Brute Force Solution

A brute-force solution would involve generating all possible subsequences of size n from the input array nums. For each subsequence, we would remove it from nums, then divide the remaining elements into two equal parts of size n. We would calculate the difference between the sums of these two parts and keep track of the minimum difference found so far.

This approach has a high time complexity due to the need to generate all possible subsequences, making it impractical for large input sizes.

Complexity Analysis

• Time Complexity: O(C(3n, n) * n), where C(3n, n) is the binomial coefficient representing the number of ways to choose n elements from 3n elements. This is because we need to generate all subsequences of size n, and for each subsequence, we need to calculate the difference between the sums of the two parts, which takes O(n) time.
• Space Complexity: O(n), as we need to store the remaining elements after removing the subsequence.

Optimal Solution

A more efficient solution can be achieved using prefix and suffix sums along with priority queues (min-heap and max-heap).

1. Calculate Prefix Sums: Iterate through the first 2n elements of nums. Maintain a max-heap of size n. For each element, add it to the heap and the prefix sum. If the heap size exceeds n, remove the largest element from the heap and subtract it from the prefix sum. This ensures we always have the sum of the smallest n elements seen so far.
2. Calculate Suffix Sums: Iterate through the last 2n elements of nums in reverse order. Maintain a min-heap of size n. For each element, add it to the heap and the suffix sum. If the heap size exceeds n, remove the smallest element from the heap and subtract it from the suffix sum. This ensures we always have the sum of the largest n elements seen so far.
3. Calculate Minimum Difference: Iterate from n to 2n. For each index i, the prefix sum up to i - 1 represents the sum of the smallest n elements in nums[0...i-1], and the suffix sum from i represents the sum of the largest n elements in nums[i...3n-1]. Calculate the difference between the prefix sum and the suffix sum, and update the minimum difference.

Edge Cases

• Empty input array: Return 0 if the input array is empty.
• n = 1: Handle the base case where we remove only one element.

Code Example (Python)

import heapq

def minDifference(nums):
    n = len(nums) // 3
    
    # Prefix sums
    prefix_sums = [0] * (2 * n + 1)
    max_heap = []
    prefix_sum = 0
    for i in range(2 * n):
        heapq.heappush(max_heap, -nums[i])
        prefix_sum += nums[i]
        if len(max_heap) > n:
            prefix_sum += heapq.heappop(max_heap)
        prefix_sums[i + 1] = prefix_sum

    # Suffix sums
    suffix_sums = [0] * (2 * n + 1)
    min_heap = []
    suffix_sum = 0
    for i in range(3 * n - 1, n - 1, -1):
        heapq.heappush(min_heap, nums[i])
        suffix_sum += nums[i]
        if len(min_heap) > n:
            suffix_sum -= heapq.heappop(min_heap)
        suffix_sums[3 * n - i] = suffix_sum

    suffix_sums.reverse()

    # Calculate minimum difference
    min_diff = float('inf')
    for i in range(n, 2 * n + 1):
        min_diff = min(min_diff, prefix_sums[i] - suffix_sums[i])

    return min_diff

Complexity Analysis

• Time Complexity: O(n log n), where n is the input size. This is because we iterate through the array twice (for prefix and suffix sums), and each heap operation takes O(log n) time.
• Space Complexity: O(n), as we need to store the prefix sums, suffix sums, and the heaps, each of size O(n).