Runtime is exponential, same as Fibonacci$$s numbers. A correct picture of recursion tree, plus a short

argument using subsitution method, shows

T$($An$)>=($

$$

$2)$kT$($An$$k$)$

and similar for the time of B$.$ We hit the base case when k $=$ n $1,$ k $=$ n $2$ depending on parity of n

hence T$($An$)>=($

$$

$2)$n

$.$