#3363

Hard

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Find the Maximum Number of Fruits Collected

Maximize the number of fruits collected by three children navigating a grid dungeon with dynamic programming.

Problem Statement

In this problem, three children must navigate an n x n grid dungeon filled with rooms containing fruits. Each child starts at one of the three corner rooms: (0, 0), (0, n-1), and (n-1, 0). The objective is for the children to move through the dungeon and collect as many fruits as possible, making exactly n-1 moves. The goal is to reach the room at position (n-1, n-1), while collecting fruits along the way.

The fruits array is a 2D array where fruits[i][j] represents the number of fruits in the room at position (i, j). The challenge lies in efficiently calculating the maximum number of fruits the children can collect, considering the state transitions between rooms and the movement restrictions of the children. The solution requires a dynamic programming approach to manage these transitions and maximize the total fruit collection.

Examples

Example 1

Input: fruits = [[1,2,3,4],[5,6,8,7],[9,10,11,12],[13,14,15,16]]

Output: 100

In this example: In total they collect 1 + 6 + 11 + 16 + 4 + 8 + 12 + 13 + 14 + 15 = 100 fruits.

Example 2

Input: fruits = [[1,1],[1,1]]

Output: 4

In this example: In total they collect 1 + 1 + 1 + 1 = 4 fruits.

Constraints

2 <= n == fruits.length == fruits[i].length <= 1000
0 <= fruits[i][j] <= 1000

Solution Approach

State Transition Dynamic Programming

Use dynamic programming to track the maximum number of fruits collected at each position in the dungeon. The state represents the positions of all three children, and the goal is to transition between states by considering valid moves. At each step, update the fruit collection count and move to the next possible position while maximizing the total.

Grid Traversal Optimization

To avoid redundant calculations, optimize the traversal by focusing on the paths that yield the highest fruit collections. The solution uses a bottom-up approach, calculating the number of fruits collected from the destination to the starting points to ensure efficient state transitions between rooms.

Handling Multiple Children

Since there are three children starting from different corners, the solution needs to account for all possible movements of the children. This requires managing the state of each child’s position separately and ensuring the moves do not overlap, while still maximizing the total fruit collection.

Complexity Analysis

Metric	Value
Time	O(n^2)
Space	O(n)

The time complexity is O(n^2), as the algorithm must traverse the grid and calculate state transitions for each child. The space complexity is O(n) because the dynamic programming state only needs to store the current positions and the accumulated fruits for each child at each step.

What Interviewers Usually Probe

Look for candidates who can efficiently handle state transitions and understand the dynamic programming pattern.
Candidates should focus on optimizing space complexity while managing the three-child scenario in the grid.
Pay attention to how candidates explain the state representation and the grid traversal approach.

Common Pitfalls or Variants

Common pitfalls

Candidates may overlook the optimization of state transitions and end up with a less efficient solution.
A common mistake is not managing the three children's movement separately, leading to incorrect fruit calculations.
Some may attempt a brute force solution that leads to time or space complexity issues, especially with larger grids.

Follow-up variants

Reduce the number of children or increase the grid size to test the scalability of the solution.
Introduce obstacles or empty rooms to modify the grid and test the adaptability of the algorithm.
Allow the children to make fewer moves and explore how the solution changes with limited movement options.

FAQ

What is the main pattern for solving the 'Find the Maximum Number of Fruits Collected' problem?

The main pattern is state transition dynamic programming, where the state represents the positions of the three children and the total fruits collected at each step.

How do the children move in the 'Find the Maximum Number of Fruits Collected' problem?

The children move through the grid by making exactly n-1 moves from their starting positions at the corners to the bottom-right room.

What is the time complexity of the 'Find the Maximum Number of Fruits Collected' problem?

The time complexity is O(n^2), as it involves traversing an n x n grid while optimizing the fruit collection path.

What is the space complexity of the 'Find the Maximum Number of Fruits Collected' problem?

The space complexity is O(n) due to the dynamic programming approach, which only needs to store the current state of the children's positions.

What are the common mistakes in solving the 'Find the Maximum Number of Fruits Collected' problem?

Common mistakes include inefficient state transitions, incorrect handling of children's movements, and using brute force solutions that lead to performance issues.

terminal

Solution

Solution 1: Dynamic Programming

According to the problem description, for the child starting from $(0, 0)$ to reach $(n - 1, n - 1)$ in exactly $n - 1$ steps, they can only move through the rooms on the main diagonal $(i, i)$, where $i = 0, 1, \ldots, n - 1$. The child starting from $(0, n - 1)$ can only move through rooms above the main diagonal, while the child starting from $(n - 1, 0)$ can only move through rooms below the main diagonal. This means that except for reaching the destination at $(n - 1, n - 1)$, no other rooms will be visited by multiple children.

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class Solution:
    def maxCollectedFruits(self, fruits: List[List[int]]) -> int:
        n = len(fruits)
        f = [[-inf] * n for _ in range(n)]
        f[0][n - 1] = fruits[0][n - 1]
        for i in range(1, n):
            for j in range(i + 1, n):
                f[i][j] = max(f[i - 1][j], f[i - 1][j - 1]) + fruits[i][j]
                if j + 1 < n:
                    f[i][j] = max(f[i][j], f[i - 1][j + 1] + fruits[i][j])
        f[n - 1][0] = fruits[n - 1][0]
        for j in range(1, n):
            for i in range(j + 1, n):
                f[i][j] = max(f[i][j - 1], f[i - 1][j - 1]) + fruits[i][j]
                if i + 1 < n:
                    f[i][j] = max(f[i][j], f[i + 1][j - 1] + fruits[i][j])
        return sum(fruits[i][i] for i in range(n)) + f[n - 2][n - 1] + f[n - 1][n - 2]

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