<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:
    vector<vector<int>> grid;
    vector<vector<vector<int>>> dp;

    bool isValid(int r, int c) {
        return r >=0 && r < grid.size() && c >= 0 && c < grid[0].size() && grid[r][c] != -1;
    }

    int getCherries(int r1, int r2, int c1) {
        int c2 = r1 + c1 - r2;

        if(!isValid(r1, c1) || !isValid(r2, c2)) {
            return INT_MIN;
        }

        if(r1 == grid.size()-1 && c1 == grid[0].size()-1) {
            return grid[r1][c1];
        }

        if(dp[r1][r2][c1] != -1) {
            return dp[r1][r2][c1];
        }

        int result = grid[r1][c1];
        if(c1 != c2) {
            result += grid[r2][c2];
        }

        vector<int> neighbours = {
            getCherries(r1+1,r2,c1),
            getCherries(r1+1, r2+1, c1),
            getCherries(r1,r2,c1+1), 
            getCherries(r1, r2+1, c1+1)};
        int temp = *max_element(neighbours.begin(), neighbours.end());
        if(temp == INT_MIN) {
            return dp[r1][r2][c1] = temp;
        }
        return dp[r1][r2][c1] = result + temp;
    }

    int cherryPickup(vector<vector<int>>& grid) {
        this->grid = grid;
        dp = vector<vector<vector<int>>> (grid.size(), vector<vector<int>>(grid.size(), vector<int>(grid.size(), -1)));
        int r =  getCherries(0, 0, 0);
        //cout << r << endl;
        return max(0,r);
    }
};
166346