The psuedo$-$code below gives a slightly different version of binary search:

RecBinarySearch$2($low$,$high,A$,$x$)$

If $(??),$ Return$(1)$

mid:$=$low$+$high$2$

If $($x$=$amid$),$ then Return$($ mid $)$

If $($xamid$),$ then low:$=$mid$+1$

Return$($ RecBinarySearch$2($low$,$high,A$,$x$))$

$($a$)$

What is the right condition for the base case?

$($b$)$

Consider running RecBinarySearch$2$ on input list:

$(2,5,8,11,15,18,23,29,31,35)$

In searching for a target item x that is in the list, the running time of the algorithm may differ depending on which item from the list is the target item. Give a target item x$,$ selected from the list, that results in the best running time. Give a target item, selected from the list, that results in the worst running time.