0461MediumArrayBinary Search

Longest Lead Cut From The Reels

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Trains the technique from

LeetCode 2226Maximum Candies Allocated to K Children

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 depot holds cable on reels, reels[i] being the whole number of metres left on reel i. An order calls for leads service leads, and every lead must be cut to the same whole number of metres.

A reel can be cut into as many leads as fit, and whatever is left over on it is scrap. A lead is one continuous run of cable, so it must come from a single reel: two short offcuts cannot be joined into one lead. Making more leads than the order asks for is fine, and reels may be left untouched.

Return the greatest length, in whole metres, that lets the order be filled. Return 0 when the depot cannot fill it even with one-metre leads.

Examples

Example 1

Input: reels = [16, 9, 27], leads = 7
Output: 6

At 6 metres the reels give 2, 1 and 4 leads, which is the 7 the order asks for. At 7 metres they give only 6.

Example 2

Input: reels = [4], leads = 5
Output: 0

The single reel holds 4 metres, so even one-metre leads come to 4, one short of the order.

Example 3

Input: reels = [100, 100], leads = 4
Output: 50

Each reel gives two 50-metre leads, filling the order of 4 exactly.

Constraints

1 <= reels.length <= 10^5
1 <= reels[i] <= 10^7
1 <= leads <= 10^12

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 longest_lead(reels: list[int], leads: int) -> int:

Java

public int longestLead(int[] reels, long leads)

The intended approach

Time O(n log M) for n reels and a longest reel of M metres.Space O(1).

If some length fills the order then every shorter length fills it too, since each reel yields at least as many shorter leads. So the lengths that work form an unbroken run from 1 up to the answer, and the answer can be hunted by halving the range between 1 and the longest reel, counting for each candidate how many leads the reels give by whole division and keeping the largest candidate that reaches the order.

Pythonentry point: longest_lead

def longest_lead(reels, leads):
    # A length that can be met is still met by any shorter length, so the
    # workable lengths form a run starting at 1. Halve the range of candidate
    # lengths, counting how many leads each candidate yields.
    # O(n log M) time for n reels of at most M metres, O(1) extra space.
    lo = 1
    hi = 0
    for r in reels:
        if r > hi:
            hi = r
    best = 0
    while lo <= hi:
        mid = (lo + hi) // 2
        made = 0
        for r in reels:
            made += r // mid
            if made >= leads:
                break
        if made >= leads:
            best = mid
            lo = mid + 1
        else:
            hi = mid - 1
    return best

The obvious version

Try every length from 1 up to the longest reel, count the leads it yields, and keep the largest length that fills the order. O(n * M).

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_longest_lead(reels, leads):
    best = 0
    top = max(reels)
    for length in range(1, top + 1):
        made = 0
        for r in reels:
            made += r // length
        if made >= leads and length > best:
            best = length
    return best

How this problem was checked

Reference solution executed against 14 test cases — 14 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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