What is the complexity of st.erase(st.begin())?
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What is the complexity of st.erase(st.begin())?
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It should be O(logn)
so this code will has O(NlogN), right? :>
If I understand correctly, inside the while cycle you are comparing a set to another set, which has roughly O(n^2) complexity
I used string
What
st
is? What cppreference tell about it?st is string
erase it's $$$O(|st|)$$$ then
For a set, the complexity would be O(logn)
Is
st
a set? If so, then it'sO(log n)
, but it seems like yourst
is a string, in which case it'sO(n)
.