[2551. Put Marbles in Bags](https://leetcode.com/problems/put-marbles-in-bags/)

### 題目描述

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.*

### 範例

**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` <= 10^5^
* 1 <= `weights[i]` <= 10^9^

### 解答
#### Python

class Solution:
    def putMarbles(self, weights: List[int], k: int) -> int:
        arr = sorted(map(sum, pairwise(weights)))
        return sum(arr[len(arr)-k+1:]) - sum(arr[:k-1])

> [name=Yen-Chi Chen][time=Sun, Jul 9, 2023]

### Reference

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