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  1. Prepare
  2. C++
  3. STL
  4. Sets-STL
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Sets-STL

  • utkarshsrivasta2
     + 4 comments

    the solution using if else is not very tangible since there are only three possible values that y can have. Here is my solution

    int main() {
    int q;
    cin>>q;
    set <long long int> s;
    for(int i=0;i<q;i++)
    {
        int y;
        long long int x;
        cin>>y>>x;
        if(y==1)
            s.insert(x);
        else if(y==2)
            s.erase(x);
        else
        {
            if(s.end()==s.find(x))
                cout<<"No"<<endl;
            else
                cout<<"Yes"<<endl;
        }
    }
    return 0;
}