1463. Cherry Pickup II

Problem Description

In this problem, we're given a 2D grid representing a field of cherries, where each cell contains a certain number of cherries. We have two robots with the goal to collect as many cherries as possible. One robot starts at the top-left corner of the grid, and the other starts at the top-right corner. The objective is to move both robots to the bottom row, collecting cherries along the way, and to maximize the total number of cherries collected by both robots combined.

The robots can only move to the cell directly below them or diagonally to the left or right below them. This means from any cell (i, j), the next move can be to cells (i+1, j-1), (i+1, j), or (i+1, j+1). Additionally, if both robots end up on the same cell, only one of them can collect the cherries there, preventing double-counting.

The challenge is to find the path for both robots such that the total amount of cherries collected is maximized, considering they must follow the grid paths and cannot move outside the grid boundaries.

Intuition

To find the maximum number of cherries both robots can collect, we can use dynamic programming. This approach typically involves breaking down the problem into smaller subproblems that we solve and combine to find the final solution.

Here, we define a three-dimensional dynamic programming array dp[i][j1][j2] where i represents the current row, and j1 and j2 represent the columns of Robot 1 and Robot 2, respectively. The value stored in dp[i][j1][j2] is the maximum number of cherries both robots can collect starting from row i to the bottom of the grid, with Robot 1 at column j1 and Robot 2 at column j2.

Since the robots can move to three different cells in the next row from their current position, we will consider all possible combinations of movements for both robots. We'll take the maximum of the valid moves, ensuring we avoid cells outside the grid boundaries.

The solution also considers that if both robots end up in the same cell, we only add the cherries from that cell once. After considering all possible movements for each row, the algorithm gradually builds up the solution, ending with the maximum count of cherries when the robots reach the bottom row.

The optimal value will be the maximum value found in dp[m-1][j1][j2], which represents the last row of the grid for any column positions j1 and j2 of the two robots.

Learn more about Dynamic Programming patterns.

Solution Approach

The solution provided revolves around the principle of dynamic programming, where the problem's complexity is reduced by breaking it down into simpler subproblems and caching intermediate results for future reference.

Here, we follow these steps to implement the solution:

1. Initialization: We create two 2D arrays f and g with dimensions n x n where n is the number of columns in the grid. They will hold the current and next state's maximum cherry counts respectively. f[0][n - 1] is initialized with the sum of cherries from the starting positions of both robots (top-left and top-right).

2. Iteration: We iterate through each row starting from the second row, as the first row is already covered by our initialization. During each iteration through rows i from 1 to m - 1 (where m is the number of rows), we iterate through all columns j1 and j2 for Robot 1 and Robot 2's positions, respectively.

3. Cherry Collection (Nested Loop): For every possible position (i, j1, j2) of the robots, we calculate the cherry count x that can be collected on row i considering both robots' positions, taking care not to double-count if they share the same cell.

4. Transitions (Nested Loop within Nested Loop): Within the same loop, we check all possible previous positions (i-1, y1, y2) where the robots could have moved from. This involves looking at the rows right above and considering three possible positions for each robot.

5. State Update: We update our next state g[j1][j2] with the maximum of the current value in g[j1][j2] and the sum of f[y1][y2] and x. This represents the maximum cherries from the previous step plus the current step's cherries.

6. Transition to Next State: Once we've computed the values for all possible positions of g, we swap f and g so that f holds our next state's values when the next row iteration begins and g is reset to be filled again.

7. Final Result: Upon completing the row iterations, we have the maximum number of cherries that can be collected stored in the f array. To find the maximum, we iterate through f using product(range(n), range(n)) to produce all combinations of j1 and j2 locations and finding the maximum entry.

It's important to understand that dynamic programming is efficient here because we avoid recomputing maximum cherry counts for each subproblem, which would be the case in a naive recursive implementation. By storing these values in the 2D arrays, we only compute each subproblem once.

We use Python's tuple assignment to efficiently swap f and g arrays. The product function from itertools module is used to generate cartesian product of the column indices for the final result calculation. The algorithm's overall time complexity is O(mn^29), considering the size of the grid and the number of possible movements.

Example Walkthrough

To illustrate the solution approach, let's consider a small example of a grid with 3 rows and 3 columns with the following number of cherries:

1grid = [
2  [3, 1, 1],
3  [2, 5, 1],
4  [1, 5, 5]
5]

Robot 1 (R1) starts at the top left (cell with 3 cherries), and Robot 2 (R2) starts at the top right (cell with 1 cherry).

Initial State:

• Initialize f and g with dimensions 3 x 3 (since we have 3 columns), and set f[0][2] with 4 (sum of cherries at R1's and R2's starting positions).

First Iteration (i=1, second row):

• Consider all combinations (j1, j2), where j1 and j2 are the column indices for R1 and R2, respectively.
• R1 can move to positions (1,0), (1,1), and (1,2), and R2 can move to positions (1,1), (1,2), and (1,3). Note: (1,3) is invalid due to grid boundaries.
• For each (j1, j2), collect cherries x without double-counting if on the same cell and update g[j1][j2] with the maximum cherry count considering the previous positions (0, y1, y2).

Second Iteration (i=2, third row):

• Similar to the first iteration, we perform the nested loops and check all combinations of R1 and R2 moves from row 2 to row 3.
• Cherry collection and state update steps are similar but now for the third row.

Transitions and Final Result:

• After iteration through all rows, swap f and g to prepare for the next row or to conclude with the final result.
• The final result is the maximum number out of all entries in f to find the maximum cherries that both robots can collect.

Example Iteration Details:

1. After the first step, f array has f[0][2] = 4 since both robots start at opposite ends of the first row, 3+1 cherries collected.
2. For the next iteration (second row), let's assume j1 = 1 and j2 = 1 (both robots are at the center). We collected 5 cherries since they both end up in the same cell.
   • When coming from f, check all possible y1, y2 to ensure we're adding the max previous value of cherries, so f[0][2] + cherries collected (5) for these positions and update g[1][1] accordingly.
   • After considering all combinations, g may look like this:

     1[0, 3, 0]
     2[7, 0, 0]
     3[0, 0, 0]

3. After iterating through all rows, the robots will have collected a maximum number of cherries. In the final iteration, robots could end up at f[2][0] and f[2][2] with the maximum sums obtained.
   • f[2][0] could have a sum of cherries collected from the first robot and f[2][2] for the second robot.
   • Find the maximum entry in f to get the result.

Through this method, we have avoided double counting, and have systematically considered each possibility to ensure that the total number of cherries collected is maximized by the end when both robots have reached the bottom row. The result we'd obtain from the f array would give us the max cherries collected.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the algorithm is determined by the number of loops and the operations that occur inside them. There are three nested loops with respect to m, n, j1, and j2, resulting in a time complexity of O(m * n^2). Within the nested loops for j1 and j2, there are two more loops for y1 and y2. Each of these run through at most 3 iterations, which contributes a constant factor, not changing the asymptotic behavior. Therefore, the overall time complexity remains O(m * n^4).

Space Complexity

The space complexity of the algorithm comes from the f and g 2D arrays that are used to store intermediate results. Both arrays have dimensions n x n, so each array requires O(n^2) space. Since there are two such arrays, you might initially think that this doubles the space requirement, but because they are swapped (f, g = g, f), the total space complexity remains O(n^2) as no additional space is required for the swapping operation.

Learn more about how to find time and space complexity quickly using problem constraints.