Introduction

In this blog post, we will explore how to solve the LeetCode problem "2163" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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

Explanation

Here's the breakdown of the solution:

• Prefix Sum Optimization: Calculate prefix sums efficiently to determine the sum of the first n elements after removing elements from the first n + i elements. Use a min-heap to track the n smallest elements seen so far and efficiently update the prefix sum when an element is removed.

• Suffix Sum Optimization: Calculate suffix sums efficiently to determine the sum of the last n elements after removing elements from the last n + i elements. Use a max-heap to track the n largest elements seen so far and efficiently update the suffix sum when an element is removed.

• Minimum Difference: Iterate through all possible removal combinations by considering all splitting points between prefix and suffix. Calculate the difference between the prefix sum and suffix sum for each split and take the minimum difference.

• Time & Space Complexity: O(n log n) time complexity due to heap operations, and O(n) space complexity for storing prefix/suffix sums and heaps.

Code

    import heapq

def minimumDifference(nums):
    n = len(nums) // 3

    # Prefix sum calculation
    prefix_sums = [0] * (2 * n + 1)
    min_heap = []
    current_sum = 0
    for i in range(2 * n):
        heapq.heappush(min_heap, nums[i])
        current_sum += nums[i]
        if len(min_heap) > n:
            current_sum -= heapq.heappop(min_heap)
        if len(min_heap) == n:
            prefix_sums[i - n + 1] = current_sum

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

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

    return min_diff