2163. Minimum Difference in Sums After Removal of Elements

Problem

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 <= 10^5

• 1 <= nums[i] <= 10^5

Solution

class Solution {
    public long minimumDifference(int[] nums) {
        int n = nums.length / 3;
        PriorityQueue<Integer> minHeap = new PriorityQueue<>();
        PriorityQueue<Integer> maxHeap = new PriorityQueue<>((a, b) -> b - a);
        long[] leftMemo = new long[nums.length];
        long[] rightMemo = new long[nums.length];
        long current = 0L;
        for (int i = 0; i <= 2 * n - 1; i++) {
            current += nums[i];
            maxHeap.add(nums[i]);
            if (maxHeap.size() > n) {
                int removed = maxHeap.poll();
                current -= removed;
                leftMemo[i] = current;
            }
            if (maxHeap.size() == n) {
                leftMemo[i] = current;
            }
        }
        current = 0;
        for (int i = nums.length - 1; i >= n; i--) {
            current += nums[i];
            minHeap.add(nums[i]);
            if (minHeap.size() > n) {
                int removed = minHeap.poll();
                current -= removed;
            }
            if (minHeap.size() == n) {
                rightMemo[i] = current;
            }
        }
        long min = Long.MAX_VALUE;
        for (int i = n - 1; i <= 2 * n - 1; i++) {
            min = Math.min(min, leftMemo[i] - rightMemo[i + 1]);
        }
        return min;
    }
}

Explain:

nope.

Complexity:

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