How comes O(N^2)?

#1

how O(N^2)? Please explain

#2

hello @chitwan_manchanda
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)+2 O(n) ==> T(1)+n O(n)==O(n^2)

#3

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