2551. Put Marbles in Bags

Description

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

 

Solutions

Solution: Greedy

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

 

JavaScript

/**
 * @param {number[]} weights
 * @param {number} k
 * @return {number}
 */
const putMarbles = function (weights, k) {
  const n = weights.length;
  const scores = [];

  for (let index = 1; index < n; index++) {
    const score = weights[index] + weights[index - 1];

    scores.push(score);
  }

  scores.sort((a, b) => a - b);

  let minScores = weights[0] + weights[n - 1];
  let maxScores = weights[0] + weights[n - 1];

  for (let index = 0; index < k - 1; index++) {
    minScores += scores[index];
    maxScores += scores[n - 2 - index];
  }

  return maxScores - minScores;
};