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Asked: May 10, 20262026-05-10T19:12:15+00:00 2026-05-10T19:12:15+00:00

While calculating the hash table bucket index from the hash code of a key,

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While calculating the hash table bucket index from the hash code of a key, why do we avoid use of remainder after division (modulo) when the size of the array of buckets is a power of 2?

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  1. 2026-05-10T19:12:15+00:00Added an answer on May 10, 2026 at 7:12 pm

    When calculating the hash, you want as much information as you can cheaply munge things into with good distribution across the entire range of bits: e.g. 32-bit unsigned integers are usually good, unless you have a lot (>3 billion) of items to store in the hash table.

    It’s converting the hash code into a bucket index that you’re really interested in. When the number of buckets n is a power of two, all you need to do is do an AND operation between hash code h and (n-1), and the result is equal to h mod n.

    A reason this may be bad is that the AND operation is simply discarding bits – the high-level bits – from the hash code. This may be good or bad, depending on other things. On one hand, it will be very fast, since AND is a lot faster than division (and is the usual reason why you would choose to use a power of 2 number of buckets), but on the other hand, poor hash functions may have poor entropy in the lower bits: that is, the lower bits don’t change much when the data being hashed changes.

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