2171. Removing Minimum Number of Magic Beans
Description
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 nonempty 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 nonempty 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 nonempty bags have an equal number of beans. There are no other solutions that removes 7 beans or fewer.
Constraints:
1 <= beans.length <= 10^{5}
1 <= beans[i] <= 10^{5}
Solutions
Solution 1: Sorting + Enumeration
We can sort all the beans in the bags in ascending order, and then enumerate the number of beans $beans[i]$ in each bag as the final number of beans in the bag. The total remaining number of beans is $beans[i] \times (n  i)$, so the number of beans that need to be taken out is $s  beans[i] \times (n  i)$, where $s$ is the total number of beans in all bags. We need to find the minimum number of beans that need to be taken out among all schemes.
The time complexity is $O(n \times \log n)$, and the space complexity is $O(\log n)$. Here, $n$ is the number of bags.









