<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>> dir = {{0, 1}, {1, 0}};
    int helper(int i1, int j1, int i2, int m, int n, vector<vector<int>> &grid, vector<vector<vector<int>>> &dp){
        int j2 = (i1 + j1) - i2; // i1 + i2 = j1 + j2
        if(i1 >= m || j1 >= n || i2 >= m || j2 >= n) return -1e9;
        if(grid[i1][j1] == -1 || grid[i2][j2] == -1) return -1e9;
        if((i1 == m-1 && j1 == n-1) || (i2 == m-1 && j2 == n-1)) return grid[m-1][n-1];

        if(dp[i1][j1][i2] != -1) return dp[i1][j1][i2];
        int maxi = -1e9;
        for(auto it1: dir){
            for(auto it2: dir){
                int ret = helper(i1+it1[0], j1+it1[1], i2+it2[0], m, n, grid, dp);
                if(i1 == i2 && j1 == j2) maxi = max(maxi, grid[i1][j1] + ret);
                else maxi = max(maxi, grid[i1][j1] + grid[i2][j2] + ret);
            }
        }
        return dp[i1][j1][i2] = maxi;
    }
    int cherryPickup(vector<vector<int>>& grid) {
        int m = grid.size();
        int n = grid[0].size();
        // if(grid[0][0] == -1 || grid[m-1][n-1] == -1) return 0;
        vector<vector<vector<int>>> dp(m, vector<vector<int>>(n, vector<int>(m, -1)));
        int ans = helper(0, 0, 0, m, n, grid, dp); // Two player start from (0, 0)
        if(ans < 0) return 0;
        return ans;
    }
};
104528