Put Marbles in Bags

Problem Description

You have k bags. You are given a 0-indexed integer array weights where weights[i] is the weight of the i-th marble. You are also given the integer k. Divide the marbles into the k bags according to the following rules:

• No bag is empty.
• If the i-th marble and j-th marble are in a bag, then all marbles with an index between the i-th and j-th indices should also be in that same bag.
• If a bag consists of all the marbles with an index from i to j inclusively, then the cost of the bag is weights[i] + weights[j].

The score after distributing the marbles is the sum of the costs of all the k bags. Return the difference between the maximum and minimum scores among marble distributions.

Key Insights

• The problem requires dividing an array into k contiguous segments while minimizing and maximizing the score based on the defined cost.
• The score of a bag is determined by the weights at the boundaries of the segment, which suggests that the extreme values (smallest and largest) will influence the score significantly.
• Sorting the weights can help to easily access the smallest and largest values necessary for calculating the scores.

Space and Time Complexity

Time Complexity: O(n log n) due to the sorting of weights. Space Complexity: O(1) if we sort in place or O(n) if we consider the space used by the sorted array.

Solution

To solve the problem, we can follow these steps:

1. Sort the weights array: This allows us to easily access the smallest and largest weights.
2. Calculate the minimum score: This can be achieved by summing the smallest k-1 weights from the beginning and the largest weights from the end.
3. Calculate the maximum score: This will involve summing the largest k-1 weights from the beginning and the smallest weights from the end.
4. Return the difference between the maximum and minimum scores.

We can use a greedy approach to select the weights that will maximize or minimize the score based on their positions.

Code Solutions