# 1463. Cherry Pickup II

#### Hard

***

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:**

![](https://assets.leetcode.com/uploads/2020/04/29/sample_1_1802.png)

```
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:**

![](https://assets.leetcode.com/uploads/2020/04/23/sample_2_1802.png)

```
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`

```python
class Solution:
    def cherryPickup(self, grid: List[List[int]]) -> int:
        visited = {}
        return max(self.dfs(grid, visited, 0, 0, 0, len(grid[0])-1), 0)
    
    def dfs(self, grid, visited, row1, col1, row2, col2):
        if (row1,col1,row2,col2) in visited:
            return visited[(row1,col1,row2,col2)]
        rows = len(grid)
        cols = len(grid[0])
        # Robots going outside of grid
        if col1 in (-1, cols) or col2 in (-1, cols):
            return -float("inf")
        # Both Reached End
        if row1 == rows and row2 == rows:
            return 0
        # Robot One Moves Below Left Position
        dfs1 = self.dfs(grid, visited, row1+1, col1-1, row2+1, col2-1)
        dfs2 = self.dfs(grid, visited, row1+1, col1-1, row2+1, col2)
        dfs3 = self.dfs(grid, visited, row1+1, col1-1, row2+1, col2+1)
        # Robot One Moves Below Only
        dfs4 = self.dfs(grid, visited, row1+1, col1, row2+1, col2-1)
        dfs5 = self.dfs(grid, visited, row1+1, col1, row2+1, col2+1)
        dfs6 = self.dfs(grid, visited, row1+1, col1, row2+1, col2)
        #Robot One Moves Below Right Position
        dfs7 = self.dfs(grid, visited, row1+1, col1+1, row2+1, col2-1)
        dfs8 = self.dfs(grid, visited, row1+1, col1+1, row2+1, col2+1)
        dfs9 = self.dfs(grid, visited, row1+1, col1+1, row2+1, col2)
        # Get Max of all possible movement
        result = max(dfs1, dfs2, dfs3, dfs4, dfs5, dfs6, dfs7, dfs8, dfs9)
        
        # Both Robot are on same grid spot
        if row1 == row2 and col1 == col2:
            result += grid[row1][col1]
        else:
            result += grid[row1][col1] + grid[row2][col2]
        visited[(row1,col1, row2,col2)] = result
        return result
        
```