0484MediumArrayGreedySortingCounting Sort

Flea Market Record Haul

Tracked in this browser only

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 collector arrives at a flea market holding budget tokens, the only currency the stalls accept.

One stall has a crate of second-hand records laid out. Record i is tagged at prices[i] tokens. The collector may take away any group of records, in any combination, as long as the tags of the records taken add up to no more than budget. Records are sold whole: there is no haggling, no swapping and no part payment, and each record in the crate can be taken at most once even when two of them carry the same tag.

Return the largest number of records the collector can walk away with.

Examples

Example 1

Input: prices = [9, 4, 12, 4, 30], budget = 17
Output: 3

Taking the two records tagged 4 and the one tagged 9 uses 4 + 4 + 9 = 17 tokens, which is the whole budget, and hands over three records.

Example 2

Input: prices = [10, 3, 3, 3], budget = 11
Output: 3

The three records tagged 3 cost 9 tokens together, leaving 2 tokens unspent and three records in hand.

Example 3

Input: prices = [40, 25], budget = 12
Output: 0

Both tags are above the whole budget, so nothing can be taken and the collector leaves empty-handed.

Constraints

prices.length == n
1 <= n <= 10^5
1 <= prices[i] <= 10^5
1 <= budget <= 10^8

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_records(prices: list[int], budget: int) -> int:

Java

public int maxRecords(int[] prices, int budget)

The intended approach

Time O(n log n)Space O(n) for the sorted copy

Sort the tags into ascending order and walk them, subtracting each tag from the remaining tokens and counting the record, until a tag no longer fits. Exchanging any record in a chosen group for an unused cheaper one never lowers the number of records and never raises the bill, so the ascending prefix is as long as any affordable group can be.

Pythonentry point: max_records

def max_records(prices, budget):
    # Cheapest first: any affordable set can be swapped item-for-item towards the
    # cheapest records without shrinking, so one ascending sweep settles it.
    # time O(n log n) for the sort, space O(n) for the sorted copy
    ordered = sorted(prices)
    bought = 0
    for price in ordered:
        if price > budget:
            break
        budget -= price
        bought += 1
    return bought

The obvious version

Keep the tags in a mutable pool. Each round scans the whole pool to locate the smallest remaining tag, removes it, and pays for it if the tokens stretch that far, stopping the moment one does not. Quadratic, but it never sorts anything.

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_records(prices, budget):
    pool = list(prices)
    left = budget
    taken = 0
    while pool:
        cheapest = 0
        for i in range(1, len(pool)):
            if pool[i] < pool[cheapest]:
                cheapest = i
        price = pool.pop(cheapest)
        if price > left:
            return taken
        left -= price
        taken += 1
    return taken

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

Create a free account to unlock the solution

See the step-by-step animation, the intuition, and clean code in every language — free, no credit card.

Create a free account Already have an account? Sign in