2551. Put Marbles in Bags - Detailed Explanation

Problem Statement

You are given an integer array weights where weights[i] is the weight of the i-th marble. You need to split all the marbles into exactly m bags. Each bag must contain a contiguous sequence of marbles. The cost of a bag is defined as the sum of the weight of the first marble and the weight of the last marble in that bag. The total cost of a distribution is the sum of the costs of all m bags.

Formally, if you split the array into m contiguous subarrays (bags), then the cost is computed as:

  cost = (first marble of bag 1 + last marble of bag 1) + (first marble of bag 2 + last marble of bag 2) + … + (first marble of bag m + last marble of bag m)

Notice that the first marble of the first bag is always weights[0] and the last marble of the last bag is always weights[n-1] (where n is the length of weights). For any split between marbles, the two marbles adjacent to the split both contribute to the total cost (one as the last marble of one bag and one as the first marble of the next bag).

Your goal is to compute the difference between the maximum total cost and the minimum total cost over all valid ways to split the marbles into m bags.

Hints

1. Observing the Fixed Base Cost:
   Since every valid split always uses weights[0] as the first marble of the first bag and weights[n-1] as the last marble of the last bag, the term weights[0] + weights[n-1] is common to every distribution.

2. Role of Splits:
   When you split the array into m bags, you make exactly m - 1 splits. For a split between indices i and i+1, the contribution to the cost is weights[i] + weights[i+1]. These contributions are “additive” in the sense that the total cost can be viewed as a fixed base cost plus the sum of contributions from the m - 1 splits.

3. Optimization by Sorting:
   To minimize the total cost, you want to choose the m - 1 splits with the smallest contributions. Conversely, to maximize the total cost, choose the m - 1 splits with the largest contributions. The answer is the difference between these two sums.

Approach

1. Compute the Base Cost:
   The base cost is the sum of the first marble in the first bag and the last marble in the last bag: [ \text{base} = \text{weights}[0] + \text{weights}[n-1] ] This part is fixed regardless of how you split the array.

2. Calculate Candidate Split Contributions:
   For each possible split between marbles (i.e. between index (i) and (i+1) for (0 \le i < n-1)), calculate: [ \text{contribution}[i] = \text{weights}[i] + \text{weights}[i+1] ] These represent the extra cost incurred if you choose that particular split.

3. Select Splits for Minimum and Maximum Cost:

   • Sort the list of contributions.
   • For the minimum total cost, sum the smallest (m - 1) contributions.
   • For the maximum total cost, sum the largest (m - 1) contributions.

4. Calculate Total Costs and Their Difference:

   • (\text{minCost} = \text{base} + \text{sum of smallest } (m-1) \text{ contributions})
   • (\text{maxCost} = \text{base} + \text{sum of largest } (m-1) \text{ contributions})
   • The answer is: [ \text{answer} = \text{maxCost} - \text{minCost} ]

Complexity Analysis

• Time Complexity:

  • Calculating the candidate contributions takes (O(n)).
  • Sorting these contributions takes (O(n \log n)).
  • Summing (m-1) elements is (O(m)), which is (O(n)) in the worst case.
  • Overall time complexity: (O(n \log n)).

• Space Complexity:

  • We use an extra array to store (n-1) candidate contributions, resulting in (O(n)) space.

Python Code