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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

Solutions

Solution 1: Dynamic Programming

We define $f[i][j_1][j_2]$ as the maximum number of cherries that can be picked when the two robots are at positions $j_1$ and $j_2$ in the $i$-th row. Initially, $f[0][0][n-1] = grid[0][0] + grid[0][n-1]$, and the other values are $-1$. The answer is $\max_{0 \leq j_1, j_2 < n} f[m-1][j_1][j_2]$.

Consider $f[i][j_1][j_2]$. If $j_1 \neq j_2$, then the number of cherries that the robots can pick in the $i$-th row is $grid[i][j_1] + grid[i][j_2]$. If $j_1 = j_2$, then the number of cherries that the robots can pick in the $i$-th row is $grid[i][j_1]$. We can enumerate the previous state of the two robots $f[i-1][y1][y2]$, where $y_1, y_2$ are the positions of the two robots in the $(i-1)$-th row, then $y_1 \in {j_1-1, j_1, j_1+1}$ and $y_2 \in {j_2-1, j_2, j_2+1}$. The state transition equation is as follows:

$$ f[i][j_1][j_2] = \max_{y_1 \in {j_1-1, j_1, j_1+1}, y_2 \in {j_2-1, j_2, j_2+1}} f[i-1][y_1][y_2] + \begin{cases} grid[i][j_1] + grid[i][j_2], & j_1 \neq j_2 \ grid[i][j_1], & j_1 = j_2 \end{cases} $$

Where $f[i-1][y_1][y_2]$ is ignored when it is $-1$.

The final answer is $\max_{0 \leq j_1, j_2 < n} f[m-1][j_1][j_2]$.

The time complexity is $O(m \times n^2)$, and the space complexity is $O(m \times n^2)$. Where $m$ and $n$ are the number of rows and columns of the grid, respectively.

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class Solution:
    def cherryPickup(self, grid: List[List[int]]) -> int:
        m, n = len(grid), len(grid[0])
        f = [[[-1] * n for _ in range(n)] for _ in range(m)]
        f[0][0][n - 1] = grid[0][0] + grid[0][n - 1]
        for i in range(1, m):
            for j1 in range(n):
                for j2 in range(n):
                    x = grid[i][j1] + (0 if j1 == j2 else grid[i][j2])
                    for y1 in range(j1 - 1, j1 + 2):
                        for y2 in range(j2 - 1, j2 + 2):
                            if 0 <= y1 < n and 0 <= y2 < n and f[i - 1][y1][y2] != -1:
                                f[i][j1][j2] = max(f[i][j1][j2], f[i - 1][y1][y2] + x)
        return max(f[-1][j1][j2] for j1, j2 in product(range(n), range(n)))

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class Solution {
    public int cherryPickup(int[][] grid) {
        int m = grid.length, n = grid[0].length;
        int[][][] f = new int[m][n][n];
        for (var g : f) {
            for (var h : g) {
                Arrays.fill(h, -1);
            }
        }
        f[0][0][n - 1] = grid[0][0] + grid[0][n - 1];
        for (int i = 1; i < m; ++i) {
            for (int j1 = 0; j1 < n; ++j1) {
                for (int j2 = 0; j2 < n; ++j2) {
                    int x = grid[i][j1] + (j1 == j2 ? 0 : grid[i][j2]);
                    for (int y1 = j1 - 1; y1 <= j1 + 1; ++y1) {
                        for (int y2 = j2 - 1; y2 <= j2 + 1; ++y2) {
                            if (y1 >= 0 && y1 < n && y2 >= 0 && y2 < n && f[i - 1][y1][y2] != -1) {
                                f[i][j1][j2] = Math.max(f[i][j1][j2], f[i - 1][y1][y2] + x);
                            }
                        }
                    }
                }
            }
        }
        int ans = 0;
        for (int j1 = 0; j1 < n; ++j1) {
            for (int j2 = 0; j2 < n;