<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 pq=0;
    
   
    int cherryPickup(vector<vector<int>>& grid) {
        vector<vector<int>> v(grid.size()*grid.size(),
                              vector<int>(grid[0].size()*grid.size(), -1));
    
    long long int ans=0;
    
     int h=findp(grid,0,0,0,0,v);
          if(ans<h) ans=h;
       return ans;
    }
    long long int findp(vector<vector<int>>& grid, int i, int j,
    int n,int m,vector<vector<int>>& v) {
            if(i>=grid.size() || j>=grid[0].size() ||
            m>=grid[0].size() || n>=grid.size() ||grid[i][j]==-1 || grid[n][m]==-1)
            {
                return INT_MIN;
            }
            if(i==grid.size()-1 && j==grid[0].size()-1 &&
            n==grid.size()-1 && m==grid[0].size()-1 ) 
           { if(grid[i][j]!=-1)
            return v[i*grid.size()+j][n*grid.size()+m]=grid[i][j];
            else return v[i*grid.size()+j][n*grid.size()+m]=INT_MIN;
           }
           if(v[i*grid.size()+j][n*grid.size()+m]!=-1) return v[i*grid.size()+j][n*grid.size()+m];
           int a=findp(grid,i+1,j,n+1,m,v);
            int b=findp(grid,i,j+1,n+1,m,v);
            int c=findp(grid,i+1,j,n,m+1,v);
            int d=findp(grid,i,j+1,n,m+1,v);
            if(a==INT_MIN && b==INT_MIN && c==INT_MIN && d==INT_MIN) return v[i*grid.size()+j][n*grid.size()+m]=INT_MIN;
            int q=max(a,max(b,max(c,d)));
            if(i==n && j==m)
            {
                q=q+grid[i][j];
            }
            else
            {
                q=q+grid[i][j]+grid[n][m];
            }
            return v[i*grid.size()+j][n*grid.size()+m]=q;
        }
};
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