15 views
int f1(int n)

{

if(n==0||n==1)

return n;

else
return (2*f1(n-1)+3*f1(n-2));

}

Time complexity of above function
| 15 views

$\text{F(n) = F(n-1) + F(n-2)}$
just that the original function call has a factor of constant multiplied to $\text{F(n-1) and F(n-2)}$
Hence the time complexity will $O(2^n)$