2681 Put Marbles In Bags

Put Marbles in Bags

You have k bags. You are given a 0-indexed integer array weights where weights[i] is the weight of the ith 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 ith marble and jth marble are in a bag, then all marbles with an index between the ith and jth 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.

 

Example 1:

**Input:** weights = [1,3,5,1], k = 2
**Output:** 4
**Explanation:** 
The distribution [1],[3,5,1] results in the minimal score of (1+1) + (3+1) = 6. 
The distribution [1,3],[5,1], results in the maximal score of (1+3) + (5+1) = 10. 
Thus, we return their difference 10 - 6 = 4.

Example 2:

**Input:** weights = [1, 3], k = 2
**Output:** 0
**Explanation:** The only distribution possible is [1],[3]. 
Since both the maximal and minimal score are the same, we return 0.

 

Constraints:

1 <= k <= weights.length <= 105
1 <= weights[i] <= 109

class Solution:
    def putMarbles(self, weights: List[int], k: int) -> int:
        # We collect and sort the value of all n - 1 pairs.
        n = len(weights)
        pairWeights = [weights[i] + weights[i + 1] for i in range(n - 1)]

        # Since python's sort function sorts the whole list, we don't limit it to the first n-1 elements here.
        pairWeights.sort()

        # Get the difference between the largest k - 1 values and the smallest k - 1 values.
        answer = 0
        for i in range(k - 1):
            answer += pairWeights[n - 2 - i] - pairWeights[i]

        return answer