Algorithm

1dim peak finder

$$\Theta (n) $$

if we adopt divide & conquer

the complexity will be

$$T(n) = T(\frac n2)+ \Theta(1)$$

and there is the base case :

$$T(1)=\Theta(1)$$

thus

$$
T(n)= \Theta(1) +\Theta(1)+…+\Theta(1)
= \Theta(\log _{2} n)
$$