Introduction

In this blog post, we will explore how to solve the LeetCode problem "2551" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

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

Explanation

Here's the solution:

• Key Idea: The cost of each bag depends only on the weights of the first and last marbles in the bag. The intermediate marbles do not directly contribute to the cost. To maximize the score, put the largest possible sum of adjacent marble weights at the boundaries of the bags. To minimize the score, put the smallest possible sum of adjacent marble weights at the boundaries of the bags.

• Calculate Adjacent Sums: Create an array of adjacent marble weights sums.

• Sort and Calculate: Sort the adjacent sums array. The k-1 largest sums will be added to maximize the cost, and the k-1 smallest sums will be added to minimize the cost. The base score (sum of first and last elements) remains same.

• Complexity:

  • Runtime: O(n log n) due to sorting.
  • Space: O(n) to store the adjacent sums.

Code

    def marble_distribution(weights, k):
    """
    Calculates the difference between the maximum and minimum scores among marble distributions.

    Args:
        weights: A list of integers representing the weights of the marbles.
        k: An integer representing the number of bags.

    Returns:
        The difference between the maximum and minimum scores.
    """

    n = len(weights)
    if k == n:
        return 0

    adjacent_sums = []
    for i in range(n - 1):
        adjacent_sums.append(weights[i] + weights[i + 1])

    adjacent_sums.sort()

    min_score = weights[0] + weights[-1]
    max_score = weights[0] + weights[-1]

    for i in range(k - 1):
        min_score += adjacent_sums[i]
        max_score += adjacent_sums[n - 2 - i]

    return max_score - min_score