LeetCode in Kotlin

2163. Minimum Difference in Sums After Removal of Elements

Hard

You are given a 0-indexed integer array nums consisting of 3 * n elements.

You are allowed to remove any subsequence of elements of size exactly n from nums. The remaining 2 * n elements will be divided into two equal parts:

• The first n elements belonging to the first part and their sum is sumfirst.
• The next n elements belonging to the second part and their sum is sumsecond.

The difference in sums of the two parts is denoted as sumfirst - sumsecond.

• For example, if sumfirst = 3 and sumsecond = 2, their difference is 1.
• Similarly, if sumfirst = 2 and sumsecond = 3, their difference is -1.

Return the minimum difference possible between the sums of the two parts after the removal of n elements.

Example 1:

Input: nums = [3,1,2]

Output: -1

Explanation: Here, nums has 3 elements, so n = 1.

Thus we have to remove 1 element from nums and divide the array into two equal parts.

• If we remove nums[0] = 3, the array will be [1,2]. The difference in sums of the two parts will be 1 - 2 = -1.

• If we remove nums[1] = 1, the array will be [3,2]. The difference in sums of the two parts will be 3 - 2 = 1.

• If we remove nums[2] = 2, the array will be [3,1]. The difference in sums of the two parts will be 3 - 1 = 2.

The minimum difference between sums of the two parts is min(-1,1,2) = -1.

Example 2:

Input: nums = [7,9,5,8,1,3]

Output: 1

Explanation: Here n = 2. So we must remove 2 elements and divide the remaining array into two parts containing two elements each.

If we remove nums[2] = 5 and nums[3] = 8, the resultant array will be [7,9,1,3]. The difference in sums will be (7+9) - (1+3) = 12.

To obtain the minimum difference, we should remove nums[1] = 9 and nums[4] = 1. The resultant array becomes [7,5,8,3]. The difference in sums of the two parts is (7+5) - (8+3) = 1.

It can be shown that it is not possible to obtain a difference smaller than 1.

Constraints:

• nums.length == 3 * n
• 1 <= n <= 105
• 1 <= nums[i] <= 105

Solution

import java.util.PriorityQueue

class Solution {
    fun minimumDifference(nums: IntArray): Long {
        val n = nums.size / 3
        val minHeap = PriorityQueue<Int>()
        val maxHeap = PriorityQueue { a: Int, b: Int -> b - a }
        val leftMemo = LongArray(nums.size)
        val rightMemo = LongArray(nums.size)
        var current = 0L
        for (i in 0..2 * n - 1) {
            current += nums[i].toLong()
            maxHeap.add(nums[i])
            if (maxHeap.size > n) {
                val removed = maxHeap.poll()
                current -= removed.toLong()
                leftMemo[i] = current
            }
            if (maxHeap.size == n) {
                leftMemo[i] = current
            }
        }
        current = 0
        for (i in nums.size - 1 downTo n) {
            current += nums[i].toLong()
            minHeap.add(nums[i])
            if (minHeap.size > n) {
                val removed = minHeap.poll()
                current -= removed.toLong()
            }
            if (minHeap.size == n) {
                rightMemo[i] = current
            }
        }
        var min = Long.MAX_VALUE
        for (i in n - 1..2 * n - 1) {
            min = Math.min(min, leftMemo[i] - rightMemo[i + 1])
        }
        return min
    }
}

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