2551. Put Marbles in Bags

Problem Statement

Hard

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.

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

Java

Source file

class Solution {

    public long putMarbles(int[] weights, int k) {
        // We collect and sort the value of all n - 1 pairs.
        int n = weights.length;
        int[] pairWeights = new int[n - 1];
        for (int i = 0; i < n - 1; ++i) {
            pairWeights[i] = weights[i] + weights[i + 1];
        }
        // We will sort only the first (n - 1) elements of the array.
        Arrays.sort(pairWeights, 0, n - 1);

        // Get the difference between the largest k - 1 values and the
        // smallest k - 1 values.
        long answer = 0l;
        for (int i = 0; i < k - 1; ++i) {
            answer += pairWeights[n - 2 - i] - pairWeights[i];
        }

        return answer;
    }
}