How can the time complexity be n^2

#1

as only n calls are made

#2

Hey @sukritbansal01
Assumming vector is defined as vector< int > v
its time complexity will be O(n^2).
because total n calls wiil be there and on every call vector get copied to new vector which is O(n) for each call . therefore O(n^2) for n calls
Its not passed by reference but by copy
so for each recursive call a copy of vector is made
So rec reln in terms of time complexity is T(n)=T(n-1) + O(n) ==>T(n-2)+2O(n) ==> T(1)+nO(n)==O(n^2)

#3

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#4

Can I have the list of answers plz? Not able to find my mistake

#5

Hey Nothing such list is available :slight_smile:
You can always open a doubt regarding the questions you are not able to understand