1011MediumArrayBinary Search

How Long the Bay Needs to Clear the Queue

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Trains the technique from

LeetCode 2594Minimum Time to Repair Cars

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 repair bay has mechanics whose ranks read ranks. A mechanic of rank r needs r minutes to finish one machine, four times r minutes to finish two, nine times r to finish three, and in general r times the square of however many machines that mechanic takes on.

The mechanics all work at the same time and none of them helps another. There are machines machines waiting, all alike, and every one has to be finished.

Return the fewest minutes after which the queue can be cleared.

Examples

Example 1

Input: ranks = [3, 7], machines = 5
Output: 28

In twenty-eight minutes the rank-three mechanic gets through three machines, needing twenty-seven of them, and the rank-seven mechanic gets through two, needing all twenty-eight. A minute less and the slower mechanic drops to a single machine, leaving the queue one short.

Example 2

Input: ranks = [2, 2], machines = 4
Output: 8

Two mechanics of the same rank split the four machines evenly, and two machines apiece takes eight minutes.

Example 3

Input: ranks = [1], machines = 1
Output: 1

One machine for the fastest rank there is takes a single minute.

Constraints

1 <= ranks.length <= 10^5
1 <= ranks[i] <= 100
1 <= machines <= 10^6

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 repair_cars(ranks: list[int], machines: int) -> int:

Java

public long repairCars(int[] ranks, int machines)

The intended approach

Time O(n + R log T) for R distinct ranks and an upper bound T on the minutes.Space O(R) for the rank tallies.

Turn the question round: given a stretch of minutes, how many machines does the bay clear? A mechanic of rank `r` takes on as many as fit, which is the largest count whose square times the rank stays inside the stretch, so the whole part of the square root of the stretch divided by the rank. Adding that over the mechanics says whether the stretch is long enough. A longer stretch never clears fewer machines, so the shortest stretch that suffices can be closed in on by halving. The fastest mechanic doing every machine alone is certainly long enough, which caps the range from the start. Ranks only go up to a hundred, and mechanics of the same rank take on the same count, so tally the ranks first. Each test of a stretch then costs a walk over at most a hundred distinct ranks rather than over every mechanic. The whole-part square root is taken from a floating estimate and then nudged into place, so nothing rests on the estimate being exact.

Pythonentry point: repair_cars

def repair_cars(ranks, machines):
    tally = {}
    quickest = ranks[0]
    for rank in ranks:
        tally[rank] = tally.get(rank, 0) + 1
        if rank < quickest:
            quickest = rank
    lo = 1
    hi = quickest * machines * machines
    while lo < hi:
        mid = (lo + hi) // 2
        done = 0
        for rank in tally:
            room = mid // rank
            count = int(room ** 0.5)
            while (count + 1) * (count + 1) <= room:
                count += 1
            while count > 0 and count * count > room:
                count -= 1
            done += count * tally[rank]
            if done >= machines:
                break
        if done >= machines:
            hi = mid
        else:
            lo = mid + 1
    return lo

The obvious version

Hand the machines out one at a time. Keep for every mechanic the minute at which their next machine would come off the bench, and always give the next machine to whichever mechanic would finish it soonest, then update that mechanic's next finishing minute. The minute the last machine comes off is the answer. Costs a step per machine rather than a halving search.

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_repair_cars(ranks, machines):
    import heapq
    heap = []
    for rank in ranks:
        heap.append((rank, rank, 1))
    heapq.heapify(heap)
    last = 0
    for _ in range(machines):
        when, rank, count = heapq.heappop(heap)
        last = when
        count += 1
        heapq.heappush(heap, (rank * count * count, rank, count))
    return last

How this problem was checked

Reference solution executed against 15 test cases — 15 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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