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What is the complexity of st.erase(st.begin())?
By Nguyen_Khac_Trung, history, 10 months ago, In English

What is the complexity of st.erase(st.begin())?

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10 months ago, # |
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It should be O(logn)

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    10 months ago, # ^ |
    ← Rev. 2   Vote: I like it 0 Vote: I do not like it 
    for(r=0;r<b.size();r++) tmp+=a[r];
if (tmp==b) res.pb(1);
while (r<a.size())
    {
        tmp.erase(tmp.begin());
        tmp+=a[r++];l++;
        if (tmp==b) res.pb(l+1);
    }

    so this code will has O(NlogN), right? :>

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      10 months ago, # ^ |
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      If I understand correctly, inside the while cycle you are comparing a set to another set, which has roughly O(n^2) complexity

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10 months ago, # |
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What st is? What cppreference tell about it?

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10 months ago, # |
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For a set, the complexity would be O(logn)

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10 months ago, # |
← Rev. 2   Vote: I like it +3 Vote: I do not like it

Is st a set? If so, then it's O(log n), but it seems like your st is a string, in which case it's O(n).

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