0471MediumArrayBinary Search

Poster Run on One Press

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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 print shop is quoting a run of identical posters. It keeps inks different inks, numbered 0 to inks - 1, and presses presses, numbered 0 to presses - 1.

Press i is plumbed for its own mix: turning out one poster on it consumes exactly recipe[i][j] grams of ink j, for every ink j. The shop already holds onhand[j] grams of ink j, which cost nothing to use, and it can order any whole number of extra grams of ink j at price[j] per gram. The total order may cost at most budget.

The whole run has to come off a single press; posters cannot be split between presses. Ink left over is simply left over, and it can never be sold or traded towards the cost of a different ink.

Return the largest number of posters the shop can turn out.

Examples

Example 1

Input: inks = 2, presses = 2, budget = 30, recipe = [[2, 1], [1, 3]], onhand = [6, 8], price = [4, 2]
Output: 7

Press 1 turns out seven posters: it needs 7 grams of ink 0 and 21 of ink 1, so the shop orders 1 gram of ink 0 for 4 and 13 grams of ink 1 for 26, spending 30 in total.

Example 2

Input: inks = 3, presses = 1, budget = 0, recipe = [[1, 2, 1]], onhand = [10, 7, 12], price = [3, 3, 3]
Output: 3

Nothing can be ordered, so the run is capped by the ink already in the cupboard; ink 1 runs out first at three posters.

Example 3

Input: inks = 2, presses = 1, budget = 6, recipe = [[1, 1]], onhand = [20, 0], price = [5, 2]
Output: 3

Ink 0 is plentiful and ink 1 is bare, so every poster needs a gram of ink 1 ordered at 2, and the budget of 6 covers three of them.

Constraints

1 <= inks, presses <= 100
0 <= budget <= 10^8
recipe.length == presses
recipe[i].length == inks
1 <= recipe[i][j] <= 100
onhand.length == price.length == inks
0 <= onhand[j] <= 10^8
1 <= price[j] <= 100

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 most_posters(inks: int, presses: int, budget: int, recipe: list[list[int]], onhand: list[int], price: list[int]) -> int:

Java

public int mostPosters(int inks, int presses, int budget, List<List<Integer>> recipe, List<Integer> onhand, List<Integer> price)

The intended approach

Time O(presses * inks * log(budget + min(onhand)))Space O(1)

For a fixed press, costing out a run of a given length is a short sum: per ink, buy only the shortfall between what the run needs and what is on hand, and charge it at that ink's price. A longer run never costs less, so for each press bisect the run length against the budget, carrying the best length found so far in as the low end so later presses only get probed for an improvement. Every press needs at least one gram of every ink, which bounds any run by the budget plus the smallest holding.

Pythonentry point: most_posters

def most_posters(inks, presses, budget, recipe, onhand, price):
    # Feasibility only ever worsens as the run grows, so bisect the run length
    # for each press separately and keep the best.
    # time O(presses * inks * log(budget + min(onhand))), space O(1)
    ceiling = budget + min(onhand)
    best = 0
    for i in range(presses):
        row = recipe[i]
        lo, hi = best, ceiling
        while lo < hi:
            mid = (lo + hi + 1) // 2
            spend = 0
            for j in range(inks):
                short = mid * row[j] - onhand[j]
                if short > 0:
                    spend += short * price[j]
                    if spend > budget:
                        break
            if spend <= budget:
                lo = mid
            else:
                hi = mid - 1
        if lo > best:
            best = lo
    return best

The obvious version

Walk every candidate run length from zero up to the budget plus the smallest holding, cost the length out on each press in turn, and remember the largest length any press can afford. No assumption is made about how cost behaves as the length grows.

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_most_posters(inks, presses, budget, recipe, onhand, price):
    ceiling = budget + min(onhand)
    best = 0
    for target in range(ceiling + 1):
        for i in range(presses):
            row = recipe[i]
            spend = 0
            for j in range(inks):
                short = target * row[j] - onhand[j]
                if short > 0:
                    spend += short * price[j]
            if spend <= budget:
                if target > best:
                    best = target
                break
    return best

How this problem was checked

Reference solution executed against 7 test cases — 7 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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