README.md

2171. 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

Solutions (Rust)

1. Solution

impl Solution {
    pub fn minimum_removal(beans: Vec<i32>) -> i64 {
        let mut beans = beans.into_iter().map(|x| x as i64).collect::<Vec<_>>();
        let mut lsum = 0;
        let mut rsum = beans.iter().sum::<i64>();
        let mut ret = i64::MAX;
        beans.sort_unstable();

        for i in 0..beans.len() {
            ret = ret.min(lsum + rsum - (beans.len() - i) as i64 * beans[i]);
            lsum += beans[i];
            rsum -= beans[i];
        }

        ret
    }
}