Can I solve my decision problem in $O(n)$ time?

Question

My algorithm solves a custom string problem but it loops twice making it $O(n^2)$ time. I'm asking here as I'm a beginner in self-learning algorithms.

Decision Problem: Given an input list $A$, can our target string $S$ be made from any of the elements of $A$?

If just one element does not exist in $A$ (which exists in $s$) then the algorithm returns false.

A = input_array
s = input_target_array

for i in s:
  if i in A:
    (all elements ∈ s) ∈ A
  else:
    break  

(if any element ∈ s) ∉ A:
  return False
else:
  return True 

Output

Enter your input for A and include spaces:
Enter letters WITH SPACES: "ab" "cd" "ac"
enter your input for s:
 "abxd"
^ False ^
 "abcd"
^ True ^

I'm looking for an algorithm that solves my made up decision problem faster than $O(n^2)$ time.

Question

Are there any methods that can be used from an algorithmic standpoint that would make this $O(n)$ time?

Answer 1

I use set difference to find out if there are any elements in $s$ that don't exist in $A$. This should be $O(n)$ in the best cases if there is a good hash. Since I'm dealing with integers it should be $O(n)$ Explained here

I then use $len(s)$ and $len(A)$ which takes $O(1)$ time according to here.

A = [1,2,3,4,9]
s = [1,2,3,4,9]

elem_not_in_A = set(s) - set(A)

if len(s) >= len(A):
    if len(elem_not_in_A) > 0:
        output False
    else:
        output True
else:
    output False

If the set difference remains $O(n)$ and all the other statements remain $O(1)$ then yes it can be solved in $O(n)$ time.