0988MediumArrayBinary SearchDynamic ProgrammingGreedySorting

Pairing Gears by Closest Tooth Count

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Trains the technique from

LeetCode 2616Minimize the Maximum Difference of Pairs

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 workshop's gears have the tooth counts teeth. A mesh joins two of the gears, and no gear may take part in more than one mesh. The slip of a mesh is how far apart its two tooth counts are.

Form exactly p meshes so that the largest slip among them is as small as it can be, and return that largest slip. When p is zero there is nothing to form and the answer is 0.

Examples

Example 1

Input: teeth = [8, 1, 2, 7, 5, 4], p = 2
Output: 1

Sorted, the counts read 1, 2, 4, 5, 7, 8. Meshing 1 with 2 and 4 with 5 gives two meshes that each slip by one, and no pair of meshes gets both slips down to nothing.

Example 2

Input: teeth = [0, 1, 2, 10, 11, 12], p = 3
Output: 8

Six gears and three meshes leaves nothing out, so one mesh has to bridge the low group and the high group. Meshing 0 with 1 and 11 with 12 leaves 2 with 10 as the narrowest bridge available.

Example 3

Input: teeth = [5, 8], p = 0
Output: 0

No meshes are asked for, so there is no slip to report.

Constraints

1 <= teeth.length <= 10^5
0 <= teeth[i] <= 10^9
0 <= p <= teeth.length / 2

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 minimize_max(teeth: list[int], p: int) -> int:

Java

public int minimizeMax(int[] teeth, int p)

The intended approach

Time O(n log n + n log M) for n gears with a spread of M: the sort, then one greedy pass per halving step.Space O(n) for the sorted counts.

Sort the tooth counts first. Then no mesh worth forming ever spans over a third count: if it did, swapping that middle count in for one of the ends would not widen the slip. So only gears that sit next to each other in sorted order need be meshed. Next, ask whether some slip is generous enough for `p` meshes. Walk the sorted counts from the left and take a neighbouring pair the moment its gap fits inside that slip, then skip past both. This greedy pass reaches as many meshes as any arrangement can, because the leftmost count either meshes with the one right after it or with nothing at all. A slip that is generous enough stays generous enough as it grows, so halve the range from zero up to the spread between the largest and smallest count until the smallest workable slip is pinned down.

Pythonentry point: minimize_max

def minimize_max(teeth, p):
    order = sorted(teeth)
    n = len(order)
    lo = 0
    hi = order[n - 1] - order[0]
    while lo < hi:
        mid = (lo + hi) // 2
        made = 0
        at = 1
        while at < n:
            if order[at] - order[at - 1] <= mid:
                made += 1
                at += 2
            else:
                at += 1
        if made >= p:
            hi = mid
        else:
            lo = mid + 1
    return lo

The obvious version

Work the sorted counts up with a table instead. For the first so many counts and so many meshes, hold the smallest largest slip reachable. The newest count either sits out, which carries the entry from one count back across, or meshes with the count before it, which costs the larger of that gap and the entry from two counts back with one mesh fewer. The last entry is the answer. Costs a cell per count and mesh rather than a pass per halving step.

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_minimize_max(teeth, p):
    order = sorted(teeth)
    n = len(order)
    far = order[n - 1] - order[0] + 1
    table = [[far] * (p + 1) for _ in range(n + 1)]
    for i in range(n + 1):
        table[i][0] = 0
    for i in range(2, n + 1):
        gap = order[i - 1] - order[i - 2]
        for k in range(1, p + 1):
            best = table[i - 1][k]
            joined = table[i - 2][k - 1]
            if gap > joined:
                joined = gap
            if joined < best:
                best = joined
            table[i][k] = best
    return table[n][p]

How this problem was checked

Reference solution executed against 15 test cases — 15 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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