Put Marbles in Bags

[iamAntimPal]

Discussion: Put Marbles in Bags

Hello everyone,

I’d like to open a discussion for the problem "2551. Put Marbles in Bags". This is a challenging problem that requires careful thought on how to split an array of weights into k contiguous bags, and then compute the difference between the maximum and minimum possible scores.

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Problem Statement

You are given:

• A 0-indexed integer array weights where weights[i] represents the weight of the ith marble.
• An integer k representing the number of bags.

You need to divide the marbles into k bags following these rules:

1. No bag is empty.
2. Contiguous grouping: If the ith marble and jth marble are in a bag, then every marble with an index between i and j must also be in that bag.
3. Bag Cost: If a bag consists of marbles from index i to j (inclusive), its cost is defined as weights[i] + weights[j].

The score of a distribution is the sum of the costs of all k bags.
Your goal is to return the difference between the maximum and minimum possible scores across all valid distributions.

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Examples

Example 1:

Input:

weights = [1, 3, 5, 1], k = 2

Output:

4

Explanation:

• One possible distribution: [1], [3, 5, 1] → Cost = (1+1) + (3+1) = 6
• Another distribution: [1, 3], [5, 1] → Cost = (1+3) + (5+1) = 10
  Thus, the difference is 10 - 6 = 4.

Example 2:

Input:

weights = [1, 3], k = 2

Output:

0

Explanation:
The only possible distribution is [1], [3], so both the maximum and minimum scores are the same, and the difference is 0.

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Constraints

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

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Discussion Points & Hints

1. Observation:
   Notice that regardless of how you split the marbles, the first marble and the last marble always contribute to the cost (since they are the endpoints of the full array). The variation in score comes only from where you decide to split the array.

2. Key Insight:
   When you split the array, the additional cost added by a split at position i (i.e., between weights[i] and weights[i+1]) is weights[i] + weights[i+1].

   • For the minimum score, you want to choose the splits that add the smallest additional costs.
   • For the maximum score, you choose the splits that add the largest additional costs.

3. Approach:
   One common approach is to:

   • Calculate the adjacent pair sums (or "cost differences") for every possible split.
   • Sort these differences.
   • Use the smallest k-1 values to compute the minimum possible additional cost and the largest k-1 values for the maximum additional cost.
   • The final answer is the difference between the maximum and minimum total costs.

4. SQL & Pandas Thoughts:

   • SQL Approach:
     While the problem is primarily algorithmic, if you wanted to simulate parts of the approach using SQL (e.g., sorting and aggregating values), you might consider window functions.
   • Pandas Approach:
     With Pandas, you can easily compute differences between adjacent rows, sort them, and then sum up the appropriate values.
     Feel free to share your implementations and optimizations!

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Invitation to Discuss

• What approaches did you try?
• How did you optimize for large input sizes?
• Any edge cases you encountered?

I’m excited to see your solutions and insights on this challenging problem! Please share your SQL and Python (Pandas) implementations and any performance considerations you took into account.

Happy coding!

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Feel free to add more details or propose alternative solutions in the discussion below.

[iamAntimPal]

any one give the answer for that and give seggtion also for that

[iamAntimPal]

ok that