<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[60][60][60][60];
int flag = false ;
int solve(vector<vector<int>>& grid,int row,int col,int r , int c){
         if (row < 0 || col < 0 || row >= grid.size() || col >= grid[0].size() || r < 0 || c < 0 || r >= grid.size() || 
         c >= grid[0].size() ||
          grid[row][col] == -1 || grid[r][c] == -1) {
            return INT_MIN;
        }
        if(row==grid.size()-1&& col == grid.size()-1){
            flag = true ;
            return  dp[row][col][r][c] = grid[row][col];
        }
        if(dp[row][col][r][c] != 0){
            return dp[row][col][r][c];
        }
         int cherry = 0;
        if(row==r && col== c){
            cherry += grid[row][col];
        }
        else {
        cherry += grid[row][col] + grid[r][c];
        }

        int f1 = solve(grid,row+1,col,r,c+1); //v,h
        int f2 = solve(grid,row,col+1,r+1,c); //h,v
        int f3 = solve(grid,row+1,col,r+1,c);
        int f4 = solve(grid,row,col+1,r,c+1);

        return  dp[row][col][r][c] =  cherry += max(max(f1,f2),max(f3,f4));

}
    int cherryPickup(vector<vector<int>>& grid) {
        int ans =  solve(grid,0,0,0,0);
        memset(dp,0,sizeof(dp));
        if(flag == false ){
            return 0;

        }
        else {
            return ans ;
        }
    }
};
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