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by adrianN 3454 days ago
AFAIK the STL is free to use any algorithm they want as long as it's O(n log n).
1 comments
niftich 3454 days ago
This is correct. The C++ standard specifies that sort must have a big-O complexity of n log(n). This can be seen on pdf page 925 of this working draft from 2014 [1].

[1] http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2014/n429...

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sqeaky 3453 days ago
So providing a faster sort would be a violation?

The standard doesn't seem to say "or better" here, but I know that in other places it does (or least used to say something similar).

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wnesensohn 3453 days ago
No, f = O(n log n) means that f doesn't grow significantly faster than n log n. That is true for f = n log n, but it's also true for f = n. The "or better" is implicit by using big-O.

Note that this wouldn't be true if the standard said that it has to be Θ(n log n).

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dietbacon 3453 days ago
Big O complexity is an upper bound.

If something is O(1) it's also O(n) and O(n!), since it's an upper bound.

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