0938MediumArrayDynamic ProgrammingGreedy

Most Units Bought From the Bundle Sale

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Trains the technique from

LeetCode 3946Maximum Number of Items From Sale I

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 sale lists its lines as items, where items[i] is [factor, price].

Line i is sold only as a whole bundle: taking it means buying exactly factor units at price each, so the bundle costs factor times price and yields factor units. A line may be taken once or left alone.

With budget to spend, return the greatest number of units that can be bought.

Examples

Example 1

Input: items = [[3, 4], [7, 2], [2, 9], [5, 3]], budget = 40
Output: 12

The four bundles cost 12, 14, 18 and 15 and yield 3, 7, 2 and 5 units. Taking the second and fourth costs 29 and yields 12 units, and no choice inside 40 does better.

Example 2

Input: items = [[10, 1], [1, 10]], budget = 10
Output: 10

One bundle costs 10 and yields 10 units, the other costs 10 and yields 1, so the budget goes on the first.

Example 3

Input: items = [[1, 2]], budget = 1
Output: 0

The only bundle costs 2, which is over budget, so nothing can be bought.

Constraints

1 <= items.length <= 1000
items[i].length == 2
1 <= items[i][0] <= 1500
1 <= items[i][1] <= 1500
1 <= budget <= 1500

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 maximum_sale_items(items: list[list[int]], budget: int) -> int:

Java

public int maximumSaleItems(int[][] items, int budget)

The intended approach

Time O(n * budget) for n lines.Space O(budget) for the figures.

Every line is all or nothing, with a fixed cost and a fixed yield, so this is a straight packing question over the budget: for each amount of money, the most units reachable. Taking the lines one at a time, a line either goes unused, leaving the figure for that amount as it was, or it is taken, which draws on the figure for the amount left after paying for it and adds its units. Sweeping the amounts downward while folding in each line is what keeps a line from being taken twice. Lines costing more than the whole budget can be skipped outright. The answer is the figure for the full budget, since spending less is always allowed and the figures only rise. Taking the cheapest bundles first does not work, because a dearer bundle can yield far more units for the money, and the two orderings pull in opposite directions.

Pythonentry point: maximum_sale_items

def maximum_sale_items(items, budget):
    best = [0] * (budget + 1)
    for factor, price in items:
        cost = factor * price
        if cost > budget:
            continue
        for amount in range(budget, cost - 1, -1):
            candidate = best[amount - cost] + factor
            if candidate > best[amount]:
                best[amount] = candidate
    return best[budget]

The obvious version

Try every choice of which lines to take, add up the cost and the units of each choice, and keep the most units among the choices that fit the budget. Exponential in the line count and it packs nothing.

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_maximum_sale_items(items, budget):
    n = len(items)
    best = 0
    for mask in range(1 << n):
        cost = 0
        units = 0
        for i in range(n):
            if mask >> i & 1:
                cost += items[i][0] * items[i][1]
                units += items[i][0]
        if cost <= budget and units > best:
            best = units
    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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