size();
        memset(dp,-1,sizeof(dp));
       int ans = cherries(0,0,0,grid);
      
    //    memset(dp,-1,sizeof(dp));
       if(ans==INT_MIN) return 0;
       else return ans;
        
    }
};
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