0955HardArrayGreedySortingHeap (Priority Queue)

Splitting a Line of Marbles Into Bags

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Trains the technique from

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

Marbles stand in a line with weights weights. They are put into exactly k bags so that every bag holds a run of neighbouring marbles, no bag is empty, and every marble goes into a bag.

A bag's cost is the weight of its first marble plus the weight of its last. The score of a way of bagging is the total of its bags' costs.

Return the difference between the largest and the smallest score any bagging can have.

Examples

Example 1

Input: weights = [14, 3, 27, 9, 41, 6], k = 3
Output: 50

Two cuts are placed among the five neighbouring pairs, whose totals are 17, 30, 36, 50 and 47. The largest score takes the two heaviest, 50 and 47, and the smallest takes 17 and 30, so the difference is 50.

Example 2

Input: weights = [5, 9], k = 2
Output: 0

There is only one place to cut, so every bagging scores the same.

Example 3

Input: weights = [5], k = 1
Output: 0

A single bag needs no cuts at all.

Constraints

1 <= weights.length <= 10^5
1 <= k <= weights.length
1 <= weights[i] <= 10^9

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def put_marbles(weights: list[int], k: int) -> int:

Java

public long putMarbles(int[] weights, int k)

The intended approach

Time O(n log n) for n marbles, the sort of the pair totals dominating.Space O(n) for the pair totals.

A bagging is settled by where the cuts go, and there are `k - 1` cuts, each between two neighbouring marbles. Cutting between two marbles means the left one ends a bag and the right one starts one, so that cut contributes exactly their two weights added together. Every score also contains the first marble and the last, whatever the cuts are, because those two always sit at the ends of the outer bags. So the score is a fixed amount plus the totals of the `k - 1` chosen neighbouring pairs, and the cuts are free to be any distinct pairs. The largest score takes the `k - 1` heaviest pair totals and the smallest takes the lightest, so the difference between them is what is left once the fixed part cancels. Sorting the pair totals and taking from both ends gives it directly, and a single bag means no cuts at all and so no difference.

Pythonentry point: put_marbles

def put_marbles(weights, k):
    n = len(weights)
    if k == 1 or n == 1:
        return 0
    pairs = []
    for i in range(n - 1):
        pairs.append(weights[i] + weights[i + 1])
    pairs.sort()
    total = 0
    for step in range(k - 1):
        total += pairs[len(pairs) - 1 - step] - pairs[step]
    return total

The obvious version

Try every way of placing the cuts, work out each bagging's score by adding up the first and last marble of every bag, and take the largest less the smallest. Exponential in the marble count and it makes no argument about what a cut is worth.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_put_marbles(weights, k):
    import itertools
    n = len(weights)
    if k == 1 or n == 1:
        return 0
    best = None
    worst = None
    for cuts in itertools.combinations(range(n - 1), k - 1):
        edges = [-1] + list(cuts) + [n - 1]
        score = 0
        for t in range(len(edges) - 1):
            start = edges[t] + 1
            end = edges[t + 1]
            score += weights[start] + weights[end]
        if best is None or score > best:
            best = score
        if worst is None or score < worst:
            worst = score
    return best - worst

How this problem was checked

Reference solution executed against 10 test cases — 10 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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