0503HardArrayHash TableMathBinary SearchCombinatoricsCountingNumber TheoryPrefix SumEuclidean AlgorithmGreatest Common Divisor

Gear Mesh Queries

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Trains the technique from

LeetCode 3312Sorted GCD Pair Queries

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 holds n gear blanks, and blank i has been cut with teeth[i] teeth. For every pair of different blanks i < j, the mesh of that pair is the largest whole number that divides both teeth[i] and teeth[j].

All n * (n - 1) / 2 meshes are written on a card and the card is then sorted into non-decreasing order. Two different pairs that share a mesh value each contribute their own entry, so the card always holds exactly n * (n - 1) / 2 entries.

Answer the batch of queries against that sorted card: for each queries[t] return the entry at position queries[t], counting positions from 0. Return the answers in the order the queries were asked.

Examples

Example 1

Input: teeth = [2, 3, 4], queries = [0, 1, 2]
Output: [1, 1, 2]

The three pairs mesh at 1, 2 and 1, so the sorted card reads 1, 1, 2 and the three queries read off its three positions in turn.

Example 2

Input: teeth = [12, 18, 24, 30], queries = [5, 0, 3, 3]
Output: [12, 6, 6, 6]

The six pairs mesh at 6, 12, 6, 6, 6 and 6, so the sorted card reads 6, 6, 6, 6, 6, 12. Position 5 holds 12, position 0 holds 6, and position 3 is asked twice and holds 6 both times.

Example 3

Input: teeth = [50000, 25000, 2], queries = [0, 1, 2]
Output: [2, 2, 25000]

The pair of large blanks meshes at 25000 and both pairs involving the small blank mesh at 2, so the sorted card reads 2, 2, 25000.

Constraints

2 <= n == teeth.length <= 10^5
1 <= teeth[i] <= 5 * 10^4
1 <= queries.length <= 10^5
0 <= queries[t] < n * (n - 1) / 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 mesh_queries(teeth: list[int], queries: list[int]) -> list[int]:

Java

public int[] meshQueries(int[] teeth, long[] queries)

The intended approach

Time O(V log V + n + q log V), where V is the largest tooth count: the multiples sweep is a harmonic sum and each query is a binary search.Space O(V): the tally, the exact-pair counts and the running totals.

Never build the card. Tally how many blanks carry each tooth count, then for each candidate mesh `d` add up the tallies of every multiple of `d` to learn how many blanks `d` divides, which turns into the number of pairs whose mesh is a multiple of `d`. Walking `d` downward and subtracting the counts already settled for `2d`, `3d` and onward leaves the number of pairs whose mesh is exactly `d`. A running total of those counts is non-decreasing in `d`, so the entry at a position is found by binary searching for the first `d` whose running total passes the position.

Pythonentry point: mesh_queries

def mesh_queries(teeth, queries):
    # Counting how many blanks are multiples of d gives the pairs whose mesh is a
    # multiple of d. Subtracting the tallies already settled for 2d, 3d and so on,
    # working downward, leaves the pairs whose mesh is exactly d. A running total
    # over d turns the sorted list into a step function, and a binary search over
    # that turns each query into a lookup.
    # O(V log V + n + q log V) time for V the largest tooth count, O(V) space.
    top = max(teeth)
    seen = [0] * (top + 1)
    for value in teeth:
        seen[value] += 1
    exact = [0] * (top + 2)
    for d in range(top, 0, -1):
        share = 0
        for m in range(d, top + 1, d):
            share += seen[m]
        pairs = share * (share - 1) // 2
        for m in range(2 * d, top + 1, d):
            pairs -= exact[m]
        exact[d] = pairs
    running = 0
    upto = [0] * (top + 1)
    for d in range(1, top + 1):
        running += exact[d]
        upto[d] = running
    out = []
    for want in queries:
        lo, hi = 1, top
        while lo < hi:
            mid = (lo + hi) // 2
            if upto[mid] > want:
                hi = mid
            else:
                lo = mid + 1
        out.append(lo)
    return out

The obvious version

Build the card literally: take the greatest common divisor of every pair of blanks with the Euclidean algorithm, sort the results and index straight into them. O(n^2 log V + n^2 log n).

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_mesh_queries(teeth, queries):
    from math import gcd
    meshes = []
    n = len(teeth)
    for i in range(n):
        for j in range(i + 1, n):
            meshes.append(gcd(teeth[i], teeth[j]))
    meshes.sort()
    return [meshes[want] for want in queries]

How this problem was checked

Reference solution executed against 12 test cases — 12 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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