set<int>s;
s.erase(s.begin());
I wanted to know whether removing the minimum element in set is a O(logN) or O(1) operation ?
I am quite confused about this situation.
Thanks in advance..
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It takes amortized constant time. See here set::begin and here set::erase.
Thanks buddy
It says just "constant" (doesn't say about amortisation), but I doubt that's true.
"For the first version (erase(position)), amortized constant." (where position is an iterator) from http://www.cplusplus.com/reference/set/set/erase/
ah, sorry, I just followed Jakube's links, but both of them apparently go to std::set::begin
Yes, I noticed that too :)
Sorry, I fixed the link.