README.md

1463. Cherry Pickup II

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

Solutions (Rust)

1. Solution

impl Solution {
    pub fn cherry_pickup(grid: Vec<Vec<i32>>) -> i32 {
        let (rows, cols) = (grid.len(), grid[0].len());
        let mut curr_row = vec![vec![None; cols]; cols];
        curr_row[0][cols - 1] = Some(grid[0][0] + grid[0][cols - 1]);

        for i in 0..rows - 1 {
            let mut next_row = vec![vec![None; cols]; cols];

            for j0 in 0..cols {
                for j1 in 0..cols {
                    if let Some(x) = curr_row[j0][j1] {
                        for c0 in j0.saturating_sub(1)..cols.min(j0 + 2) {
                            for c1 in j1.saturating_sub(1)..cols.min(j1 + 2) {
                                next_row[c0][c1] = Some(next_row[c0][c1].unwrap_or(0).max(
                                    x + grid[i + 1][c0] + grid[i + 1][c1] * (c0 != c1) as i32,
                                ));
                            }
                        }
                    }
                }
            }

            curr_row = next_row;
        }

        curr_row
            .iter()
            .flatten()
            .map(|x| x.unwrap_or(0))
            .max()
            .unwrap()
    }
}