0196MediumArrayMathEnumerationNumber TheoryPrimality TestSieve TheoryPrime Number Sieve

Indivisible Catalogue Numbers

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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 parts depot gives every stock item a whole-number catalogue code. The stocktaking team calls a code indivisible when the code is 2 or larger and no whole number other than 1 and the code itself divides it exactly.

A clerk is auditing the codes below a cut-off limit. Report how many indivisible codes are smaller than limit. Codes equal to limit are outside the audit window and must not be counted, and when limit is so small that the window holds nothing, the tally is 0.

Examples

Example 1

Input: limit = 25
Output: 9

The indivisible codes under the cut-off are 2, 3, 5, 7, 11, 13, 17, 19 and 23, so the clerk records nine of them.

Example 2

Input: limit = 3
Output: 1

Only the code 2 sits under the cut-off. The code 3 equals the cut-off, so it stays out of the window.

Example 3

Input: limit = 97
Output: 24

Twenty-four codes below 97 have no divisor other than 1 and themselves; 97 itself is indivisible but is excluded by the cut-off.

Constraints

0 <= limit <= 5 * 10^6

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_indivisible_codes(limit: int) -> int:

Java

public int countIndivisibleCodes(int limit)

The intended approach

Time O(n log log n)Space O(n)

Testing each code on its own repeats the same divisibility work over and over. Instead build one flag table for the whole window and strike out multiples: once a code survives as indivisible, every multiple of it starting at its square is composite, because any smaller multiple already carries a smaller factor that was handled earlier. Striking out from the square onward means the outer walk only has to run while the square still fits inside the window, and the answer is the number of surviving flags.

Pythonentry point: count_indivisible_codes

def count_indivisible_codes(limit):
    # Time: O(n log log n) where n = limit. Space: O(n) for the flag table.
    if limit < 3:
        return 0
    flags = bytearray([1]) * limit
    flags[0] = 0
    flags[1] = 0
    step = 2
    while step * step < limit:
        if flags[step]:
            start = step * step
            flags[start::step] = bytearray(len(range(start, limit, step)))
        step += 1
    return sum(flags)

The obvious version

Walk every code in the window and trial-divide it by each candidate factor up to its square root, counting the codes that survive. No shared table, no striking out.

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_indivisible_codes(limit):
    tally = 0
    for code in range(2, limit):
        factor = 2
        indivisible = True
        while factor * factor <= code:
            if code % factor == 0:
                indivisible = False
                break
            factor += 1
        if indivisible:
            tally += 1
    return tally

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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