I can't understand it

#1

How can it be O(N^2) ??

#2

Please paste the question here.

#3

int sum = 0;
void calcSum(vector v, int i)
{
if(i == v.size())
return;
sum += v[i];
calcSum(v, i+1);
}

How can it’s time complexity be O(N^2) ?

#4

Assuming vector is defined as vector< int > v ,its time complexity will be O(n^2).
Because total n calls will be there and on every call vector gets copied to new vector which is O(n) for each call . Therefore O(n^2) for n calls.
If the vector was passed by reference using & operator then the complexity would be O(n) because there no copying would take place.

#5

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