0240MediumArrayHash TablePrefix Sum

Trim the Ballast Tanks

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Trains the technique from

LeetCode 1674Minimum Moves to Make Array Complementary

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 survey ship carries an even number of ballast tanks in one line from bow to stern. Write n for the number of tanks. Tank i is mirrored by tank n - 1 - i on the far side of midships. You are given levels, where levels[i] is how many units tank i holds right now, and an integer capacity giving the largest load any tank may hold.

A refill takes one tank and resets it to any load from 1 to capacity inclusive. Each refill counts as one, you may refill the same tank more than once, and a refill is free to leave a tank at the load it already had.

The ship is trimmed when every mirrored pair carries the same combined load, that is, levels[i] + levels[n - 1 - i] is one and the same value for every i.

Return the fewest refills that leave the ship trimmed.

Examples

Example 1

Input: levels = [2, 7, 6, 3, 5, 2], capacity = 9
Output: 2

Resetting the fifth tank to 2 and the sixth tank to 7 gives loads [2, 7, 6, 3, 2, 7]. The mirrored pairs are 2 + 7, 7 + 2 and 6 + 3, all totalling 9, which takes 2 refills.

Example 2

Input: levels = [2, 5, 3, 6], capacity = 7
Output: 0

The mirrored pairs are 2 + 6 and 5 + 3, both totalling 8, so the ship is already trimmed.

Example 3

Input: levels = [2, 4, 4, 2], capacity = 4
Output: 2

Resetting the third tank to 2 and the fourth tank to 4 gives loads [2, 4, 2, 4], whose mirrored pairs are 2 + 4 and 4 + 2, both totalling 6, which takes 2 refills.

Constraints

n == levels.length
2 <= n <= 10^5
1 <= levels[i] <= capacity <= 10^5
n is even

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 min_refills(levels: list[int], capacity: int) -> int:

Java

public int minRefills(int[] levels, int capacity)

The intended approach

Time O(n + capacity)Space O(capacity) for the difference array over pair totals

Fix the shared pair total `T`, which must lie in `[2, 2 * capacity]`. A pair holding `low` and `high` then costs 0 refills when `T` equals `low + high`, 1 refill for every `T` from `low + 1` to `high + capacity` because resetting either side reaches those totals, and 2 refills otherwise. Instead of scoring each `T` against each pair, the code writes each pair's step function onto a difference array indexed by `T`: plus 2 over the whole legal range, minus 1 over the one-refill span, and minus 1 more at the pair's current total. One prefix sweep over the totals then reads the cost of every `T` and keeps the smallest.

Pythonentry point: min_refills

def min_refills(levels, capacity):
    n = len(levels)
    delta = [0] * (2 * capacity + 2)
    for i in range(n // 2):
        bow = levels[i]
        stern = levels[n - 1 - i]
        low = bow if bow < stern else stern
        high = bow + stern - low
        delta[2] += 2
        delta[low + 1] -= 1
        delta[high + capacity + 1] += 1
        delta[bow + stern] -= 1
        delta[bow + stern + 1] += 1
    best = None
    running = 0
    for total in range(2, 2 * capacity + 1):
        running += delta[total]
        if best is None or running < best:
            best = running
    return best

The obvious version

Try every legal pair total and, for each one, price each mirrored pair directly by asking whether it already matches, can be fixed by resetting one side, or needs both. O(n * capacity).

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_min_refills(levels, capacity):
    n = len(levels)
    best = None
    for total in range(2, 2 * capacity + 1):
        cost = 0
        for i in range(n // 2):
            bow = levels[i]
            stern = levels[n - 1 - i]
            if bow + stern == total:
                continue
            low = min(bow, stern)
            high = max(bow, stern)
            if low + 1 <= total <= high + capacity:
                cost += 1
            else:
                cost += 2
        if best is None or cost < best:
            best = cost
    return best

How this problem was checked

Reference solution executed against 11 test cases — 11 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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