Read the pseudocode below for a sorting algorithm as well as the $($incorrect$)$ proof claiming that the algorithm has expected running time

E$[$T$($n$)]=$ O$($n log$($n$)).$

Identify and$$describe the error with the proof.

Sort$($A$)$:

if len$($A$)<=1$:

return

# Pick the pivot x

x $=0$

if random.choice$([$True$,$ False$])$:

x $=$ max$($A$)$

else:

x $=$ min$($A$)$

L$,$ R $=$ Partition$($A$,$ x$)$

A $=$ L $+[$x$]+$ R

Sort$($L$)$

Sort$($R$)$

Partition$($A$,$ x$)$:

L $=[]$

R $=[]$

for element in A:

if element $<$ x:

L$.$append$($element$)$

elif element $>$ x:

R$.$append$($element$)$

return L$,$ R

# Initial call

A $=[$some array of elements$]$

Sort$($A$)$

$$Claim$$: The algorithm runs in time T$($n$)=$ O$($n log$($n$)).$

$$Proof$$: At the top level, we split the array by choosing a pivot x and partitioning the array$$into two parts, L and R$.$ Choosing the pivot and partitioning the array each take O$($n$)$ time. At$$each subsequent level, we increase the number of subproblems by a constant factor $(2),$ and the$$subproblems only become smaller. Therefore, the running time per level can only increase by a$$constant factor. By induction, since for the first level it is O$($n$),$ it will be O$($n$)$ for all levels. There$$are O$($log n$)$ levels, so in total, the running time is T$($n$)=$ O$($n log n$).[$We are expecting: A clear and concise description, in plain English, of the error with this proof.$]$