Problem Statement:

You are given a rows x cols matrix grid representing a field of cherries where grid[i][j] represents the number of cherries that you can collect from the (i, j) cell.

You have two robots that can collect cherries for you:

Robot #1 is located at the top-left corner (0, 0), and
Robot #2 is located at the top-right corner (0, cols - 1).

Return the maximum number of cherries collection using both robots by following the rules below:

From a cell (i, j), robots can move to cell (i + 1, j - 1), (i + 1, j), or (i + 1, j + 1).
When any robot passes through a cell, It picks up all cherries, and the cell becomes an empty cell.
When both robots stay in the same cell, only one takes the cherries.
Both robots cannot move outside of the grid at any moment.
Both robots should reach the bottom row in grid.

Example 1:

Input: grid = [[3,1,1],[2,5,1],[1,5,5],[2,1,1]]
Output: 24
Explanation: Path of robot #1 and #2 are described in color green and blue respectively.
Cherries taken by Robot #1, (3 + 2 + 5 + 2) = 12.
Cherries taken by Robot #2, (1 + 5 + 5 + 1) = 12.
Total of cherries: 12 + 12 = 24.

Example 2:

Input: grid = [[1,0,0,0,0,0,1],[2,0,0,0,0,3,0],[2,0,9,0,0,0,0],[0,3,0,5,4,0,0],[1,0,2,3,0,0,6]]
Output: 28
Explanation: Path of robot #1 and #2 are described in color green and blue respectively.
Cherries taken by Robot #1, (1 + 9 + 5 + 2) = 17.
Cherries taken by Robot #2, (1 + 3 + 4 + 3) = 11.
Total of cherries: 17 + 11 = 28.

Constraints:

rows == grid.length
cols == grid[i].length
2 <= rows, cols <= 70
0 <= grid[i][j] <= 100

Solution:

I see many solutions using 3D DP but it is not necessary. We can simplify the logic to 2D by storing only result of one row.

At any point while traversing from bottom to top, let us store the answer in a 2-D DP table A of size (cols,cols). A[i][j] denotes the position i for robot 1 and j for robot 2.

Further let us only consider situations where robot 2 is not to the left of robot 1 ie we want j >= i.

 
class Solution {
public:
    int cherryPickup(vector<vector<int>>& grid) 
    {
        int rows = grid.size(), cols = grid[0].size();
        vector<vector<int>> A(cols, vector<int>(cols));
        for(int i=0; i<cols; i++) for(int j=i; j<cols; j++)
            A[i][j] = grid[rows-1][i] + (i==j ? 0 : grid[rows-1][j]);
        for(int row=rows-2; row>=0; row--)
        {
            vector<vector<int>> temp(cols, vector<int>(cols));
            for(int i=0; i<cols; i++) for(int j=i; j<cols; j++)
            {
                int a = grid[row][i], b = grid[row][j];
                int cur = a + (i==j ? 0 : b);
                int nextRow = INT_MIN;
                for(int ii=max(i-1,0); ii<=min(i+1,cols-1); ii++)
                for(int jj=max(j-1,0); jj<=min(j+1,cols-1); jj++)
                    nextRow = max(nextRow, A[ii][jj]);
                temp[i][j] = nextRow + cur;
            }
            A = temp;
        }
        return A[0][cols-1];
    }
};