LeetCode-in-Java

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:

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

public class Solution {
    public int cherryPickup(int[][] grid) {
        int m = grid.length;
        int n = grid[0].length;
        int[][][] dp = new int[n][n][m];
        dp[0][n - 1][0] = grid[0][0] + grid[0][n - 1];
        for (int k = 1; k < m; k++) {
            for (int i = 0; i <= Math.min(n - 1, k); i++) {
                for (int j = n - 1; j >= Math.max(0, n - 1 - k); j--) {
                    dp[i][j][k] = maxOfLast(dp, i, j, k) + grid[k][i] + ((i == j) ? 0 : grid[k][j]);
                }
            }
        }
        int result = 0;
        for (int i = 0; i <= Math.min(n - 1, m); i++) {
            for (int j = n - 1; j >= Math.max(0, n - 1 - m); j--) {
                result = Math.max(result, dp[i][j][m - 1]);
            }
        }
        return result;
    }

    private int maxOfLast(int[][][] dp, int i, int j, int k) {
        int result = 0;
        for (int x = -1; x <= 1; x++) {
            for (int y = -1; y <= 1; y++) {
                int r = i + x;
                int c = j + y;
                if (r >= 0 && r < dp[0].length && c >= 0 && c < dp[0].length) {
                    result = Math.max(result, dp[r][c][k - 1]);
                }
            }
        }
        return result;
    }
}

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