0224MediumMathBrainteaser

Greenhouse Vent Passes

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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.

A greenhouse roof carries vent_count vents in one long row, numbered 1 through vent_count. Every vent starts shut.

The controller then runs vent_count sweeps over the row. On sweep k it visits each vent whose number is a multiple of k and flips that vent: a shut vent opens and an open vent shuts. Sweep 1 therefore touches every vent, sweep 2 touches vents 2, 4, 6 and so on, up to sweep vent_count, which touches only the last vent.

Return how many vents are open after the final sweep. If vent_count is 0 the roof has no vents and no sweeps are run.

Examples

Example 1

Input: vent_count = 8
Output: 2

Sweep 1 opens all eight vents, sweep 2 flips the even-numbered ones, and each later sweep flips its own multiples. Two vents are open once sweep 8 has run.

Example 2

Input: vent_count = 27
Output: 5

Five of the twenty-seven vents are open after the twenty-seventh sweep.

Example 3

Input: vent_count = 49
Output: 7

Seven of the forty-nine vents are open once every sweep has run.

Constraints

0 <= vent_count <= 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 count_open_vents(vent_count: int) -> int:

Java

public int countOpenVents(int ventCount)

The intended approach

Time O(log n)Space O(1)

A vent ends open exactly when it is flipped an odd number of times, and vent v is flipped once per sweep that divides it. Divisors pair off into (a, v // a) except when the two halves coincide, so the vents that survive are the ones whose number is a square, and the answer is the count of squares up to vent_count. The code finds that count with a binary search for the largest integer whose square is within bounds, which avoids floating-point rounding at the top of the range.

Pythonentry point: count_open_vents

def count_open_vents(vent_count):
    low = 0
    high = vent_count
    while low < high:
        mid = (low + high + 1) // 2
        if mid * mid <= vent_count:
            low = mid
        else:
            high = mid - 1
    return low

The obvious version

Hold the state of every vent and actually run all the sweeps, flipping multiples one at a time, then count what is open. O(n 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_count_open_vents(vent_count):
    state = [False] * (vent_count + 1)
    for sweep in range(1, vent_count + 1):
        for vent in range(sweep, vent_count + 1, sweep):
            state[vent] = not state[vent]
    return sum(1 for vent in range(1, vent_count + 1) if state[vent])

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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