n the below algorithm, increment $($of i$)$ and comparison $($A$[$i$]$

A$[$n$-1-$i$])$ operations are executed the same number of times, with their counts differing by a constant, either positively or negatively.

ALGORITHM mystery$($A$[0..$n$-1])$

for i$=0$ to the floor of n$/2$

$-1$ do

if A$[$i$]$ does not equal

A$[$n$-1-$i$]$ then

return false

return true

Question $15$ options:

True

False