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1463. Cherry Pickup II

Difficulty Topics

Description

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

cherry-pickup-ii.py
class Solution:
    def cherryPickup(self, grid: List[List[int]]) -> int:
        rows, cols = len(grid), len(grid[0])

        @cache
        def go(row, col1, col2):
            if row == rows: return 0

            cherries = grid[row][col1] if col1 == col2 else grid[row][col1] + grid[row][col2]
            res = 0

            for nc1 in range(col1 - 1, col1 + 2):
                for nc2 in range(col2 - 1, col2 + 2):
                    if 0 <= nc1 < cols and 0 <= nc2 < cols:
                        res = max(res, go(row + 1, nc1, nc2))

            return res + cherries

        return go(0, 0, cols - 1)