Cherry Pickup II

Given a rows x cols matrix grid representing a field of cherries. Each cell in grid represents the number of cherries that you can collect.

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) of the grid.

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 is passing through a cell, It picks it up all cherries, and the cell becomes an empty cell (0).

• When both robots stay on the same cell, only one of them takes the cherries.

• Both robots cannot move outside of the grid at any moment.

• Both robots should reach the bottom row in the 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.

Example 3:

Input: grid = [[1,0,0,3],[0,0,0,3],[0,0,3,3],[9,0,3,3]]
Output: 22

Example 4:

Input: grid = [[1,1],[1,1]]
Output: 4

Constraints:

• rows == grid.length

• cols == grid[i].length

• 2 <= rows, cols <= 70

• 0 <= grid[i][j] <= 100

class Solution {
    // This is dp solution. At each step, we move both robot1 and robot2 to next
    // row, and with all possibles columns (j-1, j, j+1). Please keep in mind that
    // if 2 robots go the same cell, we only collect cherries once.
    public int cherryPickup(int[][] grid) {
        int m = grid.length, n = grid[0].length;
        Integer[][][] dp = new Integer[m][n][n];
        // dp[row][c1][c2] -> max cherries with both robots at row and c1 and c2 col
        // respectively
        return dfs(grid, m, n, 0, 0, n - 1, dp);
    }

    int dfs(int[][] grid, int m, int n, int row, int c1, int c2, Integer[][][] dp) {
        if (row == m)
            return 0; // Reach to bottom row
        if (dp[row][c1][c2] != null)
            return dp[row][c1][c2];
        int ans = 0;
        for (int i = -1; i <= 1; i++) {
            for (int j = -1; j <= 1; j++) {
                int nc1 = c1 + i, nc2 = c2 + j;
                if (nc1 >= 0 && nc1 < n && nc2 >= 0 && nc2 < n) {
                    ans = Math.max(ans, dfs(grid, m, n, row + 1, nc1, nc2, dp));
                }
            }
        }
        int cherries = c1 == c2 ? grid[row][c1] : grid[row][c1] + grid[row][c2];
        return dp[row][c1][c2] = ans + cherries;
    }
}