1035MediumArrayBinary Search

The Longest Equal Lengths From the Reels

Tracked in this browser only

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.

Reels of tape have the lengths reels. A reel may be cut into any number of pieces of whole-number lengths and any offcut may be thrown away, but pieces can never be joined together.

Obtain at least k pieces, all of the same whole-number length. Return the greatest length they can have, or 0 when not even a length of one can be managed.

Examples

Example 1

Input: reels = [9, 7, 5], k = 3
Output: 5

At a length of five each reel yields at least one piece, which is three in all. At six the reel of five yields nothing and the other two yield one apiece, leaving two pieces.

Example 2

Input: reels = [7, 5, 9], k = 4
Output: 4

At a length of four the reels yield one, one and two pieces, four in all. At five they yield one, one and one, which is one short.

Example 3

Input: reels = [5, 7, 9], k = 22
Output: 0

Even at a length of one the reels yield only twenty-one pieces between them, so no length can reach twenty-two.

Constraints

1 <= reels.length <= 10^5
1 <= reels[i] <= 10^5
1 <= k <= 10^9

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 max_length(reels: list[int], k: int) -> int:

Java

public int maxLength(int[] reels, int k)

The intended approach

Time O(m log L) for m reels and a longest reel of L.Space O(1).

For a chosen length, a reel yields its own length divided by that and rounded down, and the pieces the chosen length yields altogether is that summed over the reels. That count never rises as the length grows, so the largest workable length can be closed in on by halving. The range runs from one up to the longest reel, since no piece can be longer than the reel it came from. Testing a length is one pass over the reels, and the pass can stop early once enough pieces have been counted. Should even a length of one fail to yield `k` pieces, no longer length can either, and the halving leaves the answer at nothing on its own.

Pythonentry point: max_length

def max_length(reels, k):
    longest = 0
    for length in reels:
        if length > longest:
            longest = length
    lo = 1
    hi = longest
    best = 0
    while lo <= hi:
        mid = (lo + hi) // 2
        pieces = 0
        for length in reels:
            pieces += length // mid
            if pieces >= k:
                break
        if pieces >= k:
            best = mid
            lo = mid + 1
        else:
            hi = mid - 1
    return best

The obvious version

Try every length from one up to the longest reel in turn, counting the pieces each one yields, and keep the largest length that yields enough. Costs the longest reel times the reels' count 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_max_length(reels, k):
    longest = 0
    for length in reels:
        if length > longest:
            longest = length
    best = 0
    for size in range(1, longest + 1):
        pieces = 0
        for length in reels:
            pieces += length // size
        if pieces >= k:
            best = size
    return best

How this problem was checked

Reference solution executed against 17 test cases — 17 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

Create a free account to unlock the solution

See the step-by-step animation, the intuition, and clean code in every language — free, no credit card.

Create a free account Already have an account? Sign in