0683MediumArrayBinary Search

Minutes Until the Order Is Fired

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Trains the technique from

LeetCode 2187Minimum Time to Complete Trips

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 pottery works runs several kilns side by side. Kiln i needs cycle[i] minutes to fire one batch, and it starts the next batch the instant it finishes one, so it never idles. Every kiln loads its first batch at minute 0, and the kilns run independently of each other.

An order calls for batchesNeeded fired batches in total; it does not matter which kiln produced them.

Return the earliest whole minute at which the number of batches finished across all kilns reaches batchesNeeded. A batch counts only once its full cycle has elapsed.

Examples

Example 1

Input: cycle = [3, 6, 4], batchesNeeded = 9
Output: 12

At the reported minute the three kilns have finished 4, 2 and 3 batches, which is 9 in total. One minute earlier the counts are 3, 1 and 2, which is only 6.

Example 2

Input: cycle = [7, 7], batchesNeeded = 5
Output: 21

Both kilns finish a batch every 7 minutes, so the finished total is 2 at minute 7, 4 at minute 14 and 6 at minute 21. The order of 5 is first met at minute 21.

Example 3

Input: cycle = [2, 3], batchesNeeded = 1
Output: 2

Nothing is finished at minute 1. At minute 2 the first kiln completes its opening batch, which already meets an order of one.

Constraints

1 <= cycle.length <= 10^5
1 <= cycle[i], batchesNeeded <= 10^7
The answer is at most 10^14.

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 firing_minutes(cycle: list[int], batchesNeeded: int) -> int:

Java

public long firingMinutes(int[] cycle, int batchesNeeded)

The intended approach

Time O(n log(min(cycle) * batchesNeeded)) for n kilns.Space O(1).

By minute `t` kiln `i` has finished `t // cycle[i]` batches, so the finished total is a non-decreasing function of `t`: once the order is met at some minute it stays met afterwards. That lets the answer be found by halving an interval whose upper end is the fastest kiln doing the whole order by itself, evaluating the total at each midpoint and keeping the smaller half whenever the midpoint already meets the order.

Pythonentry point: firing_minutes

def firing_minutes(cycle, batchesNeeded):
    # the slowest possible schedule is the fastest kiln doing the whole order
    # alone, so that product is a safe upper end for the search
    lo = 1
    hi = min(cycle) * batchesNeeded
    # hi is at most 10^7 * 10^7 = 10^14, so every value stays exact in 64 bits
    while lo < hi:
        mid = (lo + hi) // 2
        done = 0
        for c in cycle:
            done += mid // c
            if done >= batchesNeeded:
                break
        if done >= batchesNeeded:
            hi = mid
        else:
            lo = mid + 1
    return lo

The obvious version

Keep the next completion minute of every kiln in a min-heap and pop them one batch at a time, pushing that kiln's following completion back in, until the popped batch is the one the order was waiting for. Its minute is the answer. Linear in the number of batches, so only usable on small orders, but it assumes nothing about monotonicity.

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

import heapq


def bf_firing_minutes(cycle, batchesNeeded):
    pending = [(c, c) for c in cycle]
    heapq.heapify(pending)
    finished = 0
    while True:
        when, span = heapq.heappop(pending)
        finished += 1
        if finished == batchesNeeded:
            return when
        heapq.heappush(pending, (when + span, span))

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

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