What is cherry pickup problem with solution

The cherry pickup problem is a combinatorial optimization problem that involves finding the longest path through a grid of cells, where each cell may contain cherries that can be collected. The path must start and end at designated cells, and the path can only move in certain directions (usually up, down, left, or right).

In the standard version of the problem, there are two players who can move through the grid at the same time, and they can collect the cherries in each cell as they pass through it. The players are not allowed to move through the same cell at the same time, and they must return to the starting cell at the end of the game. The goal is to maximize the total number of cherries collected by both players.

Here is an example of the cherry pickup problem:

[ [0, 1, -1],

 [1, 0, -1],

 [-1, 1,  0]

]{codeBox}

In this example, the grid has three rows and three columns. The two players start at the top left cell (0,0) and can move either right or down. The numbers in the cells represent the number of cherries that can be collected by the players. A value of -1 indicates that the cell is blocked and cannot be accessed.

The optimal solution, in this case, would be for the players to both moves down, collecting a total of two cherries. They would then both move right, collecting a total of three cherries. Finally, they would both return to the starting cell, collecting a total of five cherries in total.

There are several ways to solve the cherry pickup problem, depending on the specific constraints of the problem and the desired performance characteristics. Here are a few approaches that may be useful:

1. Dynamic programming: One way to solve the cherry pickup problem is to use dynamic programming. This involves defining a recursive function that takes as input the current positions of the players and the remaining time steps and returns the maximum number of cherries that can be collected in the remaining time. The function can be implemented using a 3D array to store the results of the subproblems, and the solutions to the subproblems can be combined to obtain the solution to the original problem.

1. Breadth-first search: Another approach is to use breadth-first search (BFS) to explore the space of possible paths through the grid. This involves maintaining a queue of positions that have not yet been explored and iteratively expanding the search to new positions that are reachable from the current position. The BFS algorithm can be modified to keep track of the total number of cherries collected at each position and to return the maximum number of cherries collected when the search is complete.

1. Depth-first search: A depth-first search (DFS) approach can also be used to solve the cherry pickup problem. This involves recursively exploring the grid, starting at the current position and moving in all possible directions until a solution is found or all paths have been exhausted. The DFS algorithm can be modified to keep track of the total number of cherries collected at each position and to return the maximum number of cherries collected when the search is complete.

 Here is some example code that solves the cherry pickup problem using dynamic programming:

def cherryPickup(grid: List[List[int]]) -> int:

    n = len(grid)

    dp = [[[-1] * n for _ in range(n)] for _ in range(n)]

    dp[0][0][0] = grid[0][0]

    for t in range(1, 2 * n - 1):

        for i in range(max(0, t - (n - 1)), min(n - 1, t) + 1):

            for j in range(max(0, t - (n - 1)), min(n - 1, t) + 1):

                for pi in (i - 1, i, i + 1):

                    for pj in (j - 1, j, j + 1):

                        if pi >= 0 and pj >= 0 and dp[pi][pj][i] != -1:

                            if grid[pi][pj] == -1 or grid[i][j] == -1:

                                continue

                            dp[i][j][pi] = max(dp[i][j][pi], dp[pi][pj][i] + grid[i][j] + (i != j) * grid[pi][pj])

    return max(0, dp[n - 1][n - 1][n - 1]){codeBox}

This code uses a 3D dynamic programming array dp to store the maximum number of cherries that can be collected from the grid, given the current positions of the two players. The outer two loops (i and j) represent the positions of the two players, and the inner loop (t) represents the time step at which they are at those positions. The code then considers all possible previous positions of the players (pi and pj), and updates the maximum number of cherries that can be collected based on the number of cherries at the current positions and the number of cherries collected at the previous positions.