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2551. Put Marbles in Bags

Difficulty Topics

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

Solution

put-marbles-in-bags.py
class Solution:
    def putMarbles(self, weights: List[int], k: int) -> int:
        N = len(weights)
        if N == k or k == 1: return 0
        candidates = []

        for a, b in zip(weights, weights[1:]):
            candidates.append(a + b)

        candidates.sort()
        mmax = mmin = 0

        for i in range(k - 1):
            mmin += candidates[i]
            mmax += candidates[-i - 1]

        return mmax - mmin