741. Cherry Pickup

741. Cherry Pickup

In a N x N grid representing a field of cherries, each cell is one of three possible integers.

• 0 means the cell is empty, so you can pass through;
• 1 means the cell contains a cherry, that you can pick up and pass through;
• -1 means the cell contains a thorn that blocks your way.

Your task is to collect maximum number of cherries possible by following the rules below:

• Starting at the position (0, 0) and reaching (N-1, N-1) by moving right or down through valid path cells (cells with value 0 or 1);
• After reaching (N-1, N-1), returning to (0, 0) by moving left or up through valid path cells;
• When passing through a path cell containing a cherry, you pick it up and the cell becomes an empty cell (0);
• If there is no valid path between (0, 0) and (N-1, N-1), then no cherries can be collected.

Example 1:

Input: grid =
[[0, 1, -1],
 [1, 0, -1],
 [1, 1,  1]]
Output: 5
Explanation: 
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.

Note:

• grid is an N by N 2D array, with 1 <= N <= 50.
• Each grid[i][j] is an integer in the set {-1, 0, 1}.
• It is guaranteed that grid[0][0] and grid[N-1][N-1] are not -1.
• 
  

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Solution #1, ref:https://zxi.mytechroad.com/blog/dynamic-programming/leetcode-741-cherry-pickup/
其实题意就是找2条从左上到右下的路径和是最大的。路径可以重叠，但是重叠部分的cherry只能计算一次。
方法是假设有2个人同时从右下出发往右上走，因为y2 = x1 + y1 - x2, 所以我们可以用(x1,y1,x2) 3个参数就能定义一个状态

class Solution {
    public int cherryPickup(int[][] grid) {
        int n = grid.length;
        int[][][] mem = new int[n][n][n];
        for(int i = 0; i < n; i++){
            for(int j = 0; j < n; j++){
                Arrays.fill(mem[i][j], Integer.MIN_VALUE);
            }
        }
        
        return Math.max(0, rec(grid, mem, n - 1, n - 1, n - 1));
    }
    
    private int rec(int[][] grid, int[][][] mem, int x1, int y1, int x2) {
        int y2 = x1 + y1 - x2;
        if (x1 < 0 || y1 < 0 || x2 < 0 || y2 < 0) return -1;
        if (grid[x1][y1] == -1 || grid[x2][y2] == -1) return -1;
        
        if (mem[x1][y1][x2] != Integer.MIN_VALUE) return mem[x1][y1][x2];
        if (x1 == 0 && y1 == 0) return grid[y1][x1];
        
        mem[x1][y1][x2] = Math.max(rec(grid,mem,x1 - 1,y1,x2 - 1), 
                             Math.max(rec(grid,mem,x1,y1 - 1,x2 - 1), 
                                      Math.max(rec(grid,mem,x1,y1 - 1,x2), rec(grid,mem,x1 - 1,y1,x2))));
        if (mem[x1][y1][x2] >= 0) {
            mem[x1][y1][x2] += grid[x1][y1];
            if (x1 != x2) mem[x1][y1][x2] += grid[x2][y2];
        }
        
        return mem[x1][y1][x2];
    } 
}

ToDo
2D空间
https://leetcode.com/problems/cherry-pickup/discuss/109903/Step-by-step-guidance-of-the-O(N3)-time-and-O(N2)-space-solution