0634MediumArrayHash TableTwo PointersSorting

Pairing Crates Onto Pallets

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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 loading bay has crates on the floor; weights[i] is the weight of crate i in kilograms. A pallet is only accepted when it carries exactly target kilograms in exactly two crates.

One loading step chooses two crates still on the floor whose weights add up to target and moves both of them onto a pallet, out of reach of later steps. Return the largest number of loading steps that can be carried out.

Examples

Example 1

Input: weights = [1, 9, 3, 7, 5, 5], target = 10
Output: 3

Load 1 with 9, then 3 with 7, then the two crates of 5. Each pallet carries exactly 10 kilograms in two crates, and the floor is empty, so 3 steps are done.

Example 2

Input: weights = [2, 2, 2, 2], target = 4
Output: 2

Two crates of 2 make one pallet of 4, and the other two make a second. Each crate is used on one pallet only, so 2 steps are done.

Example 3

Input: weights = [1, 3, 3, 3], target = 4
Output: 1

The crate of 1 pairs with a crate of 3 for a pallet of 4. The two crates of 3 still on the floor add up to 6, not 4, so no further step is possible and the answer is 1.

Constraints

1 <= weights.length <= 10^5
1 <= weights[i] <= 10^9
1 <= target <= 10^9

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 max_operations(weights: list[int], target: int) -> int:

Java

public int maxOperations(int[] weights, int target)

The intended approach

Time O(n) — one pass with constant expected work per crate.Space O(n) for the tally of unpaired crates.

Keep a tally, keyed by weight, of crates that have been seen and are still unpaired. For each crate work out the weight its partner would need and look that up in the tally: when a crate of that weight is waiting, decrease its tally entry and count a loading step, otherwise add this crate to the tally. Because the tally stores counts rather than a plain membership flag, several crates of the same weight are all available in turn.

Pythonentry point: max_operations

def max_operations(weights, target):
    # O(n) time, O(n) space: sweep the floor once holding a tally of crates that
    # are still waiting for a partner. Each crate asks whether a crate of weight
    # target minus its own is waiting; if one is, the pair leaves together and
    # that tally drops, otherwise this crate joins the waiting tally.
    waiting = {}
    steps = 0
    for w in weights:
        need = target - w
        if waiting.get(need, 0) > 0:
            waiting[need] -= 1
            steps += 1
        else:
            waiting[w] = waiting.get(w, 0) + 1
    return steps

The obvious version

Repeatedly scan every remaining pair of crates, take the first pair found whose weights reach the target, delete both from the list, and start scanning again. Stop when a full scan finds no such pair.

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_max_operations(weights, target):
    left = list(weights)
    steps = 0
    while True:
        found = False
        for i in range(len(left)):
            for j in range(i + 1, len(left)):
                if left[i] + left[j] == target:
                    left.pop(j)
                    left.pop(i)
                    steps += 1
                    found = True
                    break
            if found:
                break
        if not found:
            return steps

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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