#2163

Hard

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Minimum Difference in Sums After Removal of Elements

Minimize the difference between sums after removing n elements from a 3n array by dividing the remaining elements into two parts.

array dynamic programming heap priority queue

Problem Statement

You are given an integer array nums consisting of 3 * n elements. You need to remove any subsequence of exactly n elements, and then divide the remaining 2 * n elements into two equal parts. The difference in sums of the two parts is denoted as sumfirst - sumsecond.

Return the minimum possible difference between the sums of the two parts after removing n elements from nums. The challenge is to find a solution that efficiently minimizes this difference using dynamic programming and heaps.

Examples

Example 1

Input: nums = [3,1,2]

Output: -1

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

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

Solution Approach

State Transition Dynamic Programming

The problem can be modeled as a state transition dynamic programming problem. You iterate through the possible subsequences to remove, calculating the sums of the remaining elements and adjusting them accordingly.

Use of Heaps

Heaps can be used to efficiently manage the largest and smallest elements in the array. By using a min-heap for the smaller part and a max-heap for the larger part, we can easily adjust the sums as we try different subsequences to remove.

Optimization and Final Calculation

After determining the subsequences to remove and splitting the remaining array into two parts, the final step involves calculating the difference in sums and optimizing the solution by selecting the minimum difference possible.

Complexity Analysis

Metric	Value
Time	O(n \log n)
Space	O(n)

The time complexity is O(n log n) due to the heap operations involved in managing the elements during the dynamic programming transitions. The space complexity is O(n) due to the space required for the heaps and state tracking.

What Interviewers Usually Probe

Tests the candidate's ability to apply dynamic programming with state transitions.
Evaluates the candidate's knowledge of heap operations for optimizing subarray sum calculations.
Checks the ability to optimize performance for large input sizes (up to 3 * 10^5 elements).

Common Pitfalls or Variants

Common pitfalls

Forgetting to handle the edge case where n is very small or large.
Inefficient heap usage which could cause time complexity to exceed the limit.
Incorrect partitioning of the remaining elements after removing the subsequence.

Follow-up variants

Variation with a constraint on the value of n for additional optimization challenges.
Problem with multiple possible subsequences to remove, requiring additional considerations for efficiency.
Variant that requires splitting the array into more than two parts instead of two.

FAQ

What is the core pattern for solving the Minimum Difference in Sums After Removal of Elements problem?

The core pattern is state transition dynamic programming combined with heap operations to manage and optimize subarray sum calculations.

How can heaps be useful in solving the problem?

Heaps allow us to efficiently manage the largest and smallest elements of the remaining array, helping in minimizing the sum difference.

Why is dynamic programming essential for this problem?

Dynamic programming helps by efficiently tracking and calculating the sums after removing subsequences, reducing the problem's complexity.

What time complexity can I expect when solving the problem?

The expected time complexity is O(n log n), primarily due to heap operations during the state transitions.

How can I handle edge cases in the Minimum Difference in Sums After Removal of Elements problem?

Ensure correct handling of small and large values of n, and be mindful of the heap and partitioning operations during dynamic programming transitions.

terminal

Solution

Solution 1: Priority Queue (Max and Min Heap) + Prefix and Suffix Sum + Enumeration of Split Points

The problem is essentially equivalent to finding a split point in $nums$, dividing the array into two parts. In the first part, select the smallest $n$ elements, and in the second part, select the largest $n$ elements, so that the difference between the sums of the two parts is minimized.

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class Solution:
    def minimumDifference(self, nums: List[int]) -> int:
        m = len(nums)
        n = m // 3

        s = 0
        pre = [0] * (m + 1)
        q1 = []
        for i, x in enumerate(nums[: n * 2], 1):
            s += x
            heappush(q1, -x)
            if len(q1) > n:
                s -= -heappop(q1)
            pre[i] = s

        s = 0
        suf = [0] * (m + 1)
        q2 = []
        for i in range(m, n, -1):
            x = nums[i - 1]
            s += x
            heappush(q2, x)
            if len(q2) > n:
                s -= heappop(q2)
            suf[i] = s

        return min(pre[i] - suf[i + 1] for i in range(n, n * 2 + 1))

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