1. first for loop runs n times
2. second for loop runs log(nlogn) :
since j = 1, 2, 4, 8, 16,........nlogn.......2^n but it runs only till nlogn
assume some 2^k = nlogn
solving we get k as log(nlogn)
By product rule: T(n) = n*log(nlogn)
n*Log(nlogn) = n[log(n) + log(logn) ] = nlog(n) + nlog(logn)
By taking Big O , nlogn > nlog(logn)
hence the answer will be nlogn