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Solve the recurrence equations:

• $T(n)= T( \frac{n}{2})+1$
• $T(1)=1$

$T(n) = T(n/2) + 1$

$=T(n/4) + 2$

$= T(n/8) + 3$

$\vdots$

$=T(n/{2^k}) + k.$

Recurrence stops when $2^k >= n$.

When $2^k = n,k = \lg n$

So, $T(n) = T(1) + \lg n \\= 1 + \lg n$

PS: Unless explicitly asked for asymptotic bound, we should give exact answers for solutions of recurrence equations.
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+1 vote
T(n)=T(n/2)+1

Using Master Theorem :

a=1,b=2,k=0,p=0

T(n)=O(logn)
+1 vote

It is a standard Recurrence for Binary Search :
T(n) = T(n/2) + Θ(1).

T(n)=T(n/2k)+k
base case T(1)=1
k=logn
T(n)= logn+1

T(n)=T(n2)+1
use master theorem we get answer as logn