Reducing the number of iterations or comparisons by a constant number may make a performance difference, depending on what gyrations you have to go through to make the reduction, but you're not changing the order of complexity. A 3-N solution is the same order as a 3/2-N solution, is the same as a single-pass (N) solution. They're all O(n). This is an important concept.

If you can reduce the order of a solution, your solution will scale better. We commonly look for O(n log n) solutions to replace O(n2) solutions, so that working with large amounts of data doesn't make our app bog down. If you merely change by a constant factor (as we're talking about in this case), you may see a constant-factor improvement at any size, but you won't alleviate any scaling problems. You're in the realm of micro-optimization.

Walking through the list is not going to be a significant portion of the computation, compared to the comparisons and math being done on the variables of interest. And it's really not worth trying to take the elements two at a time, because that ends up being a very inefficient way to walk through the list, compared to Perl's built-in for.

If our OP were to translate the algorithm into Inline::C, the reduced number of comparisons might compete well with three List::Util calls, with the difference being some constant factor on any size list.

Caution: Contents may have been coded under pressure.

In reply to Re^3: minimum, maximum and average of a list of numbers at the same time by Roy Johnson
in thread minimum, maximum and average of a list of numbers at the same time by LucaPette

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