0553HardArrayDynamic ProgrammingHeap (Priority Queue)

Auditor Discards from the Reading Ledger

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Trains the technique from

LeetCode 2163Minimum Difference in Sums After Removal of Elements

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A meter has written down 3 * n readings in the order they were taken, given as nums.

An auditor strikes out exactly n of the readings, chosen freely. The 2 * n readings still standing keep their original order: the earlier half of them, meaning the first n that survive, is the opening block, and the later half, the last n that survive, is the closing block.

The score of a set of strike-outs is the total of the opening block minus the total of the closing block. This score may be negative. Return the smallest score the auditor can reach.

Examples

Example 1

Input: nums = [4, 9, 2]
Output: -5

Strike out the last reading, 2. The opening block is 4 and the closing block is 9, so the score is 4 - 9 = -5.

Example 2

Input: nums = [6, 2, 8, 3, 9, 4]
Output: -9

Strike out the 3 at position 3 and the 4 at position 5. The survivors in order are 6, 2, 8, 9, so the opening block totals 8 and the closing block totals 17, giving a score of -9.

Example 3

Input: nums = [5, 5, 5, 5, 5, 5]
Output: 0

Every reading is 5, so whichever two are struck out both blocks total 10 and the score is 0.

Constraints

nums.length == 3 * n
1 <= n <= 10^5
1 <= nums[i] <= 10^5
The score therefore lies between -10^10 and 10^10.

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def minimum_difference(nums: list[int]) -> int:

Java

public long minimumDifference(int[] nums)

The intended approach

Time O(n log n): each of the two sweeps does one heap operation per reading.Space O(n) for the two tables and the two heaps.

Since the two blocks keep the ledger's order, there is always a cut position `i` such that the whole opening block comes from the first `i` readings and the whole closing block comes from the readings at position `i` onward. The cut can sit anywhere from `n` to `2 * n`, because each side must be able to supply `n` readings. For a fixed cut the two sides are chosen independently, so the opening block wants the `n` smallest readings on the left and the closing block wants the `n` largest on the right. Build a table of the left-hand totals by sweeping left to right while a max-heap holds the `n` smallest seen so far, replacing its largest whenever a smaller reading arrives, and build the right-hand table by the mirror sweep with a min-heap. Then take the smallest difference across the allowed cuts.

Pythonentry point: minimum_difference

def minimum_difference(nums):
    # O(n log n). Whatever is discarded, some cut position i splits the ledger so
    # the opening block is drawn entirely from the first i readings and the
    # closing block entirely from the rest, with i running from n to 2n. For each
    # cut, the best opening block is the n smallest readings on the left and the
    # best closing block is the n largest on the right, so sweep left to right
    # with a max-heap holding the n smallest so far, sweep right to left with a
    # min-heap holding the n largest so far, then pair the two tables up.
    import heapq
    total = len(nums)
    n = total // 3
    pre = [0] * (2 * n + 1)
    heap = []
    running = 0
    for i in range(2 * n):
        v = nums[i]
        if i < n:
            heapq.heappush(heap, -v)
            running += v
        else:
            biggest = -heap[0]
            if v < biggest:
                heapq.heapreplace(heap, -v)
                running += v - biggest
        if i + 1 >= n:
            pre[i + 1] = running
    suf = [0] * (2 * n + 1)
    heap2 = []
    running = 0
    for i in range(total - 1, n - 1, -1):
        v = nums[i]
        if total - i <= n:
            heapq.heappush(heap2, v)
            running += v
        else:
            smallest = heap2[0]
            if v > smallest:
                heapq.heapreplace(heap2, v)
                running += v - smallest
        if total - i >= n:
            suf[i] = running
    best = pre[n] - suf[n]
    for i in range(n + 1, 2 * n + 1):
        gap = pre[i] - suf[i]
        if gap < best:
            best = gap
    return best

The obvious version

Enumerate every set of 2 * n positions that could survive, in increasing order. The first n of those positions form the opening block and the rest the closing block, so add each up, subtract, and keep the smallest result.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_minimum_difference(nums):
    from itertools import combinations
    total = len(nums)
    n = total // 3
    best = None
    for keep in combinations(range(total), 2 * n):
        opening = 0
        for i in keep[:n]:
            opening += nums[i]
        closing = 0
        for i in keep[n:]:
            closing += nums[i]
        gap = opening - closing
        if best is None or gap < best:
            best = gap
    return best

How this problem was checked

Reference solution executed against 13 test cases — 13 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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