<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 {
private:
    int N, M;
    int dp[2][100][100];
public:
    bool is_valid(int i, int ind) {
        return 0 <= i - ind and i - ind < N and 0 <= ind and ind < M;
    }

    int cherryPickup(vector<vector<int>>& grid) {
        N = grid.size(), M = grid[0].size();
        int prev = 1, cur = 0;
        dp[cur][0][0] = grid[0][0];

        for(int i=1; i<=N+M-2; i++) {
            swap(cur, prev);

            for(int i1=0;i1<=i;i1++)
            for(int j1=0;j1<=i;j1++)
            dp[cur][i1][j1] = -1;

            for(int i1=0;i1<i;i1++)
            for(int j1=0;j1<i;j1++) 
            if(dp[prev][i1][j1] != -1) {
                dp[cur][i1][j1] = max(dp[cur][i1][j1], dp[prev][i1][j1]);
                dp[cur][i1+1][j1] = max(dp[cur][i1+1][j1], dp[prev][i1][j1]);
                dp[cur][i1][j1+1] = max(dp[cur][i1][j1+1], dp[prev][i1][j1]);
                dp[cur][i1+1][j1+1] = max(dp[cur][i1+1][j1+1], dp[prev][i1][j1]);
            }

            for(int i1=0;i1<=i;i1++)
            for(int j1=0;j1<=i;j1++)
            if(dp[cur][i1][j1] == -1 or !is_valid(i, i1) or !is_valid(i, j1) or grid[i - i1][i1] == -1 or grid[i - j1][j1] == -1)
            dp[cur][i1][j1] = -1;
            else
            dp[cur][i1][j1] += grid[i - i1][i1] + (i1!=j1? grid[i - j1][j1]: 0);
        }
        return max(0, dp[cur][M - 1][M - 1]);
    }
};
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