0282HardArrayHash TableSortingHeap (Priority Queue)Simulation

Busiest Dye Vat

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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 dye house owns vats interchangeable vats, numbered 0 through vats - 1.

batches[i] = [begin, finish] is a booked batch that asks for a vat at time begin and would hand it back at time finish. The batches are not listed in any particular order, but no two of them ask for a vat at the same time.

The floor supervisor deals with the batches in increasing order of their begin time. When a batch's turn comes:

if one or more vats are standing idle at that moment, the batch goes into the idle vat with the smallest number;
if every vat is occupied, the batch queues and starts the instant a vat is next handed back, still running for its booked length of finish - begin. Should more than one vat be handed back at that same instant, the batch takes the smallest of their numbers.

A vat counts as idle again at the very instant a batch hands it back, so a batch asking for a vat at that instant may use it.

Return the number of the vat that handles the most batches. When several vats handle that same number of batches, return the smallest of their numbers.

Examples

Example 1

Input: vats = 2, batches = [[0, 5], [1, 3], [6, 7]]
Output: 0

The batch beginning at 0 goes into vat 0 and hands it back at 5. The batch beginning at 1 finds vat 0 occupied and goes into vat 1, handing it back at 3. At time 6 both vats are idle, so the last batch goes into vat 0. Vat 0 handled two batches and vat 1 handled one.

Example 2

Input: vats = 3, batches = [[0, 100], [1, 100], [2, 3], [4, 5], [6, 7]]
Output: 2

Vats 0 and 1 are taken until time 100 by the first two batches. Each of the three short batches finds those two occupied and vat 2 idle, so vat 2 handles three batches while the others handle one each.

Example 3

Input: vats = 2, batches = [[0, 5], [1, 6]]
Output: 0

One batch goes into each vat, so both handled one batch and the smaller vat number is reported.

Constraints

1 <= vats <= 100
1 <= batches.length <= 10^5
batches[i].length == 2
0 <= begin_i < finish_i <= 5 * 10^5
All the values of begin_i are unique.

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 busiest_dye_vat(vats: int, batches: list[list[int]]) -> int:

Java

public int busiestDyeVat(int vats, int[][] batches)

The intended approach

Time O(m log m + m log vats)Space O(m + vats)

Order the batches by begin time and keep two min-heaps: one of idle vat numbers and one of busy vats keyed on release time with the vat number as the second key. Before placing a batch, move every busy vat whose release time has already arrived back into the idle heap. If the idle heap has anything, popping it yields the smallest idle number; otherwise popping the busy heap yields the earliest release with the smaller number winning a tie, and that vat is pushed back with its release time advanced by the batch's booked length. Count placements per vat and scan once with a strict comparison so ties keep the smaller number.

Pythonentry point: busiest_dye_vat

import heapq


def busiest_dye_vat(vats, batches):
    # Settle batches in begin-time order.  One min-heap holds the numbers of the
    # free vats, so the lowest free number pops first.  A second min-heap holds
    # (release time, vat) for the busy ones; before each batch every vat whose
    # release has already come is moved back to the free heap, and when nothing is
    # free the earliest release pops, with the lower vat number breaking a tie
    # because the number is the second key.
    # Time: O(m log m) to order the batches plus O(m log vats) of heap work.
    # Space: O(m + vats).
    free = list(range(vats))
    heapq.heapify(free)
    busy = []
    handled = [0] * vats
    for begin, finish in sorted(batches):
        while busy and busy[0][0] <= begin:
            _, v = heapq.heappop(busy)
            heapq.heappush(free, v)
        if free:
            v = heapq.heappop(free)
            heapq.heappush(busy, (finish, v))
        else:
            release, v = heapq.heappop(busy)
            heapq.heappush(busy, (release + finish - begin, v))
        handled[v] += 1
    best = 0
    for v in range(1, vats):
        if handled[v] > handled[best]:
            best = v
    return best

The obvious version

Order the batches by begin time and hold one release time per vat in a plain array. For each batch, scan the vats from number 0 and take the first whose release time has arrived; if none has, scan again for the smallest release time and add the booked length onto it. Tally placements and scan for the largest tally.

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_busiest_dye_vat(vats, batches):
    ready = [0] * vats
    handled = [0] * vats
    for begin, finish in sorted(batches):
        pick = -1
        for v in range(vats):
            if ready[v] <= begin:
                pick = v
                break
        if pick >= 0:
            ready[pick] = finish
        else:
            pick = 0
            for v in range(1, vats):
                if ready[v] < ready[pick]:
                    pick = v
            ready[pick] = ready[pick] + finish - begin
        handled[pick] += 1
    best = 0
    for v in range(1, vats):
        if handled[v] > handled[best]:
            best = v
    return best

How this problem was checked

Reference solution executed against 10 test cases — 10 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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