It's been long since I learned the big O notation but IMHO your code can only be O(1) if and only if there are no gaps within the data.

Just consider the worst case of IDs within (1, (1/2 * m) .. m) i.e. (n= 1/2 m + 2) and your looking for the second entry x=m/2 right after the maximum possible gap. You'll start with approximating 3/4 m and continue to seek linearly till m/2. That are 1/2 n steps which means O(n).

If you improve your code to recursively interpolating the next position between the narrowed range instead going linearly you'll make it with O(log(n)) like in binary search.

Furthermore you're only comparing the entries from the first column aren't you?

Cheers Rolf

In reply to Re^2: Rapid text searches ( O(log n) time) by LanX
in thread Rapid text searches by joomanji

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