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.

class Solution {
    // Take 2 persons starting from top-left together.
    // Now we can easily see that a state is defined by 4 variables:
    // R1, C1, R2, C2
    // But if they start from same position, we can say that they will be
    // at T manhattan distance from (0, 0).
    // The 2 points will follow this equation: R + C = T
    // Time Complexity -> O(N^3)
    // Space Complexity -> O(N^3)
    Integer[][][] dp;

    public int cherryPickup(int[][] grid) {
        int n = grid.length;
        dp = new Integer[n][n][2 * n];
        int ans = helper(grid, 0, 0, 0);
        return ans == Integer.MIN_VALUE ? 0 : ans;
    }

    int[][] options = { { 1, 0 }, { 0, 1 }, { 0, 0 }, { 1, 1 } };

    public int helper(int[][] grid, int R1, int R2, int T) {
        int C1 = T - R1, C2 = T - R2, n = grid.length;
        if (C1 >= n || C2 >= n || R1 >= n || R2 >= n || grid[R1][C1] == -1 || grid[R2][C2] == -1)
            return Integer.MIN_VALUE;
        // At this point it is guaranteed, that both points will be at (n-1, n-1)
        if (R1 == n - 1 && C1 == n - 1)
            return grid[R1][C1];
        if (dp[R1][R2][T] != null)
            return dp[R1][R2][T];
        int ans = Integer.MIN_VALUE;
        for (int[] option : options)
            ans = Math.max(ans, helper(grid, R1 + option[0], R2 + option[1], T + 1));
        if (ans != Integer.MIN_VALUE) {
            ans += grid[R1][C1];
            if (R1 != R2 || C1 != C2)
                ans += grid[R2][C2];
        }
        dp[R1][R2][T] = ans;
        return ans;
    }
}