0469MediumArrayGreedyHeap (Priority Queue)

Compactor Passes at the Scrap Yard

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Trains the technique from

LeetCode 1962Remove Stones to Minimize the Total

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 scrap yard weighs the shredded metal sitting in each of its bins: bins[i] is the load of bin i in kilos. A booked slot on the yard's compactor gives the crew exactly passes passes, and every one of them has to be used.

A single pass is aimed at one bin of the crew's choosing. It takes away half of that bin's current load, rounded down, and leaves the rest in the bin. A bin holding one kilo therefore loses nothing when a pass is aimed at it, and the same bin may be chosen on as many passes as the crew likes.

Return the smallest total load, across all bins, that can be left in the yard once the booked passes have all been used.

Examples

Example 1

Input: bins = [12, 5], passes = 3
Output: 6

The loads can be walked down to 3 and 3 by aiming two passes at the first bin and one at the second, leaving 6 kilos in the yard.

Example 2

Input: bins = [7, 7, 7], passes = 2
Output: 15

Two of the bins drop from 7 to 4 kilos and the third is untouched, so 15 kilos remain.

Example 3

Input: bins = [26, 4, 18, 11], passes = 5
Output: 22

One workable slot ends with loads 7, 4, 5 and 6, which add up to the figure returned.

Constraints

1 <= bins.length <= 10^5
1 <= bins[i] <= 10^4
1 <= passes <= 10^5

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_scrap_left(bins: list[int], passes: int) -> int:

Java

public int leastScrapLeft(int[] bins, int passes)

The intended approach

Time O(n + passes * log n)Space O(n) for the heap

Each pass is independent of the others once its target is fixed, and the amount a pass sheds grows with the load it lands on, so keep the loads in a heap ordered largest first and always spend the next pass on the load at the top. Replace that load with what is left of it and subtract the shed kilos from a running total. Once the top of the heap is down to a single kilo no remaining pass can shed anything, so the loop can stop there.

Pythonentry point: least_scrap_left

import heapq


def least_scrap_left(bins, passes):
    # A max-heap keyed on the negated load; one pop-and-push per pass.
    # time O(n + passes * log n), space O(n) for the heap
    heap = [-load for load in bins]
    heapq.heapify(heap)
    total = sum(bins)
    for _ in range(passes):
        top = -heap[0]
        shed = top // 2
        if shed == 0:
            break                      # every bin is down to one kilo
        total -= shed
        heapq.heapreplace(heap, -(top - shed))
    return total

The obvious version

Hold the loads in a plain list and simulate the booking pass by pass, each time walking the whole list to find the heaviest bin and updating it in place. Add the list up at the end.

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_scrap_left(bins, passes):
    load = list(bins)
    for _ in range(passes):
        pick = 0
        for i in range(1, len(load)):
            if load[i] > load[pick]:
                pick = i
        load[pick] -= load[pick] // 2
    return sum(load)

How this problem was checked

Reference solution executed against 8 test cases — 8 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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