0793EasyArrayGreedySorting

Least Paid at the Clamp Stall

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Trains the technique from

LeetCode 2144Minimum Cost of Buying Candies With Discount

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 hardware stall prices every clamp on its table separately; price[i] is the price of the i-th clamp. You want to take all of them home, and the stall runs this offer.

Split the clamps into groups however you like, so that every clamp belongs to exactly one group. In a group of exactly three clamps, the cheapest clamp of that group is free and you pay for the other two. In a group of any other size, you pay for every clamp in it.

Return the smallest total you can pay for all of the clamps.

Examples

Example 1

Input: price = [8, 3, 6, 1]
Output: 15

Put the clamps priced 8, 6 and 3 in one group: 3 is the cheapest of that group, so it is free and you pay 8 and 6. The clamp priced 1 sits in a group of its own and is paid for, giving 8 + 6 + 1 = 15.

Example 2

Input: price = [40, 40, 40, 40]
Output: 120

Group three of the clamps, taking one of them free and paying for the other two, and leave the fourth in a group of its own, giving 40 + 40 + 40 = 120.

Example 3

Input: price = [9]
Output: 9

One clamp cannot form a group of three, so it is paid for.

Constraints

1 <= price.length <= 100
1 <= price[i] <= 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 least_paid(price: list[int]) -> int:

Java

public int leastPaid(int[] price)

The intended approach

Time O(n log n) for n clamps: one sort, then one pass.Space O(n) for the sorted copy of the prices.

Sort the prices from dearest to cheapest and read them off in threes: the first two of each block of three are paid for and the third one is taken free, and any one or two prices left at the tail are paid for. The running sum skips every position whose index leaves remainder two on division by three.

Pythonentry point: least_paid

def least_paid(price):
    # dearest first, then every third item along is the free one of its group
    ordered = sorted(price, reverse=True)
    total = 0
    for i, p in enumerate(ordered):
        if i % 3 != 2:
            total += p
    return total

The obvious version

Try every subset of clamps as the set taken free. A subset is reachable when the paid clamps can be handed out two per free clamp with each paid clamp priced at least as high as the free clamp it covers, which is checked by pairing the dearest free clamps with the dearest paid clamps. Keep the smallest bill over all reachable subsets. Exponential.

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_least_paid(price):
    n = len(price)
    whole = sum(price)
    best = whole
    for mask in range(1 << n):
        free = [price[i] for i in range(n) if (mask >> i) & 1]
        paid = [price[i] for i in range(n) if not ((mask >> i) & 1)]
        if len(paid) < 2 * len(free):
            continue
        free.sort(reverse=True)
        paid.sort(reverse=True)
        ok = True
        k = 0
        for f in free:
            if paid[k] < f or paid[k + 1] < f:
                ok = False
                break
            k += 2
        if ok:
            bill = whole - sum(free)
            if bill < best:
                best = bill
    return best

How this problem was checked

Reference solution executed against 12 test cases — 12 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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