LeetCode #2551 — HARD

Put Marbles in Bags

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

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.

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Greedy

Example 1

[1,3,5,1]
2

Example 2

[1,3]
2

Step 02

Core Insight

What unlocks the optimal approach

Each bag will contain a subarray, and only the endpoints of these subarrays matter.
Each subarray only contributes two numbers to the sum. Use this property to choose the subarrays optimally.
Try to use a priority queue.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2551: Put Marbles in Bags
class Solution {
    public long putMarbles(int[] weights, int k) {
        int n = weights.length;
        int[] arr = new int[n - 1];
        for (int i = 0; i < n - 1; ++i) {
            arr[i] = weights[i] + weights[i + 1];
        }
        Arrays.sort(arr);
        long ans = 0;
        for (int i = 0; i < k - 1; ++i) {
            ans -= arr[i];
            ans += arr[n - 2 - i];
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2551: Put Marbles in Bags
func putMarbles(weights []int, k int) (ans int64) {
	n := len(weights)
	arr := make([]int, n-1)
	for i, w := range weights[:n-1] {
		arr[i] = w + weights[i+1]
	}
	sort.Ints(arr)
	for i := 0; i < k-1; i++ {
		ans += int64(arr[n-2-i] - arr[i])
	}
	return
}

# Accepted solution for LeetCode #2551: Put Marbles in Bags
class Solution:
    def putMarbles(self, weights: List[int], k: int) -> int:
        arr = sorted(a + b for a, b in pairwise(weights))
        return sum(arr[len(arr) - k + 1 :]) - sum(arr[: k - 1])

// Accepted solution for LeetCode #2551: Put Marbles in Bags
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2551: Put Marbles in Bags
// class Solution {
//     public long putMarbles(int[] weights, int k) {
//         int n = weights.length;
//         int[] arr = new int[n - 1];
//         for (int i = 0; i < n - 1; ++i) {
//             arr[i] = weights[i] + weights[i + 1];
//         }
//         Arrays.sort(arr);
//         long ans = 0;
//         for (int i = 0; i < k - 1; ++i) {
//             ans -= arr[i];
//             ans += arr[n - 2 - i];
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #2551: Put Marbles in Bags
function putMarbles(weights: number[], k: number): number {
    const n = weights.length;
    const arr: number[] = [];
    for (let i = 0; i < n - 1; ++i) {
        arr.push(weights[i] + weights[i + 1]);
    }
    arr.sort((a, b) => a - b);
    let ans = 0;
    for (let i = 0; i < k - 1; ++i) {
        ans += arr[n - i - 2] - arr[i];
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × log n)

Space

O(n)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.