Problem Description

You are given an integer array nums with 3 * n elements. You are allowed to remove a subsequence of n elements from nums. After removing these n elements, you divide the remaining array into two equal parts, each containing n elements. The first part's sum is sum_first and the second part's sum is sum_second. Your goal is to minimize the absolute difference between these two sums.

In essence, you are trying to balance the sums of two subsequences by carefully choosing which n elements to remove to achieve the smallest possible difference between the sums of these two parts.

Intuition

To solve this problem, we need to use a strategy that allows us to calculate the sums of potential subarrays efficiently after removing a subsequence of elements. One way to approach the solution is to consider both the prefix sum and the suffix sum:

1. Find max prefix sums for the first 2n elements: Starting from the left, we calculate the maximum sum of any n elements in the first 2n elements. We want to know the largest possible sum we could have on the left side when removing n elements.

2. Find max suffix sums for the last 2n elements: Similarly, starting from the right, we calculate the maximum sum of any n elements in the last 2n elements. It gives us the same but for the right side, telling us the largest possible sum we could end up with on the right side, again after removing n elements.

3. Use heaps to calculate these prefix and suffix sums: Since we are interested in the n largest sums at any point, we use a max heap for the prefix sum and a min heap for the suffix sum to keep track of the largest and smallest elements, respectively.

4. Determine the minimum difference: We calculate the difference between the prefix sum and the suffix sum at every point possible (where the remaining 2n elements can be split into two equal parts after removing n elements). The smallest such difference would be our answer.

By breaking the problem into prefix and suffix calculations, we efficiently narrow down the sums we need to compare to determine the minimum possible difference.

Learn more about Dynamic Programming and Heap (Priority Queue) patterns.

Solution Approach

The solution uses a two-heap approach and prefix/suffix sums to keep track of the potential maximum sum of n numbers in two segments of the array nums.

Here is a step-by-step breakdown of the solution approach:

1. Initialization: Define two empty heaps, q1 for the prefix part and q2 for the suffix part. Also, create arrays pre and suf to store prefix and suffix sums respectively, each initialized with zeros and of size m + 1, where m is the length of the array nums.

2. Prefix Sums Calculation:

   • Iterate through the first 2n elements of nums and push the negative of each element into heap q1 (max heap since Python's heapq is a min heap by default).
   • After each insertion, if the heap size exceeds n, remove the smallest element (which is the largest negative number) from the heap and adjust the running sum s.
   • Store the running sum s in pre at the current index to represent the maximum sum of n numbers for the first part up to that index.

3. Suffix Sums Calculation:

   • Iterate through the last 2n elements of nums in reverse and push each element into heap q2 (min heap).
   • Similarly, if the heap size exceeds n, remove the smallest (top) element from the heap and adjust the running sum s.
   • Store the running sum s in suf at the current index to represent the maximum sum of n numbers for the second part starting at that index.

4. Finding the Minimum Difference:

   • We can split nums into two parts after any index i in the range [n, 2n]. To achieve this, we calculate and compare pre[i] - suf[i + 1] for each i in this range.
   • The reason for the subtraction is that pre[i] stores the maximum sum of n numbers from the start to the ith index, whereas suf[i + 1] stores the maximum sum of n numbers from the i + 1th index to the end.
   • By comparing these values, we are effectively considering the best sums achievable for both parts of the array after removing a subsequence of n elements. This way, the minimum difference calculated represents the closest balance we can reach between the sums of the two parts, considering every possible split point within the range.

Throughout the implementation, heaps are utilized to keep track of the largest or smallest n elements in a running window, which allows us to efficiently calculate the desired maximum prefix and suffix sums. The approach makes sure that we consider every split point for the remaining 2n elements and thus guarantee that the minimum absolute difference is found.

Example Walkthrough

Let's say we have an integer array nums with 9 elements (n=3):

nums = [1, 2, 3, 4, 5, 6, 7, 8, 9]

We can follow the solution approach outlined above:

1. Initialization:

   • Create two empty heaps, q1 (as a max heap) and q2 (as a min heap).
   • Initialize the prefix sums array pre and the suffix sums array suf with zeros and of size 4 (3n + 1).

   q1 = []
   q2 = []
   pre = [0, 0, 0, 0]
   suf = [0, 0, 0, 0]

2. Prefix Sums Calculation:

   • Iterate through the first 6 elements (2n) of nums and push the negative of each element into q1.
   • If the heap q1 exceeds size 3, pop and subtract the smallest element (which is the largest negative number).
   • Update the pre array with the running sum s at each index.

3. Suffix Sums Calculation:

   • Iterate through the last 6 elements (2n) of nums in reverse order and push each element into q2.
   • If the heap size q2 exceeds size 3, pop and subtract the smallest element.
   • Update the suf array with the running sum s at the corresponding index.

4. Finding the Minimum Difference:

   • Calculate and compare pre[i] - suf[i + 1] for i ranging from 3 to 6.
   • The smallest difference is the answer.

Let's perform these calculations step by step with the example array:

Step 1: Prefix Sums

• After iteration and maintaining the max heap:

  pre = [0, 1, 3, 6, 6, 10, 15]

• We obtain the maximum sum of n numbers in the first 2n numbers of the array.

Step 2: Suffix Sums

• Reverse iterating and maintaining the min heap:

  suf = [30, 23, 18, 14, 14, 9, 3, 0, 0, 0]

• The suf array is filled similarly but in the reverse order from the end of the array.

Step 3: Calculate Minimum Difference

• With our pre and suf arrays, we start comparing sums for the range [3, 6]:

For i = 3: pre[i] = 6, suf[i + 1] = 14, difference = |6 - 14| = 8
For i = 4: pre[i] = 6, suf[i + 1] = 9, difference = |6 - 9| = 3
For i = 5: pre[i] = 10, suf[i + 1] = 3, difference = |10 - 3| = 7

The smallest difference is when i = 4 yielding an absolute difference of 3.

Therefore, to minimize the difference between sum_first and sum_second, we would remove three elements such that the remaining elements after index 4 and up to index 5 (inclusive) are balanced. This would give us:

First part: [1, 2, 3, 4], sum_first = 10
Second part: [6, 7, 8], sum_second = 21 - 12 (removed element at index 5) = 9

And the minimum absolute difference between the sums of these two parts is 3, as calculated.

Solution Implementation

Time and Space Complexity

The given Python code aims at finding the minimum possible difference between the sum of the elements of three equally-sized partitions of the array. It makes use of heaps to maintain the n-largest elements seen so far, where n = m / 3 and m is the length of the input list nums.

Time Complexity

The time complexity of the code is a result of several operations:

1. Enumerating and processing the first 2/3 of the list: Each insertion into a heap (implemented as a priority queue) is O(log n), and we do this for the first 2n elements of the list, giving a total of O(2n log n) for this section.

2. Iterating and processing the last 1/3 of the list backwards: Similarly, we apply a heap operation for each of the last n elements, resulting in O(n log n) time.

3. Finding the minimum difference: This involves iterating through the middle third of the list to find the minimum difference. This is a linear operation with complexity O(n).

Therefore, the dominant term is the heap operations, summing up to a total time complexity of O(2n log n + n log n) which simplifies to O(m log(m/3)) since n = m / 3.

Space Complexity

The space complexity of the code is determined by:

1. Heap storage: Two heaps are used, each storing up to n elements, thus requiring O(n) space.

2. Precomputed sums: Two arrays, pre and suf, each of size m + 1, are used to store precomputed sums of heaps, contributing O(m) space.

3. Variables and constants: A constant amount of additional space is used for variables and iteration, which does not depend on n or m and thus is O(1).

Considering the heaps and the arrays for precomputed sums, the total space complexity is O(m + 2n) which simplifies to O(m) since m is larger than 2n.

In conclusion, the given code has a time complexity of O(m log(m/3)) and a space complexity of O(m).

Learn more about how to find time and space complexity quickly using problem constraints.