0 0 votes f(n)=$2^n$ g(n)=n! h(n)=$n^{logn}$ which one is true? A) f(n)=O(g(n)) and g(n)=O(h(n)) B) f(n)=$\Omega(g(n)))$ and g(n)=O(h(n)) C) g(n)=O(f(n)) and h(n)=O(f(n)) D) h(n)=O(f(n)) and g(n)=$\Omega(f(n))$ Algorithms algorithms time-complexity + – aditi19 980 views answer comment Share Follow Print See all 5 Comments 5 5 Comments reply Show 2 previous comments aditi19 commented Nov 6, 2018 reply Follow flag pls give a detailed solution. i'm unable to understand 0 0 replyShare SameekshaGupta commented Nov 7, 2018 reply Follow flag Take two functions at a time, compare them by taking log and draw conclusion from them. You will get f(n)<g(n), h(n)<g(n) and h(n)<f(n) . So, the answer is (d). 0 0 replyShare aditi19 commented Nov 7, 2018 reply Follow flag https://www.cs.nmsu.edu/~ipivkina/Fall08cs372/Fall05/asympNot.pdf 0 0 replyShare Please log in or register to add a comment.