2 votes 2 votes Solve using Masters theorem 2T (n/2) + n log n Algorithms algorithms master-theorem + – sh!va asked Oct 29, 2016 sh!va 2.1k views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply mcjoshi commented Oct 29, 2016 reply Follow Share $a = b^k$ and $p \gt -1$ $\color{navy}{\tt T(n) = (n^{log_2(2)}*log^{1+1}(n) = \theta(nlog^2n)}$ 4 votes 4 votes Prashant. commented Oct 29, 2016 reply Follow Share Simple master theorm not apllied here. You can use extended master theorm. which gives (n log2n ). 6 votes 6 votes Please log in or register to add a comment.
0 votes 0 votes we know, T(n) = a t(n/b) + Θ(nk .logpn) a=2, b=2 ,k=1 and p=1 so the condition is a=bk and p>=-1 so T(n)=(nlog 22)*log p+1n so it is n*log2n Rajesh Raj answered Oct 29, 2016 Rajesh Raj comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes here a=2,b=2,k=1,p=1; now, here a=b so we will use the formula T(n)=θ(n^log a to the base b log n^p+1) so answer is θ(n log^2n) here we will use the formula for masters theorem as Shashank Kumar Mishr answered Mar 7, 2017 Shashank Kumar Mishr comment Share Follow See all 0 reply Please log in or register to add a comment.