1463. Cherry Pickup II

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

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Solutions

Solution 1

Intuition

store max cherries in each row

Code

class Solution {
    private static final int[] DIRS = { -1, 0, 1 };

    public int cherryPickup(int[][] grid) {
        int rows = grid.length, cols = grid[0].length;
        int[][][] dp = new int[rows][cols][cols];
        dp[0][0][cols - 1] = grid[0][0] + grid[0][cols - 1];
        int res = 0;

        // from row 1 to bottom
        for (int row = 1; row < rows; row++) {

            // robot#1 from left top \
            for (int col1 = 0; col1 < cols && col1 <= row; col1++) {
                // robot#2 form right top /
                for (int col2 = cols - 1; col2 > col1 && col2 >= (cols - 1 - row); col2--) {

                    int prevCherries = 0;
                    for (int dir1 : DIRS) {
                        for (int dir2 : DIRS) {
                            int prevRow = row - 1, prevCol1 = col1 + dir1, prevCol2 = col2 + dir2;
                            if (prevCol1 >= 0 && prevCol1 < cols && prevCol2 >= 0 && prevCol2 < cols
                                    && prevCol1 != prevCol2) {
                                prevCherries = Math.max(prevCherries, dp[prevRow][prevCol1][prevCol2]);
                            }

                        }
                    }
                    int cherries = grid[row][col1] + grid[row][col2];
                    dp[row][col1][col2] = cherries + prevCherries;
                    res = Math.max(res, dp[row][col1][col2]);
                }

            }

        }
        return res;
    }
}