LeetCode in Kotlin

2171. Removing Minimum Number of Magic Beans

Medium

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 {
    fun minimumRemoval(beans: IntArray): Long {
        beans.sort()
        val n = beans.size
        var sum: Long = 0
        for (bean in beans) {
            sum += bean.toLong()
        }
        var minbeans = Long.MAX_VALUE
        var prefix: Long = 0
        var suffix: Long
        var count: Long
        for (i in 0 until n) {
            prefix += beans[i].toLong()
            suffix = sum - prefix
            count = prefix - beans[i] + (suffix - beans[i] * (n - i - 1L))
            if (count < minbeans) {
                minbeans = count
            }
        }
        return minbeans
    }
}

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