What is the complexity of st.erase(st.begin())?
# | User | Rating |
---|---|---|
1 | tourist | 4009 |
2 | jiangly | 3839 |
3 | Radewoosh | 3646 |
4 | jqdai0815 | 3620 |
4 | Benq | 3620 |
6 | orzdevinwang | 3612 |
7 | Geothermal | 3569 |
8 | ecnerwala | 3494 |
9 | Um_nik | 3396 |
10 | gamegame | 3386 |
# | User | Contrib. |
---|---|---|
1 | Um_nik | 163 |
1 | maomao90 | 163 |
3 | -is-this-fft- | 162 |
4 | atcoder_official | 158 |
4 | cry | 158 |
6 | awoo | 157 |
7 | nor | 155 |
7 | adamant | 155 |
9 | TheScrasse | 153 |
10 | maroonrk | 152 |
What is the complexity of st.erase(st.begin())?
Name |
---|
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)
.