0899EasyArrayBinary SearchGreedySortingPrefix Sum

Parcels Within Each Allowance

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Trains the technique from

LeetCode 2389Longest Subsequence With Limited Sum

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.

Parcel weights are given as nums, and each entry of queries is a weight allowance.

For each allowance, work out the largest number of parcels that can be chosen so their weights total at most that allowance. Parcels may be chosen from anywhere in the list.

Return those counts, one per allowance, in the order the allowances are given.

Examples

Example 1

Input: nums = [17, 4, 23, 9], queries = [30, 4, 3, 53]
Output: [3, 1, 0, 4]

Sorted, the weights run 4, 9, 17, 23 with running totals 4, 13, 30, 53. An allowance of 30 reaches three parcels, one of 4 reaches one, one of 3 reaches none, and one of 53 takes all four.

Example 2

Input: nums = [1, 1, 1, 1, 1], queries = [1, 2, 3, 4, 5, 6]
Output: [1, 2, 3, 4, 5, 5]

Every parcel weighs one, so each allowance fits that many parcels, capped at the five available.

Example 3

Input: nums = [100, 200, 300], queries = [50]
Output: [0]

Even the lightest parcel is over the allowance, so none fit.

Constraints

1 <= nums.length <= 1000
1 <= queries.length <= 1000
1 <= nums[i] <= 10^6
1 <= queries[i] <= 10^6

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 answer_queries(nums: list[int], queries: list[int]) -> list[int]:

Java

public int[] answerQueries(int[] nums, int[] queries)

The intended approach

Time O(n log n + m log n) for n parcels and m allowances.Space O(n) for the running totals.

To fit as many parcels as possible under an allowance, take the lightest ones: for any chosen count, the lightest that many parcels have the smallest total, so if any choice of that size fits then the lightest ones do. So sort the weights and build running totals, where the `i`-th running total is the least weight any `i` parcels can come to. The running totals rise, so answering one allowance is a matter of finding how far along them it reaches, which a binary search settles. Every allowance is answered against the same totals, so the sorting is done once.

Pythonentry point: answer_queries

def answer_queries(nums, queries):
    order = sorted(nums)
    running = []
    total = 0
    for v in order:
        total += v
        running.append(total)
    out = []
    for allowance in queries:
        low = 0
        high = len(running)
        while low < high:
            mid = (low + high) // 2
            if running[mid] <= allowance:
                low = mid + 1
            else:
                high = mid
        out.append(low)
    return out

The obvious version

For each allowance, try every possible selection of parcels, keep the ones that fit, and take the largest count among them. Exponential in the parcel count and it never argues that the lightest parcels are the ones worth taking.

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_answer_queries(nums, queries):
    n = len(nums)
    out = []
    for allowance in queries:
        best = 0
        for mask in range(1 << n):
            total = 0
            count = 0
            for i in range(n):
                if mask >> i & 1:
                    total += nums[i]
                    count += 1
            if total <= allowance and count > best:
                best = count
        out.append(best)
    return out

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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