#741

Hard

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Cherry Pickup

Maximize cherries collected on a grid, employing state transition dynamic programming with careful navigation across obstacles.

array dynamic programming matrix

Problem Statement

You are given a square grid, grid, where each cell contains one of three values: 1 (cherry), -1 (obstacle), or 0 (empty). The task is to determine the maximum number of cherries that can be collected by moving through the grid from the top-left corner to the bottom-right corner, and returning back to the start. During the trip, you must avoid obstacles and cannot step outside the grid.

The movement is restricted to up, down, left, or right directions. You need to design a solution that maximizes the cherry count by efficiently navigating the grid using dynamic programming, leveraging state transitions for optimal pathfinding while respecting the constraints of obstacles.

Examples

Example 1

Input: grid = [[0,1,-1],[1,0,-1],[1,1,1]]

Output: 5

The player started at (0, 0) and went down, down, right right to reach (2, 2). 4 cherries were picked up during this single trip, and the matrix becomes [[0,1,-1],[0,0,-1],[0,0,0]]. Then, the player went left, up, up, left to return home, picking up one more cherry. The total number of cherries picked up is 5, and this is the maximum possible.

Example 2

Input: grid = [[1,1,-1],[1,-1,1],[-1,1,1]]

Output: 0

Example details omitted.

Constraints

n == grid.length
n == grid[i].length
1 <= n <= 50
grid[i][j] is -1, 0, or 1.
grid[0][0] != -1
grid[n - 1][n - 1] != -1

Solution Approach

State Transition Dynamic Programming

The solution revolves around using dynamic programming (DP) to track the state of the two players as they move through the grid. Each state represents the positions of the two players, and for each state, we calculate the maximum cherries that can be collected from that configuration. The state transitions depend on the possible moves of both players, and the goal is to maximize the total cherry collection while avoiding obstacles.

Tracking Maximum Cherries Collected

We maintain a 3D DP array where each entry holds the maximum cherries collected up to a given state. The states are defined by the positions of the two players on the grid. For each move, we compute the maximum number of cherries collected based on previous states, ensuring we select valid movements and add the cherries from the grid if reachable.

Avoiding Obstacles and Boundaries

While navigating, we must carefully handle obstacles and boundaries. If either player encounters a cell with -1, the state transition for that configuration becomes invalid. Special care must be taken to ensure that the DP array is updated only for valid positions, ensuring optimal collection without violating the grid's constraints.

Complexity Analysis

Metric	Value
Time	O(N^3)
Space	O(N^2)

The time complexity of the solution is O(N^3) due to the three-dimensional DP array and the fact that we examine all combinations of player positions on the grid. The space complexity is O(N^2) because we need to store the DP table, which depends on the size of the grid, where N is the grid's dimension.

What Interviewers Usually Probe

Does the candidate clearly explain how dynamic programming is applied to track the states of both players and their transitions?
Can the candidate handle the constraints effectively, especially managing obstacles in the grid?
Is the candidate able to explain how they would optimize for both time and space complexity in this problem?

Common Pitfalls or Variants

Common pitfalls

Failing to handle edge cases, such as when obstacles block all possible paths, leading to incorrect results.
Incorrectly implementing the state transition logic, which could result in suboptimal paths or missing valid paths.
Neglecting to properly update the DP table, which may cause inefficient state transitions and affect the maximum cherry count.

Follow-up variants

What if the grid has more than two players? Can the problem be generalized for multiple agents collecting cherries?
How would you adjust the solution for non-square grids or grids of varying dimensions?
What would happen if the player could only move in one direction at a time, or if diagonal movement was allowed?

FAQ

How do I solve Cherry Pickup using dynamic programming?

You solve this by using a 3D DP array that tracks the state of two players' positions, maximizing cherries collected while avoiding obstacles.

What is the time complexity of the Cherry Pickup problem?

The time complexity is O(N^3) due to the 3D DP array and state transitions across the grid.

Can I solve Cherry Pickup without dynamic programming?

While it's possible, dynamic programming offers the most efficient solution by reducing redundant calculations and optimizing state transitions.

How does GhostInterview help with Cherry Pickup?

GhostInterview assists by providing a structured environment to practice the state transition dynamic programming approach, highlighting key issues and improving efficiency.

What are the common pitfalls in solving Cherry Pickup?

The common pitfalls include improper state transitions, handling obstacles incorrectly, and failing to update the DP table accurately for optimal results.

terminal

Solution

Solution 1: Dynamic Programming

According to the problem description, the player starts from $(0, 0)$, reaches $(n-1, n-1)$, and then returns to the starting point $(0, 0)$. We can consider the player as starting from $(0, 0)$ to $(n-1, n-1)$ twice.

class Solution:
    def cherryPickup(self, grid: List[List[int]]) -> int:
        n = len(grid)
        f = [[[-inf] * n for _ in range(n)] for _ in range((n << 1) - 1)]
        f[0][0][0] = grid[0][0]
        for k in range(1, (n << 1) - 1):
            for i1 in range(n):
                for i2 in range(n):
                    j1, j2 = k - i1, k - i2
                    if (
                        not 0 <= j1 < n
                        or not 0 <= j2 < n
                        or grid[i1][j1] == -1
                        or grid[i2][j2] == -1
                    ):
                        continue
                    t = grid[i1][j1]
                    if i1 != i2:
                        t += grid[i2][j2]
                    for x1 in range(i1 - 1, i1 + 1):
                        for x2 in range(i2 - 1, i2 + 1):
                            if x1 >= 0 and x2 >= 0:
                                f[k][i1][i2] = max(f[k][i1][i2], f[k - 1][x1][x2] + t)
        return max(0, f[-1][-1][-1])

Continue Practicing

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