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:
n elements belonging to the first part and their sum is sumfirst.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.
sumfirst = 3 and sumsecond = 2, their difference is 1.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 * n1 <= n <= 1051 <= nums[i] <= 105This approach uses a combination of prefix and suffix analysis with the help of priority queues to find the minimum difference between the sums of two parts after removing n elements.
We maintain two arrays for prefix and suffix information: one to track the maximum sum of n elements from the beginning (prefix) and one for the minimum sum of n elements from the end (suffix). By continuously tracking these sums in a priority queue (max-heap for prefix and min-heap for suffix), we iterate to find the minimum difference possible between them.
This Python solution uses the prefix and suffix arrays to manage the sums efficiently. The prefix array stores the maximum sum of n numbers from the beginning of the array using a min-heap. Conversely, the suffix analysis maintains the minimum sum from the back using a max-heap. The difference between these prefix and suffix sums is calculated, and the minimum difference is returned.
JavaScript
Time Complexity: O(n log n), where n is the length of the array divided by 3.
Space Complexity: O(n) due to the storage of prefix and suffix data structures.
In this approach, we leverage a sliding window technique to efficiently evaluate every possible subarray of length 2n using the 'unordered removal' strategy. The idea is to use prefix and suffix evaluations within a limited range in order to balance the parts optimally.
The C++ implementation proposes a more direct manipulation of the problem using two priority queues to track and update necessary elements. By processing elements in window-like advances, you can guarantee that the window will contain the necessary average values to balance itself for two parts after n removals.
Java
Time Complexity: O(n log n) due to heap operations over splits.
Space Complexity: O(n) for auxiliary storage.
| Approach | Complexity |
|---|---|
| Prefix and Suffix Analysis with Priority Queues | Time Complexity: O(n log n), where n is the length of the array divided by 3. |
| Sliding Window with Unordered Removal | Time Complexity: O(n log n) due to heap operations over splits. |
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