Minimum Difference in Sums After Removal of Elements

Problem Description

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 belong to the first part, and their sum is sum_first. The next n elements belong to the second part, and their sum is sum_second. The difference in sums of the two parts is denoted as sum_first - sum_second. Return the minimum difference possible between the sums of the two parts after the removal of n elements.

Key Insights

• The task involves removing n elements from an array of size 3 * n, resulting in 2 * n remaining elements.
• The remaining elements must be split into two equal parts, and the goal is to minimize the difference between their sums.
• Utilizing heaps (priority queues) can help efficiently manage the selection of the largest and smallest elements for both parts.
• Dynamic programming approaches can also be applied to consider various ways of partitioning the elements.

Space and Time Complexity

Time Complexity: O(n log n) - due to sorting and priority queue operations. Space Complexity: O(n) - for storing elements in heaps.

Solution

To solve this problem, we can utilize two priority queues (heaps) to maintain the largest and smallest sums of the two parts after removing n elements. The algorithm follows these steps:

1. Sort the array to easily access the largest and smallest elements.
2. Use a max-heap to keep track of the largest n elements for sum_first and a min-heap for the smallest n elements for sum_second.
3. Iterate through possible removals of elements and calculate the two sums accordingly.
4. Update the minimum difference encountered during these calculations.

This approach ensures that we efficiently determine the best possible sums after removing the required elements.

Code Solutions