a.Θ(n)

b.Θ(nLogn)

c.Θ(n*n)

d.Θ(log n)

a. Θ(n) is the correct answer, just use the master's theorem and you would find the answer.

T(n) = 2T(n/2) + log n T(1) = 1 Substitute n = 2^k T(2^k) = k + 2T(2^(k-1)) T(2^k) = k + 2(k-1) + 4T(2^(k-2)) = k + 2(k-1) + 4(K-2) + 8T(2^(k-3)) = k + 2(k-1) + 4(K-2) + 8(k-3) + 16T(2^(k-4)) = k + 2(k-1) + 4(K-2) + 8(k-3) + ...... + 2^kT(2^(k-k)) = k + 2(k-1) + 4(K-2) + 8(k-3) + .......+ 2^kT(1) = k + 2(k-1) + 4(K-2) + 8(k-3) + .......+ 2^k --------(1) 2T(2^k) = 2k + 4(k-1) + 8(K-2) + ...... + 2*2^k + 2^(k+1) --------(2) Subtracting 1 from 2, we get below T(2^k) = - k + 2 + 4 ...... 2^(k-2) + 2^(k-1) + 2^k + 2^(k+1) = - k + 2 * (1 + 2 + 4 + ..... 2^k) = -k + [2*(2^k - 1)] / [2-1] = -k + [2*(2^k - 1)] T(n) = -Logn + 2*(n - 1) T(n) = Θ(n) Sources : http://quiz.geeksforgeeks.org/gate-gate-cs-2014-set-2-question-23/