2551. Put Marbles in Bags

Problem

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

Solution (Java)

class Solution {
    public long putMarbles(int[] A, int k) {
        int n = A.length - 1;
        long B[] = new long[n], res = 0;
        for (int i = 0; i < B.length; i++) {
            B[i] = A[i] + A[i + 1];
        }
        Arrays.sort(B);
        for (int i = 0; i < k - 1; i++) {
            res += B[n - 1 - i] - B[i];
        }
        return res;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).