0273MediumArrayBinary Search

Smallest Binder Comb for the Pass Budget

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Trains the technique from

LeetCode 1283Find the Smallest Divisor Given a Threshold

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 print shop binds jobs on a comb machine. pages[i] is the page count of job i, and every job is bound on its own.

A comb of size d grips at most d pages per pass, so job i occupies ceil(pages[i] / d) passes: the machine keeps taking full bites of d pages and a final short bite still uses up a whole pass.

Combs are stocked in every whole size from 1 upward, and the shop is prepared to run at most budget passes in total across all the jobs. Return the smallest comb size that keeps the total number of passes at or below budget.

budget is never smaller than the number of jobs, so some comb size always works.

Examples

Example 1

Input: pages = [5, 7], budget = 4
Output: 4

A comb of 4 takes job 0 in two passes, of 4 pages then 1, and job 1 in two passes, of 4 pages then 3. That is 4 passes altogether, which the budget allows.

Example 2

Input: pages = [4, 8, 12], budget = 6
Output: 4

A comb of 4 grips job 0 in a single pass, job 1 in two and job 2 in three, for exactly 6 passes.

Example 3

Input: pages = [17], budget = 1
Output: 17

A comb of 17 binds the one 17-page job in a single pass, which fits a budget of 1 pass.

Constraints

1 <= pages.length <= 5 * 10^4
1 <= pages[i] <= 10^6
pages.length <= budget <= 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 smallest_binder_comb(pages: list[int], budget: int) -> int:

Java

public int smallestBinderComb(int[] pages, int budget)

The intended approach

Time O(n log M), M the largest page countSpace O(1)

Fitting a wider comb can only reduce the passes a job needs, so once a comb size comes in under the budget every larger size does too. Bisect the interval from 1 to the largest job, costing each candidate comb with one sweep that sums the rounded-up pass counts, and keep the low end of the interval as the answer when it closes.

Pythonentry point: smallest_binder_comb

def smallest_binder_comb(pages, budget):
    # A larger comb never needs more passes on any job, so the comb sizes that
    # come in under the budget are exactly the sizes at or above the answer.
    # That lets the range 1..max(pages) be bisected, testing one candidate comb
    # with a single sweep over the jobs.
    # Time: O(n log M), M being the largest job.  Space: O(1).
    lo = 1
    hi = max(pages)
    while lo < hi:
        mid = (lo + hi) // 2
        passes = 0
        for p in pages:
            passes += -(-p // mid)
        if passes <= budget:
            hi = mid
        else:
            lo = mid + 1
    return lo

The obvious version

Try comb sizes 1, 2, 3, ... in order and return the first one whose total pass count is within the budget.

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_smallest_binder_comb(pages, budget):
    d = 1
    while True:
        passes = 0
        for p in pages:
            passes += -(-p // d)
        if passes <= budget:
            return d
        d += 1

How this problem was checked

Reference solution executed against 9 test cases — 9 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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