Reduce memory yes, speed-up depends.

If he needs the number of different(!) elements, there would be a tremendous speed-up but the hash would be superfluous.

If he needs number of elements, he would have to sum up all different elements between $beg and $end instead of just subtracting $beg from $end. Depending on the data set this additional step could eat away any savings from performing a few steps more in the binary search.

To be precise: It must be 2*log2(average number of duplicates) > (average number of different elements in a search) to get a speed-up.

In reply to Re^2: Modified Binary Search by jethro
in thread Modified Binary Search by Limbic~Region

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