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

/**
 * @param {number[]} weights
 * @param {number} k
 * @return {number}
 */
var putMarbles = function(weights, k) {
    var arr = [];
    for (var i = 0; i < weights.length - 1; i++) {
        arr.push(weights[i] + weights[i + 1]);
    }
    arr.sort((a, b) => a - b);
    var min = arr.slice(0, k - 1).reduce((sum, num) => sum + num, 0);
    var max = arr.slice(arr.length - k + 1).reduce((sum, num) => sum + num, 0);
    return max - min;
};

Explain:

nope.

Complexity:

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