Question

This is a question Im trying to figure out for a job pre-interview questioniar. and i dont know python that well so im struggling with this problem.

the real questions for the job are big-o performance and making it better but i can do that i just need to understand this code

question A: in is_bar they compare the iterator to the array (if i == b). is this the equivalent of (if i == b.length)?

question B:
so following the flow of the function and assuming the above and (a = [1,2,3] b=[3,4,5]) is true the first comparison is comparing two iterators????

so it would be:

1. is_bar(b, 0)
2. for i in b (says a but its b from above)
3. if 0 == 0
4. so it evaluates to true and does c.append(0)
5. does this 3 more times and since its comparing iterators it would append 1,2 than unique (c)
   6.which makes a new array the same size as c and makes every values = 1

but that wouldnt make sense. Appearently there is somthing im missing because a function that takes 2 arrays, then creates an array of equal size of the second array and makes all values = 1 would be stupid.

question C: what does b.keys() return exactly (i looked up the function but it says “list of all the keys used” but that doesnt make sense since it is a one-dimension array)

 def foo(a, b):
         """ a and b are both lists """
         c = []
         for i in a:
             if is_bar(b, i):
                 c.append(i)
         return unique(c)

      def is_bar(a, b):
         for i in a:
             if i == b:
                 return True
         return False

      def unique(arr):
         b = {}
         for i in arr:
             b[i] = 1

         return b.keys()

Answer 1

A: No. It compares each element of a to b. So if b is an element of a it returns True.

B: Change based on the above.

C: b is not an array. It’s a dictionary (the b = {}).