2171. Removing Minimum Number of Magic Beans

Problem

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 <= 10^5

• 1 <= beans[i] <= 10^5

Solution (Java)

class Solution {
    public long minimumRemoval(int[] beans) {
        Arrays.sort(beans);
        int n = beans.length;
        long sum = 0;
        for (int bean : beans) {
            sum += bean;
        }
        long minbeans = Long.MAX_VALUE;
        long prefix = 0;
        long suffix;
        long count;
        for (int i = 0; i < n; i++) {
            prefix += beans[i];
            suffix = sum - prefix;
            count = (prefix - beans[i]) + (suffix - beans[i] * (n - i - 1L));
            if (count < minbeans) {
                minbeans = count;
            }
        }
        return minbeans;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).