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"Given a number N, find the numbers below N that are divisible by 3 or 5?" Is there any faster way to solve it without using bruteforce?

By Farnan.PUST, history, 7 years ago, In English

I can do it in O(N) time. but can it be more faster?

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7 years ago, # |
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Use inclusion/exclusion principle.

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7 years ago, # |
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floor(N / 3) + floor(N / 5) — floor(N / 15)

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7 years ago, # |
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As you said below N. So N cannot be accounted for. The formula will be

floor((N-1)/3) + floor((N-1)/5) — floor((N-1)/LCM(3,5))

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