Answer:
0(n)
Step-by-step explanation:
Result previous exercise:
procedure count(a1a2...an : string with n > 1)
i:=0
for k:=1 to n
if ak =1 then i:=i + I
return i
Note: If you use a different algorithm, then you could possible get different results.
SOLUTION
There is only one part of the code that contains an operation (comparison), namely if a_k =1
This comparison is executed in every iteration of the for-loop
k can take on the values from 1 to n (for k:=I to n), thus k can take on n values.
Thus in total there are then n comparisons, while n is 0(n).