2551. Put Marbles in Bags #

Problem Statement #

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: #

impl Solution {
    pub fn put_marbles(weights: Vec<i32>, k: i32) -> i64 {
        let n = weights.len();
        // pair weights of adjacent bags
        let mut pair_weights: Vec<_> = (0..n - 1).map(|i| weights[i] + weights[i + 1]).collect();

        // sort the pair weights
        pair_weights.sort();

        // sum the difference between the largest and smallest pair weights
        (0..(k - 1) as usize).fold(0_i64, |ans, i| {
            ans + (pair_weights[n - 2 - i] - pair_weights[i]) as i64
        })
    }
}

... #