Removing Minimum Number of Magic Beans

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

SOLUTION:

class Solution:
    def minimumRemoval(self, beans: List[int]) -> int:
        n = len(beans)
        beans.sort()
        total = sum(beans)
        cost = total
        for i in range(n):
            currcost = total - (n - i) * beans[i]
            cost = min(cost, currcost)
        return cost