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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 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

  • class Solution {
    public long minimumRemoval(int[] beans) {
        Arrays.sort(beans);
        long s = 0;
        for (int v : beans) {
            s += v;
        }
        long ans = s;
        int n = beans.length;
        for (int i = 0; i < n; ++i) {
            ans = Math.min(ans, s - (long) beans[i] * (n - i));
        }
        return ans;
    }
}

  • class Solution {
public:
    long long minimumRemoval(vector<int>& beans) {
        sort(beans.begin(), beans.end());
        long long s = accumulate(beans.begin(), beans.end(), 0ll);
        long long ans = s;
        int n = beans.size();
        for (int i = 0; i < n; ++i) ans = min(ans, s - 1ll * beans[i] * (n - i));
        return ans;
    }
};

  • class Solution:
    def minimumRemoval(self, beans: List[int]) -> int:
        beans.sort()
        ans = s = sum(beans)
        n = len(beans)
        for i, v in enumerate(beans):
            ans = min(ans, s - v * (n - i))
        return ans


  • func minimumRemoval(beans []int) int64 {
	sort.Ints(beans)
	s := 0
	for _, v := range beans {
		s += v
	}
	ans := s
	n := len(beans)
	for i, v := range beans {
		ans = min(ans, s-v*(n-i))
	}
	return int64(ans)
}

  • function minimumRemoval(beans: number[]): number {
    const n = beans.length;
    let sum = beans.reduce((a, c) => a + c, 0);
    beans.sort((a, b) => a - b);
    let ans = sum;
    for (let i = 0; i < n; i++) {
        let num = beans[i];
        ans = Math.min(sum - num * (n - i), ans);
    }
    return ans;
}

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