<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 func(vector<vector<int>> &v, int m, int n, int i1, int j1, int j2, vector<vector<vector<int>>>&dp){
        // i1+j1 == i2+j2
        int i2 = i1+j1-j2;
        if(i1 >= m || i2 >=m || i1 < 0 || i2 < 0 || j1 >= n || j2 >= n || j1 < 0 || j2 < 0){
            return INT_MIN;
        }

        if(v[i1][j1] == -1 || v[i2][j2] == -1){
            return INT_MIN;
        }

        if(dp[i1][j1][j2] != -1){
            return dp[i1][j1][j2];
        }

        int score = 0;
        if(i1 == i2 && j1 == j2){
            score += v[i1][j1];
        }
        else{
            score += v[i1][j1] + v[i2][j2];
        }
        if(i1 == m-1 && j1 == n-1){
            return dp[i1][j1][j2] = score;
        }

        int maxi = INT_MIN;
        maxi = max(maxi,func(v,m,n,i1+1,j1,j2,dp)); //Both down;
        maxi = max(maxi,func(v,m,n,i1+1,j1,j2+1,dp)); //Down right
        maxi = max(maxi,func(v,m,n,i1,j1+1,j2,dp)); //Right Down
        maxi = max(maxi,func(v,m,n,i1,j1+1,j2+1,dp)); //right right
        return dp[i1][j1][j2] = score + maxi;
    }
    int cherryPickup(vector<vector<int>>& grid) {
        int m=grid.size(), n=grid[0].size();
        vector<vector<vector<int>>> dp(m+1,vector<vector<int>>(n+1,vector<int>(n+1,-1)));
        int ans = func(grid,m,n,0,0,0,dp);
        if(ans < 0){
            return 0;
        }
        return ans;
    }
};
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