Cherry Pickup II

class Solution {

    public int collectCherries(int[][] field) {
        int rows = field.length;
        int cols = field[0].length;
        int cache[][][] = new int[rows][cols][cols];

        for (int r = rows - 1; r >= 0; r--) {
            for (int c1 = 0; c1 < cols; c1++) {
                for (int c2 = 0; c2 < cols; c2++) {
                    int sum = 0;
                    // Collect cherries from current positions
                    sum += field[r][c1];
                    if (c1 != c2) {
                        sum += field[r][c2];
                    }
                    // Calculate maximum possible from next rows
                    if (r != rows - 1) {
                        int maxVal = 0;
                        for (int nextC1 = c1 - 1; nextC1 <= c1 + 1; nextC1++) {
                            for (int nextC2 = c2 - 1; nextC2 <= c2 + 1; nextC2++) {
                                if (nextC1 >= 0 && nextC1 < cols && nextC2 >= 0 && nextC2 < cols) {
                                    maxVal = Math.max(maxVal, cache[r + 1][nextC1][nextC2]);
                                }
                            }
                        }
                        sum += maxVal;
                    }
                    cache[r][c1][c2] = sum;
                }
            }
        }
        return cache[0][0][cols - 1];
    }
}

The provided Java solution for the Cherry Pickup II problem uses dynamic programming to determine the maximum number of cherries that can be collected by two robots moving on a grid. The method collectCherries is designed to handle a grid (int[][] field) where rows represent the different positions from top to bottom of the grid and columns represent the positions from left to right.

Key concepts of the solution:

• Initialize a 3D array cache to store maximum cherries collected up to each position for robot1 at column c1 and robot2 at column c2 for every row r.
• Iterate from the bottom of the grid (rows - 1) to the top (0), considering all possible positions of the two robots (c1 and c2 for columns).
• For each position combination of c1 and c2, compute the sum of cherries collected by both robots. Robot located at the same column collects cherries only once.
• If not at the last row, update the sum by adding the maximum possible cherries collected from the next row for any valid move of robot1 and robot2 (both can move to their current, left, or right adjacent columns if within grid boundary).
• Update the cache array with the calculated cherry sum for that position pair.
• The result, maximum cherries collected starting from the top row with robots starting at the first and last column, is stored in cache[0][0][cols - 1].

Implementation Steps:

1. Initialize the cache 3D array to store intermediary results.
2. Use nested loops to iterate over each row from bottom to top and each possible pair of column positions for the two robots.
3. Inside the innermost loop, calculate the initial sum of cherries from the current column positions of robots, taking care not to double-count if both robots are in the same column.
4. If not in the last row, loop through all possible next positions of both robots and update the sum with the maximum collected cherries from valid positions.
5. Store the maximum collectible cherries for each combination of positions in the cache array.
6. Return the maximum cherries collectible with starting positions at the first and last columns of the top row.

This approach ensures time-efficient cherry collection computation for each possible robot position as they traverse the grid, making optimal use of dynamic programming to avoid redundant calculations.

class Solution:
    def collectCherries(self, matrix: List[List[int]]) -> int:
        rows = len(matrix)
        cols = len(matrix[0])
        memo = [[[0]*cols for _ in range(cols)] for __ in range(rows)]

        for level in reversed(range(rows)):
            for first in range(cols):
                for second in range(cols):
                    cherries = 0
                    # pick cherries from current position
                    cherries += matrix[level][first]
                    if first != second:
                        cherries += matrix[level][second]
                    # transition to the next row
                    if level != rows-1:
                        cherries += max(memo[level+1][next_first][next_second]
                                        for next_first in [first, first+1, first-1]
                                        for next_second in [second, second+1, second-1]
                                        if 0 <= next_first < cols and 0 <= next_second < cols)
                    memo[level][first][second] = cherries
        return memo[0][0][cols-1]

Cherry Pickup II is a problem where you aim to maximize the number of cherries collected by two robots moving through a grid from the top to the bottom row. Each cell in the grid contains a certain number of cherries that the robots can collect. The objective is achieved using dynamic programming to store and compute maximum cherries that can be collected at each step, with both robots starting from different positions.

In the provided Python3 solution:

• Start by defining a function collectCherries in the Solution class that takes a 2D list matrix as input.
• Determine the number of rows and columns in the matrix.
• Initialize a 3D list memo to store the cherry picks for every position and state of the robots.
• Begin from the last row and iterate upwards, considering all possible positions of the two robots (axes first for the first robot and second for the second robot).
• For each position, compute the number of cherries collected if both robots collect cherries from their respective positions (avoid double counting if both robots end up in the same column).
• Additionally, factor in the cherries collected from potential positions in the next row where the robots could move: directly below, one step to the left, or one step to the right, ensuring the positions are within the grid bounds.
• Return the maximum cherries collected starting from the first row, first and last columns as the initial positions of the two robots.

This approach ensures that by the time you compute the values for the top row in memo, you have already considered all possible paths and accumulated cherry counts for the robots moving through all rows of the grid. This method returns the maximum number of cherries that can be collected by both robots when they traverse the grid optimally from top to bottom.