What is the complexity of given problem?

Question

I have found the solution to be nLogn but i dont know if i am correct!

for i in range(n):
     for j in range(log(i)):
         #some operation O(1)

Answer 1

Its O(log(n!)).Which can also be written as O(log(n^n)) or O(n*log(n))
As you can see the number of operations will be log(1)+log(2)+...log(n)=log(n!)
MORE EXPLANATION
-when i=1, j will run upto log(i)=log(1)
-similarly when i=2,j will run upto log(i)=log(2)
Thats why total operations are log(1)+log(2)+...log(n)=log(n!)

Answer 2

As already stated, your algorithm has complexity $$ log(1)+log(2)+...+log(n) = log(n!) $$ and by Stirling Approximation you have $$ log(n!) = O(n\cdot log(n))$$