Introduction

In this blog post, we will explore how to solve the LeetCode problem "2171" 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 are given an array of positive integers beans, where each integer represents the number of magic beans found in a particular magic bag. Remove any number of beans (possibly none) from each bag such that the number of beans in each remaining non-empty bag (still containing at least one bean) is equal. Once a bean has been removed from a bag, you are not allowed to return it to any of the bags. Return the minimum number of magic beans that you have to remove. Example 1: Input: beans = [4,1,6,5] Output: 4 Explanation: - We remove 1 bean from the bag with only 1 bean. This results in the remaining bags: [4,0,6,5] - Then we remove 2 beans from the bag with 6 beans. This results in the remaining bags: [4,0,4,5] - Then we remove 1 bean from the bag with 5 beans. This results in the remaining bags: [4,0,4,4] We removed a total of 1 + 2 + 1 = 4 beans to make the remaining non-empty bags have an equal number of beans. There are no other solutions that remove 4 beans or fewer. Example 2: Input: beans = [2,10,3,2] Output: 7 Explanation: - We remove 2 beans from one of the bags with 2 beans. This results in the remaining bags: [0,10,3,2] - Then we remove 2 beans from the other bag with 2 beans. This results in the remaining bags: [0,10,3,0] - Then we remove 3 beans from the bag with 3 beans. This results in the remaining bags: [0,10,0,0] We removed a total of 2 + 2 + 3 = 7 beans to make the remaining non-empty bags have an equal number of beans. There are no other solutions that removes 7 beans or fewer. Constraints: 1 <= beans.length <= 105 1 <= beans[i] <= 105

Explanation

Here's the solution to the problem:

• Sort the array: Sorting allows us to efficiently consider each bean count as the potential target value for the remaining bags.
• Iterate and calculate: For each bean count, we calculate the total number of beans that need to be removed to make all remaining non-empty bags have that count.

• Minimize: We keep track of the minimum number of removed beans encountered so far.

• Runtime Complexity: O(n log n) due to sorting. Storage Complexity: O(1) excluding the input array (in-place sort if possible).

Code

    def minimum_removal(beans):
    beans.sort()
    n = len(beans)
    total_sum = sum(beans)
    min_removal = float('inf')

    for i in range(n):
        removal = total_sum - beans[i] * (n - i)
        min_removal = min(min_removal, removal)

    return min_removal