0136MediumArrayTwo PointersSorting

Three-Weight Ballast Trim

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Trains the technique from

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 submersible carries a rack of ballast pods. Releasing a pod shifts the craft's trim by a signed amount, listed in pods: a positive amount lifts the nose, a negative amount drops it.

The pilot has to release exactly three pods, taken from three different rack positions, and wants the shifts to add up as near as possible to goal.

Return that combined shift, not the positions. Two different choices can land the same distance away on opposite sides of goal; when that happens, return the smaller of the two totals.

Examples

Example 1

Input: pods = [-4, 3, 9, 12], goal = 7
Output: 8

Releasing the pods worth -4, 3 and 9 shifts trim by 8, one short of the goal. The other three choices land at 11, 17 and 24, all further off.

Example 2

Input: pods = [-7, -4, -3, 6, 9], goal = 5
Output: 2

Two choices sit three away: -4, -3 and 9 total 2, while -7, 6 and 9 total 8. Nothing lands nearer, so the tie rule hands back the smaller total.

Example 3

Input: pods = [-1000, -1000, -1000, 1000], goal = -10000
Output: -3000

The rack cannot drop the nose anywhere near that far, so the lowest reachable total, -3000, is the closest it can get.

Constraints

3 <= pods.length <= 500
-1000 <= pods[i] <= 1000
-10^4 <= goal <= 10^4
The three pods must come from three distinct rack positions; equal amounts at different positions are different choices
If two reachable totals are equally far from goal, the smaller total is the answer

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 closest_trim_total(pods: list[int], goal: int) -> int:

Java

public int closestTrimTotal(int[] pods, int goal)

The intended approach

Time O(n^2)Space O(n) for the sorted copy

Sort a copy of the shifts, fix the first pod, then sweep the rest with two pointers closing in from both ends. A total under the goal can only improve by raising the low pointer, a total over it by lowering the high pointer, so every pointer step discards choices that are no nearer than the one just measured. Rank candidates by distance first and by smaller total second, which bakes the tie rule straight into the scan.

Pythonentry point: closest_trim_total

def closest_trim_total(pods, goal):
    order = sorted(pods)
    n = len(order)
    best = None
    for i in range(n - 2):
        lo = i + 1
        hi = n - 1
        while lo < hi:
            total = order[i] + order[lo] + order[hi]
            if best is None:
                best = total
            else:
                gap = abs(total - goal)
                held = abs(best - goal)
                if gap < held or (gap == held and total < best):
                    best = total
            if total < goal:
                lo += 1
            elif total > goal:
                hi -= 1
            else:
                return total
    return best

The obvious version

Walk all three-position choices with three nested loops and keep the winner under the same distance-then-smaller-total ranking, O(n^3).

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_closest_trim(pods, goal):
    n = len(pods)
    best = None
    for i in range(n):
        for j in range(i + 1, n):
            for k in range(j + 1, n):
                total = pods[i] + pods[j] + pods[k]
                if best is None:
                    best = total
                    continue
                gap = abs(total - goal)
                held = abs(best - goal)
                if gap < held or (gap == held and total < best):
                    best = total
    return best

How this problem was checked

Reference solution executed against 10 test cases — 10 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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