0556EasyMath

Reachable Batch Size

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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 crating line assembles an order by starting from one single unit and then applying a sequence of doublers, triplers and quintuplers: each device in the line multiplies the running count by 2, by 3 or by 5. Devices may be used in any order and any number of times, and the line may also be left empty.

A count n is reachable when some arrangement of those devices produces exactly n units. The count 1 is reachable, since the empty line already delivers one unit.

Given an integer n, return true when n is reachable and false when it is not. The line never produces zero or a negative count, so any n that is not strictly positive is unreachable.

Examples

Example 1

Input: n = 4050
Output: true

One doubler, four triplers and two quintuplers turn the single starting unit into 2 * 81 * 25 = 4050 units, so the count is reachable.

Example 2

Input: n = 385
Output: false

385 units would need a device that multiplies by 7 and another that multiplies by 11, and the line has neither.

Example 3

Input: n = -720
Output: false

The line only ever reports positive counts, so a negative target cannot be produced.

Constraints

-2^31 <= n <= 2^31 - 1

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 is_ugly(n: int) -> bool:

Java

public boolean isUgly(int n)

The intended approach

Time O(log n), since every division at least halves the value.Space O(1).

Reject anything that is not strictly positive up front. Then divide `n` by 2 as long as it divides evenly, then by 3 the same way, then by 5, so that what remains carries none of the three device factors. If the leftover is the starting unit the arrangement existed; any other leftover holds a factor no device can contribute.

Pythonentry point: is_ugly

def is_ugly(n):
    # O(log n) time, O(1) space: strip every factor of 2, 3 and 5 from n and
    # see whether anything other than the starting unit is left behind.
    if n <= 0:
        return False
    for step in (2, 3, 5):
        while n % step == 0:
            n //= step
    return n == 1

The obvious version

Factor `n` completely by trial division with every candidate divisor up to the square root, collect the prime factors in a list, and then confirm that every collected factor is one of 2, 3 and 5.

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_is_ugly(n):
    if n <= 0:
        return False
    if n == 1:
        return True
    factors = []
    rest = n
    d = 2
    while d * d <= rest:
        while rest % d == 0:
            factors.append(d)
            rest //= d
        d += 1
    if rest > 1:
        factors.append(rest)
    for f in factors:
        if f not in (2, 3, 5):
            return False
    return True

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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