# Solving Leetcode Interviews in Seconds with AI: Removing Minimum Number of Magic Beans


	# 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](https://www.chatmagic.app), 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
	```python
	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
	```