1018EasyArrayHash TableBinary SearchSorting

Swapping One Crate Each to Even the Loads

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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.

Two depots hold crates whose weights read mine and yours. The two depots carry different totals.

Swap exactly one of my crates for exactly one of yours so that afterwards both depots carry the same total. At least one such swap is possible.

Return the pair [a, b], where a is the weight of the crate leaving my depot and b the weight of the crate leaving yours. Where several swaps work, return the one whose a is smallest, and among those the one whose b is smallest.

Examples

Example 1

Input: mine = [35, 17, 4, 24, 10], yours = [63, 21]
Output: [24, 21]

My depot carries ninety and yours eighty-four, so the crate I give up has to be three heavier than the one I take. Handing over 24 for 21 leaves both depots at eighty-seven. My lighter crates, 4, 10 and 17, have no partner three lighter among yours.

Example 2

Input: mine = [2, 4, 6], yours = [2, 4, 2]
Output: [4, 2]

Mine total twelve and yours eight, so I hand over a crate two heavier than the one I take. Giving up 4 for 2 leaves both at ten. Giving up my 2 would need a crate of nothing to come back.

Example 3

Input: mine = [3, 7], yours = [1, 5]
Output: [3, 1]

Mine total ten and yours six, so again the outgoing crate is two heavier. Both swaps work here, 3 for 1 and 7 for 5, and the rule asks for the one whose outgoing crate is lighter.

Constraints

1 <= mine.length <= 10^4
1 <= yours.length <= 10^4
1 <= mine[i] <= 10^5
1 <= yours[i] <= 10^5
The two depots carry different totals.
At least one swap works.

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 fair_candy_swap(mine: list[int], yours: list[int]) -> list[int]:

Java

public int[] fairCandySwap(int[] mine, int[] yours)

The intended approach

Time O(a log a + b) for the sort of my crates and the lookup over yours.Space O(b) for the lookup.

Sending `a` across and taking `b` back moves my total by `b - a` and yours by the same amount the other way. For the two to land together, my total has to end up at half of what the two carry between them, so `b - a` must come to half the amount yours currently leads mine by. Call that the shift; it is settled before a single crate is examined. So gather your crate weights into a lookup, then walk mine from lightest to heaviest and take the first weight whose partner, that weight plus the shift, turns up among yours. Walking lightest first gives the smallest `a` outright, and once `a` is fixed the partner is forced, so there is nothing to compare for `b`. Were the two totals to differ by an odd amount, no swap could work at all, since the shift would not be a whole weight — but the promise of an answer rules that out.

Pythonentry point: fair_candy_swap

def fair_candy_swap(mine, yours):
    sum_mine = 0
    for weight in mine:
        sum_mine += weight
    sum_yours = 0
    for weight in yours:
        sum_yours += weight
    shift = (sum_yours - sum_mine) // 2
    theirs = set(yours)
    for weight in sorted(mine):
        if weight + shift in theirs:
            return [weight, weight + shift]
    return []

The obvious version

Try every crate of mine against every crate of yours, working out both totals after that swap and keeping the pair when they match, then take the smallest such pair. Costs every pair rather than one shift and a lookup.

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_fair_candy_swap(mine, yours):
    sum_mine = 0
    for weight in mine:
        sum_mine += weight
    sum_yours = 0
    for weight in yours:
        sum_yours += weight
    best = None
    for a in mine:
        for b in yours:
            if sum_mine - a + b == sum_yours - b + a:
                pair = [a, b]
                if best is None or pair < best:
                    best = pair
    return best if best is not None else []

How this problem was checked

Reference solution executed against 14 test cases — 14 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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